2022 Staar Algebra I Math Rationales

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

1 Option D is correct To determine the x-intercepts (values where a graph crosses the x-axis [horizontal axis]) from the

graph of a quadratic function, the student should have determined that the graph of function r

intersects (crosses) the x-axis where x = −10 and x = 2, so the x-intercepts are (−10, 0) and (2, 0).

To find the y-intercept (value where a graph crosses the y-axis [vertical axis]), the student should

have determined that the graph of function r intersects the y-axis where y = −5, so the y-intercept is

at (0, −5).

Option A is incorrect The student switched the values of the x- and y-intercepts and reversed the signs of the intercepts, likely

because of confusion about the relationship between the intercepts and the factored form of a quadratic

equation. In addition, the student switched the x- and y-coordinates of the ordered pairs. In choosing

(5, 0) as the x-intercept, the student confused the x-intercept with the y-intercept (y = −5), reversed the

sign, and placed the intercept in the x-coordinate position. In choosing (0, 10) and (0, −2) as

y-intercepts, the student confused the y-intercept (y = −5) with the x-intercepts (x = −10 and x = 2),

reversed the signs, and placed the intercepts in the y-coordinate positions. The student needs to focus on

understanding how to identify the intercepts of a quadratic function when given a graph.

Option B is incorrect The student chose the correct values for the intercepts but switched the x- and y-coordinates of the

ordered pairs. The x-intercepts are located at −10 and 2, but the correct ordered pairs are (−10, 0) and

(2, 0) instead of (0, −10) and (0, 2). Similarly, the student chose the correct location of the y-intercept

but transposed the values of the coordinates; the correct ordered pair is (0, −5) instead of (−5, 0). The

student needs to focus on understanding the correct order of the values in an ordered pair.

Option C is incorrect The student switched the values of the x- and y-intercepts and reversed the signs of the intercepts, likely

because of confusion about the relationship between the intercepts and the factored form of a quadratic

equation. In choosing (0, 5) as the x-intercept, the student confused the x-intercept with the y-intercept

and reversed the sign. In choosing (10, 0) and (−2, 0) as y-intercepts, the student confused the

y-intercept with the x-intercepts and reversed the signs. The student needs to focus on understanding

how to identify the intercepts of a quadratic function when given a graph and on understanding the

correct order of the values in an ordered pair.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

2 Option H is correct To determine the rate of change (constant rate of increase or decrease) of the number of items the

worker packed in boxes with respect to the number of minutes the worker has been packing items in

boxes, the student could have chosen two points from the table and calculated the amount of change. The

student could have used the first two sets of values in the table and applied the slope formula,

−

, resulting in

28 20 8

7 5 2

−

= = =

−

Therefore, the rate of change is 4 items packed per

minute. This is an efficient way to solve the problem; however, other methods could be used to solve the

problem correctly.

Option F is incorrect The student likely subtracted the first two values in the “Number of Items Packed” column (28 – 20 = 8),

disregarding the corresponding change in the number of minutes. The student needs to focus on

understanding that the rate of change of a linear relationship is equal to the change in the values of the

dependent variable divided by the corresponding change in the values of the independent variable.

Option G is incorrect The student likely found the average of the values in the “Number of Items Packed” column,

20 28 44 56 148

+ + +

. The student needs to focus on understanding that the rate of change of a

linear relationship is equal to the change in the values of the dependent variable divided by the

corresponding change in the values of the independent variable.

Option J is incorrect The student likely used the first row of values from the table and subtracted 5 from 20 (20 – 5 = 15). The

student needs to focus on understanding that the rate of change of a linear relationship is equal to the

change in the values of the dependent variable divided by the corresponding change in the values of the

independent variable.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

3 Option B is correct To determine the system of equations that is best represented by the graph, the student could have

found the equation of each line shown in the graph in slope-intercept form (y = mx + b, where

−

represents the slope of the line and b represents the value of the y-intercept). To find the

equation of the line that is increasing from left to right, the student could have recognized that the

graph intersects the y-axis at (0, −3), so the value of the y-intercept, b, is −3. The student could

have substituted the x- and y-coordinates of (0, −3) and (−5, −5) into the slope formula, resulting in

( )

5 0 5 5

− − −

−

= = =

− − −

. Since b = −3 and

m =

, the equation for the line that is increasing from left

to right is

yx=−

. Similarly, to find the equation of the line that is decreasing from left to right,

the student could recognize that the graph intersects the y-axis at (0, −8), so the value of the

y-intercept, b, is −8. The student could have substituted the x- and y-coordinates of (0, −8) and

(−5, −5) into the slope formula, resulting in

( )

5 0 5 5

− − −

= = = −

− − −

. Since b = −8 and

m =−

, the

equation for the line that is increasing from left to right is

yx= − −

This is an efficient way to

solve the problem; however, other methods could be used to solve the problem correctly.

Option A is incorrect The student likely found the correct slope of each line but reversed the values of the y-intercepts,

using −8 for the value of b in the line that is increasing and −3 for the value of b in the line that is

decreasing. The student needs to focus on correctly identifying the y-intercept of a linear equation

when given a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option C is incorrect

The student likely used the reciprocal of the slope formula,

−

to calculate the slopes as

( )

5 0 5 5

− − −

= = =

−

− − −

for the line that is increasing and

( )

5 0 5 5

− − −

= = = −

− − −

for the line that is

decreasing. In addition, the student likely switched the values of the y-intercepts, using −8 for the

value of b in the line that is increasing and −3 for the value of b in the line that is decreasing. The

student needs to focus on correctly applying the slope formula and correctly identifying the

y-intercept of a linear equation when given a graph.

Option D is incorrect

The student likely used the reciprocal of the slope formula,

−

, to calculate the slopes as

( )

5 0 5 5

− − −

= = =

−

− − −

for the line that is increasing and

( )

5 0 5 5

− − −

= = = −

− − −

for the line that is

decreasing. The student correctly identified the values of the y-intercepts. The student needs to focus

on correctly applying the slope formula.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

4 Option J is correct To determine whether a graph represents y as a function of x, the student could have recalled that a

function represents a relationship where each input, x, has a single output, y. Also, the student could

have recalled that a function could have repeated output values if all the input values are different.

The student could have analyzed the graph and noticed that each value of x on the continuous

function has only one corresponding value of y. This is an efficient way to solve the problem;

however, other methods could be used to solve the problem correctly.

Option F is incorrect The student likely concluded that any parabola represents a function. A parabola represents a function

only when it opens vertically, either up or down. The student needs to focus on understanding

whether a relation (relationship between the x- and y-values of ordered pairs) represented in a graph

defines a function.

Option G is incorrect The student likely concluded that any parabola represents a function. A parabola represents a function

only if it opens vertically, either up or down. The student needs to focus on understanding whether a

relation (relationship between the x- and y-values of ordered pairs) represented in a graph defines a

function.

Option H is incorrect The student likely concluded that any continuous graph represents a function, failing to observe that,

for example, when x = 0, there are three corresponding values of y at y = 6, y = 0, and y = −6. The

student needs to focus on understanding whether a relation (relationship between the x- and y-values

of ordered pairs) represented in a graph defines a function.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

5 Option A is correct To determine the solution to the system of linear equations, the student could have used the

elimination method. Multiplying the first equation by 2 results in the equation 4x + 2y = 80. The

student could have added this to the second equation, x – 2y = −20, to get the result 5x = 60. The

student then could have divided both sides of the resulting equation by 5, obtaining x = 12. Next, to

find the corresponding value of y, the student could have substituted x = 12 into the second equation,

resulting in 12 – 2y = −20. Subtracting 12 from both sides results in −2y = −32. Finally, the student

could have divided both sides of the equation by −2, resulting in y = 16. Since x = 12 and y = 16, the

ordered pair that is a solution to the system of equations is (12, 16). This is an efficient way to solve

the problem; however, other methods could be used to solve the problem correctly.

Option B is incorrect The student likely switched the coefficient of y in the second equation with the coefficient of x to solve

the system using the elimination method, resulting in the equations 2x + y = 40 and 2x – y = −20.

Next, the student likely found the sum of the two equations, but when adding 40 + (−20), the

student likely ignored the negative sign and found the sum to be 60, resulting in the equation

4x = 60. Next, the student likely divided both sides of the equation by 4 to isolate the x, resulting in

x = 15. Finally, the student likely substituted 15 for x in the second equation to solve for y, resulting

in x – 2y = −20; (15) – 2y = −20; −2y = −35; y = 17.5. The student needs to focus on correctly

using the elimination method to solve a system of equations.

Option C is incorrect The student likely used the elimination method incorrectly by multiplying the y-term in the first

equation by 2 and multiplying the x-term in the second equation by −2, understanding that the

coefficients must be opposites of each other. The student likely obtained 2x + 2y = 40 and

−2x – 2y = −20 and added these equations together to obtain the result 0 = 20 and concluded that

there was no solution. The student needs to focus on correctly using the elimination method to solve a

system of equations.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option D is incorrect The student likely used the elimination method incorrectly by multiplying the y-term in the first

equation by 2 and by multiplying the x-term and the constant term in the second equation by −2,

understanding that the coefficients must be opposites of each other. The student likely obtained

2x + 2y = 40 and −2x – 2y = 40. Next, the student likely added x- and y-terms together but

subtracted the constant terms, resulting in 0 = 0. The student then concluded that there are an

infinite number of solutions to the system of equations. The student needs to focus on correctly using

the elimination method to solve a system of equations.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

6 Option J is correct To determine the graph that best represents a quadratic function with a range (all possible y-values)

of all real numbers greater than or equal to 3, the student could have analyzed the graph and

determined that it represents a parabola with vertex at (−2, 3). The student could have observed that

the parabola opens upward and that the least y-value occurs at 3 and that all other y-values on the

parabola are greater than 3. The student then could have concluded that the range of the quadratic

function represented in this graph is all real numbers greater than or equal to 3. This is an efficient

way to solve the problem; however, other methods could be used to solve the problem correctly.

Option F is incorrect The student likely analyzed the graph and determined that it represents a parabola with vertex at

(2, −3). The student likely reversed the sign of the y-coordinate and concluded that 3 is the least

y-value instead of the greatest, perhaps because it is a negative number. The student needs to focus

on identifying the correct graph when given the range of a quadratic function.

Option G is incorrect The student likely analyzed the graph and determined that the graph represents a parabola with

vertex at (−2, 3). The student then likely incorrectly concluded that since 3 is the greatest y-value,

the range must be all numbers greater than 3. The student needs to focus on identifying the correct

graph when given the range of a quadratic function.

Option H is incorrect The student likely analyzed the graph and determined that the graph represents a parabola with

vertex at (2, −3). The student then likely reversed the sign of the y-coordinate and concluded that

since the parabola opens in an upward direction, the graph must represent a quadratic function with a

range of all real numbers greater than or equal to 3. The student needs to focus on identifying the

correct graph when given the range of a quadratic function.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

7 Option C is correct To determine the value of y when x = 64 in a directly proportional relationship, the student could

have recognized that the relation can be represented by the equation y = kx, where k is the constant

of proportionality. To determine the value of k, the student could have substituted x = 512 and

y = 128 into the equation and solved for k. The result would be 128 = 512k. Then the student could

have divided both sides by 512, obtaining

k =

. Next, the student could have rewritten the equation

yx=

. Finally, the student could have substituted x = 64 into the equation and solved for y,

resulting in

( )

64 16

y ==

. This is an efficient way to solve the problem; however, other methods

could be used to solve the problem correctly.

Option A is incorrect The student likely transposed the values of x and y when substituting into the direct variation

equation y = kx, resulting in 512 = 128k. The student then likely divided both sides by 128, obtaining

k = 4. Finally, the student likely multiplied k, the constant of proportionality, by x = 64, resulting in

y = 256. The student needs to focus on understanding how to set up and solve direct variation

problems.

Option B is incorrect The student likely correctly solved the variation equation y = kx by substituting x = 512 and y = 128

into the equation and solving for k, resulting in

k =

. Next, the student likely rewrote the equation

yx=

. Finally, the student likely incorrectly multiplied the constant of proportionality by 128

instead of 64, resulting in

( )

128 32

y ==

. The student needs to focus on understanding how to set

up and solve direct variation problems.

Option D is incorrect The student likely calculated the answer by incorrectly dividing the two given values of x, resulting in

512

y ==

. The student needs to focus on understanding how to set up and solve direct variation

problems.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

8 Option G is correct To determine which graph best represents the solution set for the inequality 8x + 5y ≤ 100 in the

given context, the student could have recognized that the “≤” symbol represents an inclusive

inequality and would be represented by a solid boundary line on the graph. To graph the boundary

line, the student could have rewritten the equation of the boundary line in slope-intercept form,

y = mx + b, where m represents the slope of the boundary line and b represents the y-intercept.

First, the student could have subtracted 8x from both sides of the equation of the boundary line,

obtaining 5y = −8x + 100. Next, the student could have divided both sides of the equation by 5, with

the result

yx= − +

. The student could then have recognized that the slope of the boundary line

m =−

and the y-intercept would be located at the point represented by the ordered pair (0, 20).

The student could have determined the correct region to be shaded by substituting the ordered pair

(0, 0) into the inequality

yx − +

to test for a true statement. Since

( )

0 0 20

 − +

equivalent to 0 ≤ 20, which is a true statement, the ordered pair (0, 0) should be included in the

shaded region of the graph. Therefore, the student should have shaded the part of the graph that

contains the origin. This is an efficient way to solve the problem; however, other methods could be

used to solve the problem correctly.

Option F is incorrect

The student likely identified the correct boundary line in slope-intercept form,

yx= − +

, with

correct slope,

−

, and y-intercept, 20. The student likely interpreted the inequality symbol “≤” as

meaning “greater than” instead of “less than or equal to,” which would be represented by a dashed

line instead of a solid line, and a shaded region that would not include the origin. The student needs

to focus on understanding how the inequality symbol affects the graph of the solution set of a linear

inequality.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option H is incorrect

The student likely identified the correct boundary line in slope-intercept form,

yx= − +

, with

correct slope,

−

, and y-intercept, 20. The student likely interpreted the inequality symbol “≤” as

meaning “greater than or equal to” instead of “less than or equal to,” which could be represented by a

solid line, and identified the shaded region of the graph as not including the ordered pair (0, 0), since

0 ≥ 20 is not a true statement. The student needs to focus on understanding how the inequality

symbol affects the graph of the solution set of a linear inequality.

Option J is incorrect

The student likely identified the correct boundary line in slope-intercept form,

yx= − +

, with

correct slope,

−

, and y-intercept, 20. The student likely interpreted the inequality symbol “≤” as

meaning “less than” instead of “less than or equal to,” which would be represented by a dashed line

instead of a solid line. The student needs to focus on understanding how the inequality symbol affects

the graph of the solution set of a linear inequality.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

9 Option A is correct To determine which function best represents the graph of an exponential function, the student could

have recognized that an exponential function is in the form p(x) = ab

, where a is the y-intercept

(value where the graph crosses the y-axis), b is the decay factor (constant rate by which successive

values decrease), and x is the variable (symbol used to represent an unknown number). From the

graph, the student could have interpreted that the y-intercept at (0, 1) means that the value of a is 1.

To write the exponential function, the student could have substituted the values from the ordered pair

(−1, 4) into the equation p(x) = ab

, using a = 1, x = −1, and p(−1) = 4, obtaining 4 = (1)(b)

–1

. To

solve for b, the student could have rewritten the equation as

and then multiplied both sides of

the equation by b, obtaining 4b = 1. Next, the student could have divided both sides of the equation

by 4, resulting in the solution b = 0.25. Substituting a = 1 and b = 0.25 into the exponential equation

p(x) = ab

, the student could have obtained p(x) = (1)(0.25)

= (0.25)

. This is an efficient way to

solve the problem; however, other methods could be used to solve the problem correctly.

Option B is incorrect The student likely interpreted the ordered pair near (2, 0) as the y-intercept, using 2 for the initial

value (a). Then the student likely determined that the value of the common factor was 0.25 but

multiplied 0.25 by the initial value to obtain b = (0.25)(2) = 0.5. Using a = 2 and b = 0.5 in the

exponential equation p(x) = ab

, the student likely obtained p(x) = (2)(0.5)

= 2(0.5)

. The student

needs to focus on understanding how to determine the initial value of an exponential function when

given a graph.

Option C is incorrect The student likely correctly identified the initial value from the graph as 1 and the decay factor as 0.25.

The student likely interpreted the value of b in the exponential equation p(x) = ab

as the sum of the

initial value and the decay factor and used b = 1 + 0.25 = 1.25. Substituting b = 1.25 into the

exponential equation, the student likely obtained p(x) = (1.25)

. The student needs to focus on

understanding how to determine the decay factor of an exponential function when given a graph.

Option D is incorrect The student likely correctly identified the initial value from the graph as being 1 and the decay factor

as being 0.25. When substituting these values into the exponential equation p(x) = ab

, the student

likely neglected the decimal point in the decay factor and obtained p(x) = (1)(25)

= (25)

. The

student needs to focus on understanding how to correctly write the value of the decay factor of an

exponential function when given a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

10 Option J is correct To determine which expression is equivalent to (n – 4)(2n + 7), the student could have multiplied

each term in the factor (n – 4) by each term in the factor (2n + 7) and then combined like terms

(terms that contain the same variables raised to the same powers or constant terms). The

multiplication steps are n(2n + 7) – 4(2n + 7), resulting in 2n

+ 7n – 8n – 28. The student could

have combined like terms to obtain 2n

– 1n – 28, which is equivalent to 2n

– n – 28. This is an

efficient way to solve the problem; however, other methods could be used to solve the problem

correctly.

Option F is incorrect The student likely found the sum of the binomials instead of the product, adding the terms containing

the variable n and the terms that are constants to obtain n + 2n + (−4) + 7 = 3n + 3. The student

needs to focus on understanding how to find the product of two binomials.

Option G is incorrect The student likely multiplied each term in the factor (n – 4) by each term in the factor (2n + 7)

but determined the product of n and 2n to be 2n instead of 2n

. The student likely multiplied

n(2n + 7) – 4(2n + 7) to get a result of 2n + 7n – 8n – 28 and then combined like terms get as a

result the expression 1n – 28, which is equivalent to n – 28. The student needs to focus on

understanding how to find the product of two binomials.

Option H is incorrect The student likely multiplied each term in the factor (n – 4) by each term in the factor (2n + 7) to

obtain n(2n + 7) – 4(2n + 7) = 2n

+ 7n – 8n – 28. Next, the student likely combined the like terms

7n and −8n incorrectly to get a result of −15n instead of −n. The student needs to focus on

understanding how to find the product of two binomials.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

11 Option B is correct To determine the situation that best shows causation (in which an event is the result of the

occurrence of another event), the student should have recognized that the number of hours an

employee works causes the amount of money an employee earns to change. If the number of hours

an employee works decreases, the amount of money the employee earns will also decrease.

Option A is incorrect The student likely did not realize that the two events are not related and that the number of people on

a bus has no effect on the number of animals in a zoo. The student needs to focus on understanding

causation in real-world problems.

Option C is incorrect The student likely identified a situation that shows possible association (relationship) but not

causation. The student likely did not realize that if the amount of a discount for a sale increases, the

number of items sold may or may not decrease. The student needs to focus on understanding

causation in real-world problems.

Option D is incorrect The student likely did not realize that the two events are not related and that the number of bike trails

in a city has no effect on the amount of rainfall in the city. The student needs to focus on

understanding causation in real-world problems.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

12 Option J is correct To determine which system of equations (two or more equations containing the same set of variables

[symbols used to represent unknown numbers]) can be used to represent line h, represented in a

table, and line j, represented by a graph, the student could write each equation in slope-intercept

form, y = mx + b, where m represents the slope of each line and b represents the y-intercept of each

line. To find the equation for line h, the student could have used the first two sets of values in the

table and applied the slope formula

−

, resulting in

( )

1 7 6 3

8 16

−−

= = = −

− − −

. To determine

the y-intercept of line h, the student could have substituted the slope and one of the ordered pairs

from the table, (−16, 7), into the point-slope equation, y – y

= m(x – x

), where m represents the

slope of the line and (x

, y

) represents a point on the line. The student could have obtained

( )

7 16



− = − − −



and then solved for y, by first distributing the

−

, resulting in

7 12

yx− = − −

. Next, the student could have added 7 to both sides to isolate the y, resulting in the

equation

yx= − −

. To find the equation of line j, which is represented by a graph, the student

could have used the slope-intercept form of a line, y = mx + b. The student could have first located

the y-intercept at (0, 1) and identified that b = 1. The student could then have applied the slope

formula,

−

, using the ordered pairs (4, −4) and (−4, 6), to get the result

( )

10 5

4 4 8 4

−−

= = = −

− − −

. Next, the student could have substituted

m =−

and b = 1 into y = mx + b

to obtain

yx= − +

. This is an efficient way to solve the problem; however, other methods could be

used to solve the problem correctly.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option F is incorrect The student likely determined the correct y-intercepts for both line h and line j but incorrectly applied

the slope formula, using

−

instead of

−

For line h, the student likely calculated

the slope using the first two ordered pairs represented in the table and then made a sign error,

resulting in

( )

8 16

1 7 6 3

− − −

−

= = = =

− − −

. The student likely substituted

m =

and b = −5 into

y = mx + b to get the result

yx=−

. For line j, the student likely calculated the slope using the

ordered pairs (4, −4) and (−4, 6) and then made a sign error, resulting in

( )

4 4 8 4

10 5

−

− − −

= = = =

−−

Next, the student likely substituted

m =

and b = 1 into

y = mx + b to obtain

yx=+

. The student needs to focus on understanding how to identify the

slope and y-intercept of a line when given a table of values or a graph.

Option G is incorrect The student likely determined the correct y-intercepts for both line h and line j but incorrectly used

the opposite of the value of each slope. For line h, the student likely substituted

m =

and b = −5

into y = mx + b, resulting in

yx=−

. For line j, the student likely substituted

m =

and b = 1

into y = mx + b, obtaining

yx=+

. The student needs to focus on understanding how to identify

the slope and y-intercept of a line when given a table of values or a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option H is incorrect The student likely determined the correct y-intercepts for both line h and line j but incorrectly applied

the slope formula, using

−

instead of

−

. For line h, the student likely calculated

the slope using the first two ordered pairs represented in the table, resulting in

( )

8 16

1 7 6 3

− − −

−

= = = = −

− − −

Next, the student likely substituted

m =−

and b = −5 into

y = mx + b, resulting in

yx= − −

. For line j, the student likely calculated the slope using the

ordered pairs (4, −4) and (−4, 6), resulting in

( )

4 4 8 4

10 5

−

− − −

= = = = −

−

−−

. Next, the student

likely substituted

m =−

and b = 1 into y = mx + b, obtaining

yx= − +

. The student needs to

focus on understanding how to identify the slope and y-intercept of a line when given a table of values

or a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

13 Option C is correct To determine which answer choice best describes the transformation of the graph of f(x) = x

to the

graph of n(x) = x

– 1, the student could have recognized that subtracting 1 from x

causes the

vertex of the parabola to shift from the origin, (0, 0), down 1 unit to (0, −1). The student then could

have realized that all the y-values on the parabola would also shift down 1 unit, representing a

vertical translation down of the entire parabola. This is an efficient way to solve the problem;

however, other methods could be used to solve the problem correctly.

Option A is incorrect The student likely determined that the “−1” in n(x) would cause the graph to translate up 1 unit

instead of down. The student needs to focus on understanding how changes to a function affect the

graph of the function.

Option B is incorrect The student likely determined that the “−1” in n(x) would cause the graph to translate left 1 unit

instead of down. The student needs to focus on understanding how changes to a function affect the

graph of the function.

Option D is incorrect The student likely determined that the “−1” in n(x) would cause the graph to translate right 1 unit

instead of down. The student needs to focus on understanding how changes to a function affect the

graph of the function.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

14 −3 and any equivalent

values are correct

To determine the value of k represented in the binomial factor (d + k), the student could have

recognized the need to find the factors (numbers or expressions that can be multiplied to get another

number or expression) of d

– d – 6. The student could have determined that d

is equal to d ∙ d and

written d as the first term in each factor. The student could have determined that the second terms in

the two factors are 2 and −3 because their product (answer when multiplying) is −6 (the last term in

the given expression) and that their sum is −1 (the coefficient of the middle term in the expression

given). The student then could have written the factors as (d + 2)(d – 3). Finally, the student could

have compared (d + 2)(d – 3) to the given factored form, (d + 2)(d + k) and recognized that the

value of k must be −3. This is an efficient way to solve the problem; however, other methods could

be used to solve the problem correctly.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

15 Option B is correct To determine the equation that is best represented by the graph, the student could have found the

equation of the line shown in the graph by using the slope-intercept form of a linear equation

(y = mx + b, where

−

represents the slope of the line and b represents the value of the

y-intercept). The student could have recognized from the graph that the line intersects the y-axis at

(0, 7) and concluded that the value of b is 7. Then the student could have used the ordered pairs (4,

0) and (8, −7) from the graph and applied the slope formula,

−

, resulting in

7 0 7 7

8 4 4 4

− − −

= = = −

−

. Next, the student could have substituted

m =−

and b = 7 into y = mx + b,

obtaining

yx= − +

. This is an efficient way to solve the problem; however, other methods could

be used to solve the problem correctly.

Option A is incorrect The student likely calculated the slope correctly but then determined that the x-intercept is at (4, 0)

and concluded that the value of b is 4. Next, the student likely substituted

m =−

and b = 4 into

y = mx + b, with the result

yx= − +

. The student needs to focus on understanding how to

determine the y-intercept of a linear function when given a graph.

Option C is incorrect The student likely concluded that the value of b is 4 by using the x-intercept of the line at (4, 0) and

then incorrectly applied the slope formula, using

−

instead of

−

. The student

likely substituted the ordered pairs (4, 0) and (8, −7) from the graph into

−

, resulting in

8 4 4 4

7 0 7 7

−

= = = = −

− − − −

. Finally, the student likely substituted b = 4 and

m =−

into

y = mx + b, resulting in

yx= − +

. The student needs to focus on understanding how to write a

linear function in slope-intercept form when given a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option D is incorrect The student likely recognized from the graph that the line intersects the y-axis at (0, 7) and correctly

concluded that the value of b is 7. Next, the student likely applied the slope formula incorrectly, using

−

instead of

−

. The student likely substituted the ordered pairs (4, 0) and (8, −7)

from the graph into

−

, resulting in

8 4 4 4

7 0 7 7

−

= = = = −

− − − −

. Finally, the student

likely substituted b = 7 and

m =−

into y = mx + b, resulting in

yx= − +

. The student needs to

focus on understanding how to find the slope of a linear function when given a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

16 Option J is correct

To determine the expression equivalent to

( )

, the student could have applied the power of a

power property, (a

)

= a

, to the factor (d

)

, obtaining

( )( )

8 8 18

c d c d







. Next, the student could

have applied the quotient of powers property,

−

, to the factors containing c, obtaining

8 18

8 2 18 6 18

c d c d

−

. This is an efficient way to solve the problem; however, other methods could be

used to solve the problem correctly.

Option F is incorrect The student likely applied the properties of exponents incorrectly by dividing the exponents of the

factors containing c and adding the exponents of the factor containing d, obtaining

6 3 4 9

c d c d

. The

student needs to focus on understanding how to apply the power of a power property and the quotient

of powers property when simplifying expressions.

Option G is incorrect The student likely applied the quotient of powers property incorrectly by dividing the exponents of the

factors containing c instead of subtracting, obtaining

( )

4 18

c d c d=

. The student needs to focus on

understanding how to apply the quotient of powers property when simplifying expressions.

Option H is incorrect The student likely applied the power of a power property incorrectly by adding the exponents of the

factor containing d, obtaining

8 2 6 3 6 9

c d c d

−+

. The student needs to focus on understanding how to

apply the power of a power property when simplifying expressions.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

17 Option A is correct To determine which value of x is a solution to the equation, the student could have first recognized

that one side of the equation must be set equal to 0. The student could have subtracted 30x and

added 45 to both sides of the equation, resulting in 5x

– 30x + 45 = 0. Next, the student could have

found the factors (numbers or expressions that can be multiplied to get another number or

expression) of 5x

– 30x + 45 = 0 and solved for the value of x. The student could have factored out

a 5 from the equation, resulting in 5(x

– 6x + 9) = 0. The student could have then found the factors

of x

– 6x + 9. The student could have recognized that x

and 9 represent perfect squares (numbers

made by squaring whole numbers). Using this, the student could have noticed that x

– 6x + 9 has

the form of a perfect square trinomial, a

– 2ab + b

, which factors as (a – b)

. In this case,

– 2ab + b

= x

– 2(3)x + 3

, which makes a = x and b = 3, so the factors can be written as

(x – 3)

. Finally, the student could have set the factor (x – 3) equal to 0 and solved for x, obtaining

x = 3. This is an efficient way to solve the problem; however, other methods could be used to solve

the problem correctly.

Option B is incorrect The student likely used x

– 6x + 9 and identified the equation as a perfect square trinomial but wrote

the factored form as (x + 3)

instead of (x – 3)

. The student then likely set the factor (x + 3) equal

to 0 and solved for x, obtaining x = −3. The student needs to focus on understanding how to find the

factors and solutions of a quadratic equation.

Option C is incorrect The student likely factored out a 5 from the equation, resulting in 5(x

– 6x + 9) = 0, and concluded

that the solution to the quadratic equation is x = 5 instead of continuing to factor the expression. The

student needs to focus on understanding how to find the factors and solutions of a quadratic equation.

Option D is incorrect The student likely factored out a 5 from the equation, resulting in 5(x

– 6x + 9) = 0, and concluded

that the solution to the quadratic equation is x = −5 instead of continuing to factor the expression.

The student needs to focus on understanding how to find the factors and solutions of a quadratic

equation.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

18 Option G is correct To determine the domain (all possible x-values) of the part of the discrete linear function shown, the

student could have identified all the x-values for which the graph has a corresponding y-value. The

x-values for the ordered pairs represented on the graph are {0, 1, 2, 3, 4, 5, 6, 7, 8}. Therefore, the

domain is the set of these numbers, which is {0, 1, 2, 3, 4, 5, 6, 7, 8}. This is an efficient way to

solve the problem; however, other methods could be used to solve the problem correctly.

Option F is incorrect The student likely used the values on the scale of the x-axis as the values of the domain. The student

needs to focus on understanding how to identify the domain of a discrete function from its graph and

express the domain using set notation.

Option H is incorrect The student likely switched the values of the domain and range and then used the values on the scale

of the y-axis as the values of the domain. The student needs to focus on understanding how to

identify the domain of a discrete function from its graph and express the domain using set notation.

Option J is incorrect The student likely switched the values of the domain and range and listed the y-values for the ordered

pairs instead of the x-values. The student needs to focus on understanding how to identify the domain

of a discrete function from its graph and express the domain using set notation.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

19 Option D is correct To determine the rate of change (constant increase or decrease) of y with respect to x, the student

could have chosen two points from the graph and calculated the amount of change. The student could

have used the ordered pairs (−3, 5) and (9, −5) and applied the slope formula,

−

, resulting

( )

5 5 10 5

12 6

− − −

= = = −

−−

. Therefore, the rate of change is

−

. This is an efficient way to solve the

problem; however, other methods could be used to solve the problem correctly.

Option A is incorrect The student likely identified the y-intercept of the line as the rate of change. The student needs to

focus on understanding the meaning of the rate of change and how to find it when given a graph.

Option B is incorrect The student likely calculated the rate of change as the change in x divided by the change in y, using

−

instead of

−

The student likely used values from the ordered pairs (−3, 5) and

(9, −5), resulting in

( )

12 6

5 5 10 5

−−

= = = −

− − −

. The student needs to focus on correctly applying the

slope formula to find the rate of change when given a graph.

Option C is incorrect The student likely used the x-intercept of the line, 3, and then likely estimated the y-intercept to be 2.

The student likely understood that rate of change is the change in y divided by the change in x,

interpreted these two values as changes from the origin, and found the rate of change to be

. The

student needs to focus on correctly applying the slope formula to find the rate of change when given a

graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

20 12.3 and any equivalent

values are correct

To determine the value of the y-intercept of the graph, the student could have realized that

h(x) = 12.3(4.9)

represents an exponential function in the form h(x) = a(b)

, where a is the

y-intercept (the value where the graph crosses the y-axis), b is the growth factor, and x is the

variable (symbol used to represent an unknown number). Therefore, the value of the y-intercept of

the graph of h(x) = 12.3(4.9)

is 12.3. This is an efficient way to solve the problem; however, other

methods could be used to solve the problem correctly.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

21 Option A is correct

To determine the equivalent expression, the student could first have found the quotient

8.8

2.2

Next, the student could have applied the quotient of powers property,

−

, to

−

, obtaining

( )

10 10

−−

−

. Finally, the student could have recognized that the simplified expression is equal

to the product of 4 and 10

, or 4 × 10

. This is an efficient way to solve the problem; however, other

methods could be used to solve the problem correctly.

Option B is incorrect

The student likely correctly determined the quotient

8.8

2.2

but did not apply the quotient of powers

property correctly, adding the exponents instead of subtracting. The student needs to focus on

understanding how to apply the quotient of powers property when simplifying expressions.

Option C is incorrect

The student likely correctly determined the quotient

8.8

2.2

but did not apply the quotient of powers

property correctly, dividing the exponents instead of subtracting. The student needs to focus on

understanding how to apply the quotient of powers property when simplifying expressions.

Option D is incorrect

The student likely correctly determined the quotient

8.8

2.2

but did not apply the quotient of powers

property correctly, adding the exponents instead of subtracting, and made a sign error. The student

needs to focus on understanding how to apply the quotient of powers property when simplifying

expressions.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

22 Option H is correct To determine which function (relationship where each input has a single output) best models the data,

the student could have used a graphing calculator to generate the function using quadratic regression

(a method of determining the quadratic function of best fit). The function that best models the data is

d(x) = 0.26x

– 3.11x. This is an efficient way to solve the problem; however, other methods could be

used to solve the problem correctly.

Option F is incorrect The student likely reversed the values of time, x, and depth, d(x), when entering the data into a

graphing calculator and disregarded the value of the constant term, c, that was generated. The

student needs to focus on understanding how to use technology to determine a quadratic function

that best fits a table of data.

Option G is incorrect The student likely reversed the values of time, x, and depth, d(x), when entering the data into a

graphing calculator. The student needs to focus on understanding how to use technology to determine

a quadratic function that best fits a table of data.

Option J is incorrect The student likely used the quadratic regression feature on a graphing calculator correctly but used

the value of the coefficient of determination, r

= 1, as the value of the constant term, c. The student

needs to focus on understanding how to use technology to determine a quadratic function that best

fits a table of data.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

23 Option B is correct To determine the equivalent expression, the student could have applied the commutative property,

a + b = b + a, to rearrange the terms, resulting in

(5rt – 3rw – 8tw) + (6rt – 4rw + 2tw) = 5rt + 6rt – 3rw – 4rw – 8tw + 2tw. Next, the student could

have combined like terms (terms that contain the same variables raised to the same powers),

obtaining (5 + 6)rt + (–3 + (–4))rw + (–8 + 2)tw = 11rt – 7rw – 6tw. This is an efficient way to solve

the problem; however, other methods could be used to solve the problem correctly.

Option A is incorrect The student likely made sign errors when combining the rw terms, using −3rw + 4rw = rw, and tw

terms, using −8tw – 2tw = −10tw. The student needs to focus on understanding how to combine like

terms in polynomials.

Option C is incorrect The student likely made a sign error when combining the rw terms, using −3rw + 4rw = rw. The

student needs to focus on understanding how to combine like terms in polynomials.

Option D is incorrect The student likely made a sign error when combining the tw terms, using −8tw – 2tw = −10tw. The

student needs to focus on understanding how to combine like terms in polynomials.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

24 Option F is correct To determine which function could be represented by the quadratic function p(x) with the given

solutions (x-values when p(x) is equal to 0), the student could have used the solutions to construct

and simplify the equation of a quadratic function using p(x) = (x – u)(x – v), where u and v represent

solutions to the equation p(x) = 0. The student could have used the values of the given solutions,

x = −7 and x = 7, letting u = −7 and v = 7, and substituted those values into p(x) = (x – u)(x – v) to

obtain p(x) = [x – (−7)](x – 7). Then the student could have found the product of [x – (−7)](x – 7)

to equal (x + 7)(x – 7) = x

– 49, so p(x) = x

– 49. This is an efficient way to solve the problem;

however, other methods could be used to solve the problem correctly.

Option G is incorrect The student likely substituted the correct solutions into p(x) = (x – u)(x – v) to obtain

p(x) = [x – (−7)](x – 7) but then made a sign error when multiplying −7 and 7, obtaining 49 instead

of −49. The student needs to focus on understanding how to multiply binomial expressions.

Option H is incorrect The student likely substituted the correct solutions into p(x) = (x – u)(x – v) to obtain

p(x) = [x – (−7)](x – 7) but then the student likely added −7 to −7 to obtain −14, instead of

multiplying −7 and 7. The student needs to focus on understanding how to multiply binomial

expressions.

Option J is incorrect The student likely substituted the correct solutions into p(x) = (x – u)(x – v) to obtain

p(x) = [x – (−7)](x – 7) but then added 7 to 7 to obtain 14, instead of multiplying −7 and 7. The

student needs to focus on understanding how to multiply binomial expressions.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

25 Option D is correct To determine the equation of the line in slope-intercept form, y = mx + b, where m represents the

slope of the line and b represents the y-intercept of the line, when given the graph of the line, the

student could have first identified the two ordered pairs shown on the graph, (−2, 6) and (1, −4). The

student then could have substituted those two points into the slope formula,

−

, resulting in

( )

4 6 10 10

− − −

= = = −

−−

. To determine the y-intercept of the line, the student could have substituted

the slope and one of the ordered pairs from the graph, (−2, 6), into the point-slope equation,

y – y

= m(x – x

), where m represents the slope of the line and (x

, y

) represents a point on the

line. The student could have obtained

( )

yx− = − − −

and then solved for y, by first distributing

the

−

, resulting in

10 20

yx− = − −

. Next, the student could have added 6 to both sides of the

equation to isolate the y, resulting in the equation

10 2

yx= − −

. This is an efficient way to solve the

problem; however, other methods could be used to solve the problem correctly.

Option A is incorrect

The student likely correctly determined the value of the y-intercept,

−

The student then likely

correctly calculated the change in x, 3, but when determining the change in y, counted the horizontal

grid lines, including the grid lines at y = 6 and y = −4, and found the change in y to be −11 instead

of −10 and the value of the slope to be

−

. Finally, the student likely reversed the values of the

slope and the y-intercept when substituting those values into the slope-intercept form of a linear

equation, y = mx + b. The student needs to focus on understanding how to write a linear function in

slope-intercept form when given a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option B is incorrect

The student likely correctly determined the value of the y-intercept,

−

The student then likely

correctly calculated the change in x, 3, but when determining the change in y, counted the horizontal

grid lines, including the grid lines at y = 6 and y = −4, and found the change in y to be −11 instead

of −10 and the value of the slope to be

−

. The student needs to focus on understanding how to

write a linear function in slope-intercept form when given a graph.

Option C is incorrect The student likely identified the correct values for the slope, m, and the y-intercept, b, but reversed

those values when substituting them into the slope-intercept form of a linear equation, y = mx + b.

The student needs to focus on understanding how to write a linear function in slope-intercept form

when given a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

26 Option G is correct To determine which exponential function models the values given in the table, the student could have

recognized that an exponential function is of the form

( )

v x ab=

where a is the y-intercept (value

where the graph crosses the y-axis), b is the common factor (constant rate by which successive

values decrease), and x is the variable (symbol used to represent an unknown number). From the

table, the student could have determined that the y-intercept at (0, 9,000) means that the value of a

is 9,000. Next, the student could have determined the common factor, b, by dividing each v(x) value

by the previous v(x) value, calculating

8,100 7,290 6,561

0.9

9,000 8,100 7,290

===

. Substituting a = 9,000 and

b = 0.9 into the exponential equation v(x) = ab

, the student could have obtained

v(x) = (9,000)(0.9)

= 9,000(0.9)

. This is an efficient way to solve the problem; however, other

methods could be used to solve the problem correctly.

Option F is incorrect The student likely correctly determined that the value of a is 9,000. When determining the value of

the common factor, b, the student likely inverted the division to obtain b = 1.1 instead of 0.9. The

student needs to focus on how to find the value of the common factor of an exponential function from

a table of values.

Option H is incorrect The student likely identified the y-intercept of the function as the value of v(x) when x = 1 instead of

when x = 0, to obtain a = 8,100. When determining the value of the common factor, b, the student

likely inverted the division to obtain b = 1.1 instead of 0.9. The student needs to focus on how to

identify the y-intercept and how to find the common factor of an exponential function from a table of

values.

Option J is incorrect The student likely identified the y-intercept of the function as the value of v(x) when x = 1 instead of

when x = 0, to obtain a = 8,100. When determining the value of the common factor, b, the student

likely correctly calculated that the value of the common factor, b, is 0.9. The student needs to focus

on how to identify the y-intercept of an exponential function from a table of values.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

27 2 and any equivalent

values are correct

To determine the positive solution to x

+ 9x – 22 = 0, the student could have recognized the need to

find the factors (numbers or expressions that can be multiplied to get another number or expression)

of x

+ 9x – 22. The student could have determined that x

is equal to x ∙ x and written x as the first

term in each factor. The student then could have determined that the second terms in the two factors

are 11 and −2, because their product (answer when multiplying) is −22 (last term in the expression

given) and their sum is 9 (coefficient of middle term in the expression given). The student could have

then written the factors as (x + 11)(x – 2). Next, the student could have set each factor equal to zero

(x + 11 = 0 and x – 2 = 0) and solved each equation for x, resulting in x = –11 and x = 2. Finally,

the student could have recognized that x = 2 is the positive solution to x

+ 9x – 22 = 0. This is an

efficient way to solve the problem; however, other methods could be used to solve the problem

correctly.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

28 Option G is correct To determine the inequality representing all possible combinations of x, the number of monitors, and

y, the number of keyboards, that can be purchased for the computer lab, the student should have

first identified that each monitor costs $250 and each keyboard costs $50, and represented those

costs by the expressions 250x and 50y. So the total cost of the monitors and keyboards for the

computer lab should be represented by the expression 250x + 50y. Then the student should have

realized that the phrase “at most” can be represented by the inequality symbol “≤”, so the phrase “at

most $4,500” can be represented by “≤ 4,500”. Therefore, all possible combinations of x and y should

be represented by the inequality 250x + 50y ≤ 4,500.

Option F is incorrect The student likely correctly identified the coefficients (numbers multiplied by a variable) representing

the costs of each monitor and each keyboard but identified the phrase “at most” as being represented

by the inequality symbol “<”, instead of “≤”, resulting in 250x + 50y < 4,500. The student needs to

focus on understanding how to write a linear inequality given a real-world situation.

Option H is incorrect The student likely reversed the coefficients (numbers multiplied by a variable) representing the costs

of each monitor and each keyboard and identified the phrase “at most” as being represented by the

inequality symbol “<”, resulting in 50x +250y < 4,500. The student needs to focus on understanding

how to write a linear inequality given a real-world situation.

Option J is incorrect The student likely reversed the coefficients (numbers multiplied by a variable) representing the costs

of each monitor and each keyboard, resulting in 50x + 250y ≤ 4,500. The student needs to focus on

understanding how to write a linear inequality given a real-world situation.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

29 Option A is correct To determine which function (relationship where each input has a single output) best models the data

in the table and graph, the student could have used a graphing calculator to generate the function

using linear regression (method of determining a linear function, y = mx + b, where m represents the

slope of the linear function, and b represents the y-intercept). The function that best models the data

is y = −13.5x + 97.8. This is an efficient way to solve the problem; however, other methods could be

used to solve the problem correctly.

Option B is incorrect The student likely correctly identified the slope of the linear function and estimated from the graph

that the x-intercept would occur near x = 7.3, but used that value for b, the y-intercept. The student

needs to focus on how to use technology to generate the equation of a function when given data in a

table or graph.

Option C is incorrect The student likely generated the correct function using linear regression but reversed the values of

the slope and y-intercept. The student needs to focus on understanding how to write a linear function

that was generated with technology using linear regression.

Option D is incorrect The student likely correctly identified the slope of the linear function and estimated from the graph

that the x-intercept would occur near x = 7.3. Next, the student likely reversed the values of the

slope and y-intercept, using the estimated x-intercept as the slope and the slope as the y-intercept.

The student needs to focus on understanding how to use technology to generate the equation of a

function when given data in a table or graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

30 Option H is correct To determine which statement is true, the student could have first found the factors (numbers or

expressions that can be multiplied to get another number or expression) of x

– 36. The student could

have recognized that x

– 36 can be rewritten as (x)

– (6)

, which represents the difference of

squares pattern, where a

– b

can be written as the product of the binomial factors (a + b) and

(a – b). Applying this pattern, the student could have rewritten the expression x

– 36 = (x)

– (6)

the product (x + 6)(x – 6). Finally, the student could have solved for the zeros (input value, x, that

produces an output value, y, of zero) by setting each factor (expression within the parentheses) equal

to zero (x + 6 = 0 and x – 6 = 0) and solving for x, resulting in x = –6 and x = 6. This is an efficient

way to solve the problem; however, other methods could be used to solve the problem correctly.

Option F is incorrect The student likely incorrectly identified the difference of squares pattern as a

– b

= (a – b)(a – b),

obtaining (x – 6)(x – 6) instead of (x + 6)(x – 6). The student needs to focus on understanding how

to factor an expression representing the difference of squares.

Option G is incorrect The student likely incorrectly identified the difference of squares pattern as a

– b

(a – b)(a – b),

and then likely divided 36 by 2 instead of taking the square root of 36, resulting in (x – 18)(x – 18).

The student needs to focus on understanding how to factor an expression representing the difference

of squares.

Option J is incorrect The student likely correctly identified the difference of squares pattern as a

– b

= (a + b)(a – b).

The student then likely divided 36 by 2 instead of taking the square root of 36, resulting in

(x + 18)(x – 18). The student needs to focus on understanding how to factor an expression

representing the difference of squares.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

31 Option A is correct To determine the value of f(−5), the student should have substituted −5 for x in the function

(relationship where each input has a single output) and then simplified the function, resulting in

f(−5) = 7 – 4(−5) = 7 – (−20) = 27.

Option B is incorrect The student likely made a sign error when multiplying 4(−5), resulting in 7 – 20 = −13. The student

needs to focus on understanding how to perform arithmetic with rational numbers.

Option C is incorrect The student likely subtracted 4 from 7 before multiplying by −5, obtaining 3(−5) = −15. The student

needs to focus on understanding how to apply the order of operations when simplifying a numeric

expression.

Option D is incorrect The student likely calculated the product 7(−4)(−5) = 140. The student needs to focus on

understanding how to apply the order of operations when simplifying a numeric expression.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

32 Option H is correct To determine which graph best represents the linear function y = −4(x + 3) – 2, the student could

have applied the point-slope equation, y – y

= m(x – x

), where m represents the slope of the line

and (x

, y

) represents a point on the line. Solving for y, the student could have obtained

y = m(x – x

) + y

and then determined that the graph of the given line has a slope of −4 and

contains the point (−3, −2). To graph the line, the student could have plotted the point (−3, −2)

and used the slope to find that the points (−2, −6) and (−4, 2) also lie on the line. Therefore, this

graph could represent y = −4(x + 3) – 2 because it has a slope equal to −4 and contains the point

(−3, −2). This is an efficient way to solve the problem; however, other methods could be used to

solve the problem correctly.

Option F is incorrect The student likely solved the point-slope equation, y – y

= m(x – x

), for y, resulting in

y = m(x – x

) + y

. The student then likely interpreted the slope of the line y = −4(x + 3) – 2 as

4 instead of –4 and identified a point on the line as (−3, 2) instead of (−3, −2). The student needs to

focus on understanding how to identify the key features of a linear graph when given an equation in

point-slope form.

Option G is incorrect The student likely solved the point-slope equation, y – y

= m(x – x

), for y, resulting in

y = m(x – x

) + y

. The student then likely interpreted the slope of the line y = −4(x + 3) – 2 as

4 instead of −4 but correctly identified that the line contains the point (−3, −2). The student needs to

focus on understanding how to identify the key features of a linear graph when given an equation in

point-slope form.

Option J is incorrect The student likely solved the point-slope equation, y – y

= m(x – x

), for y, resulting in

y = m(x – x

) + y

. The student then likely identified a point on the line as (−3, 2) instead of

(−3, −2) but correctly identified the slope of the line as −4. The student needs to focus on

understanding how to identify the key features of a linear graph when given an equation in point-

slope form.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

33 Option D is correct To determine a factor of the given expression, 10x

– 19x + 6, the student could have found the

factors (numbers or expressions that can be multiplied to get another number or expression) of the

expression. The student could have first multiplied 10x

by 6, resulting in 60x

. The student then

could have identified two terms that have a product of 60x

and a sum of −19x, which are −15x and

−4x. Then the student could have rewritten the expression in expanded form using these two terms,

resulting in 10x

– 15x – 4x + 6. The student could have grouped the first two terms and last two

terms of the expression and factored out the greatest (largest) common factor from each group of

terms, resulting in 5x(2x – 3) – 2(2x – 3). Next, the student could have factored out the binomial

(2x – 3) from the expression, resulting in the factored form (5x – 2)(2x − 3). Finally, the student

could have recognized that (5x – 2) is one of the factors of the given expression. This is an efficient

way to solve the problem; however, other methods could be used to solve the problem correctly.

Option A is incorrect The student likely determined that two factors of 60x

are 10x and 6x, and that two factors of 6 are

−3 and −2 but disregarded the value of the linear term of the quadratic equation. The student needs

to focus on understanding how to factor an expression of the form ax

+ bx + c.

Option B is incorrect The student likely determined that two factors of 60x

are 10x and 6x, and that two factors of 6 are

−1 and −6 but disregarded the value of the linear term of the quadratic equation. The student needs

to focus on understanding how to factor an expression of the form ax

+ bx + c.

Option C is incorrect The student likely determined the correct expanded form of the expression, 10x

– 15x – 4x + 6, but

likely switched the constant terms when factoring out the common factor from each group. The

student needs to focus on understanding how to factor an expression of the form ax

+ bx + c.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

34 8 and any equivalent

values are correct

To determine the rate of change (constant rate of increase or decrease) of the distance in feet below

sea level with respect to time in seconds the submarine traveled, the student could have chosen two

points from the table and calculated the amount of change. The student could have used the first two

sets of values in the table and applied the slope formula,

−

, resulting in

604 460 144

18 0 18

−

= = =

−

Therefore, the student could have concluded that the rate of change is

8 feet per second. This is an efficient way to solve the problem; however, other methods could be

used to solve the problem correctly.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

35 Option B is correct To determine which equation best represents the line shown on the grid, the student could have

recognized that because the line is horizontal, the equation of the line can be written as y = c,

where c is the value through which the line intersects (crosses) the y-axis, resulting in y = −6. This is

an efficient way to solve the problem; however, other methods could be used to solve the problem

correctly.

Option A is incorrect The student likely recognized that because the line is horizontal, the equation of the line can be

written as y = c, and that the slope of a horizontal line is 0. The student then likely used the value of

the slope, 0, as the constant in the equation, obtaining y = 0. The student needs to focus on

understanding how to write the equation of a horizontal line.

Option C is incorrect The student likely recognized that the line is horizontal and that the slope of a horizontal line is 0.

Then the student likely used the value of the slope, 0, and used x = 0 since a horizontal line is parallel

to the x-axis. The student needs to focus on understanding how to write the equation of a horizontal

line.

Option D is incorrect The student likely recognized that the line is horizontal and has a y-intercept of −6. Then the student

likely used x = −6 since a horizontal line is parallel to the x-axis. The student needs to focus on

understanding how to write the equation of a horizontal line.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

36 Option H is correct To determine the best interpretation of one of the values in the function g(x) = 18(1.3)

, the student

should have recognized that in an exponential function g(x) = ab

, a represents the initial value of the

function (when x = 0), b is the common factor (constant rate by which successive values increase or

decrease), and x is the variable (symbol used to represent an unknown number). In this situation, the

variable x represents the number of months. In g(x) = 18(1.3)

, the initial value of the population is

18 insects, and the value of the common factor is b = 1.3. The student should have recognized that

since 1.3 > 1, this situation represents exponential growth, with a growth factor of 1.3 or 130% and a

growth rate of 0.3 or 30% per month. The student should then have concluded that the insect

population increased at a rate of 30% per month.

Option F is incorrect The student likely misinterpreted b = 1.3 as representing that the population increases by 13 insects

per month, instead of recognizing that 1.3 is a factor of growth and not a constant rate of change of

the population. The student needs to focus on interpreting the meaning of b in exponential function in

the form g(x) = ab

Option G is incorrect The student likely misinterpreted b = 1.3 as representing that the population decreases by 13 insects

per month, instead of recognizing that 1.3 is a factor of growth and not a constant rate of change of

the population. The student needs to focus on interpreting the meaning of b in exponential function in

the form g(x) = ab

Option J is incorrect The student likely misinterpreted b = 1.3 as representing that the insect population decreases at a

rate of 30% per month, instead of recognizing that 1.3 is a factor of growth. The student needs to

focus on interpreting the meaning of b in exponential function in the form g(x) = ab

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

37 Option A is correct

To determine which ordered pair is in the solution set of

yx − −

the student should have

recognized that the graph of the solution set of the inequality would have a boundary line that is

dashed because the “>” symbol indicates that the solution set of the inequality does not include the

points that lie on the boundary line. Next, the student could have used the test point (0, 0) to

determine which half-plane is included in the solution set. Substituting (0, 0) into

yx − −

, the

student could have obtained

( )

0 0 4

 − −

and then 0 > −4. Since that is a true statement, the

student could have then concluded that the solution set of the inequality is the half-plane that

contains (0, 0), not including the points on the boundary line. Finally, the student could have realized

that the point (−8, 8) lies in that half-plane. This is an efficient way to solve the problem; however,

other methods could be used to solve the problem correctly.

Option B is incorrect

The student likely determined that the point (6, −5) lies on the line

yx= − −

, not understanding

that the points on the boundary line are not included in the solution set of the inequality

yx − −

The student needs to focus on understanding how to determine whether an ordered pair is in the

solution set of an inequality.

Option C is incorrect

The student likely used (4, −6) as the test point and substituted it into

yx − −

to obtain

( )

6 4 4

−  − −

and then

−  −

. Next, the student likely incorrectly concluded that the inequality

is true and that the solution set of the inequality is the half-plane that contains (4, −6). The student

needs to focus on understanding how to determine whether an ordered pair is in the solution set of an

inequality.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option D is incorrect The student likely reversed the coordinates of (−2, −7) when using it as test point and substituting

into

yx − −

obtaining

( )

2 7 4

−  − − −

and then

−  −

. Next, the student likely concluded

that the solution set of the inequality is the half-plane that contains that point. The student needs to

focus on understanding how to determine whether an ordered pair is in the solution set of an

inequality.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

38 Option J is correct To determine the equation that best represents the line shown in the graph, the student could have

used the slope-intercept form of a linear equation (y = mx + b, where

−

represents the

slope of the line and b represents the value of the y-intercept). The student first could have

recognized that the graph intersects the y-axis at (0, 32), so the value of the y-intercept, b, is 32.

Next, the student could have substituted the x- and y-coordinates of (0, 32) and (10, 104) into the

slope formula, resulting in

104 32 72 9

40 0 40 5

−

= = =

−

. Since b = 32 and

m =

, the equation for the line

yx=+

. This is an efficient way to solve the problem; however, other methods could be used

to solve the problem correctly.

Option F is incorrect The student likely used the change in x divided by the change in y to find the slope of the line,

ignoring the y-intercept of 32. The student needs to focus on understanding how to write a linear

function in slope-intercept form when given a graph.

Option G is incorrect The student likely determined the slope of the line correctly but disregarded the y-intercept of 32. The

student needs to focus on understanding how to write a linear function in slope-intercept form when

given a graph.

Option H is incorrect The student likely used the change in x divided by the change in y to find the slope of the line, using

the correct value for b, the y-intercept. The student needs to focus on understanding how to write a

linear function in slope-intercept form when given a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

39 Option C is correct  To determine which graph best represents the solution set of x + 2y < −2 and y – x < 3, the student

could have graphed each inequality and its solution set and determined where the two solution sets

overlap with each other. To graph the first inequality, the student could have isolated the y on one

side of the inequality and subtracted x from both sides of the inequality, resulting in 2y < −x – 2.

Next, the student could have divided both sides of the inequality by 2, to obtain

yx − −

. The

boundary line for this inequality,

yx= − −

, will cross the y-axis (vertical axis) at the point (0, −1)

and will have a slope (steepness of a straight line graphed on a coordinate grid;

−

) of

m =−

. The student could have then determined that the “<” symbol means “less than,” indicating

that the boundary line is represented by a dashed line. Next, to determine where to shade, the

student could have used the test point (0, 0) to determine which half-plane represents the solution to

the inequality. Substituting (0, 0) into the original inequality, the student could have obtained

(0) + 2(0) < −2 and then simplified, resulting in 0 < −2. Since this statement is not true, the student

could have concluded that the solution set for the inequality x + 2y < −2 is the half-plane that does

not contain (0, 0) and does not include the points on the boundary line.

For the second inequality, y – x < 3, the student could have added x to both sides of the inequality to

obtain y < x + 3. The boundary line for this inequality will cross the y-axis at the point (0, 3) and will

have a slope of 1. The student could have then determined that the “<” symbol means “less than,”

indicating that the boundary line is represented by a dashed line. Next, to determine where to shade,

the student could have used the test point (0, 0) to determine which half-plane represents the

solution to the inequality. Substituting (0, 0) into the original inequality, the student could have

obtained (0) – (0) < 3 and then simplified, resulting in 0 < 3. Since this statement is true, the student

could have concluded that the solution set for the inequality y – x < 3 is the half-plane that contains

(0, 0) and does not include the points on the boundary line. Finally, the student could have verified

the region where the shaded half-planes overlap. This is an efficient way to solve the problem;

however, other methods could be used to solve the problem correctly.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option A is incorrect  The student likely found and graphed the correct equation for each boundary line but misinterpreted

the “<” symbols as meaning that the boundary lines are included in the solution set. The student

needs to focus on understanding how to graph the solution set of a system of inequalities.

Option B is incorrect  The student likely found and graphed the correct equation for each boundary line but chose the

incorrect half-plane to shade for the inequality

yx − −

. The student needs to focus on

understanding how to graph the solution set of a system of inequalities.

Option D is incorrect  The student likely found and graphed the correct equation for each boundary line but chose the

incorrect half-plane to shade for the inequality

yx − −

and misinterpreted the “<” symbols as

meaning that the boundary lines are included in the solution set. The student needs to focus on

understanding how to graph the solution set of a system of inequalities.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

40 Option F is correct  To determine which graph best represents the function h(x) = (x + 1)(x – 3), the student could have

identified that the zeros of h(x) could be obtained by setting each factor equal to 0 and solving for x.

Setting x + 1 = 0, the student could have solved for x by subtracting 1 from both sides of the

equation to obtain x = −1. Setting x – 3 = 0, the student could have solved for x by adding 3 to both

sides of the equation to obtain x = 3. Next, the student could have recognized that the zeros of a

function are the x-intercepts (values where the graph of a function crosses the x-axis) of the graph of

the function. The student then could have identified that the graph has x-intercepts of −1 and 3.

Finally, the student could have determined the value of the y-intercept by substituting x = 0 into the

function and solving for y, resulting in h(0) = (0 + 1)(0 – 3) = −3. The student could have chosen the

graph with x-intercepts of −1 and 3 and y-intercept of −3. This is an efficient way to solve the

problem. However, other methods could be used to solve the problem correctly.

Option G is incorrect  The student likely used the values of the constants in the binomial factors (x + 1) and (x – 3) and

interpreted that the x-intercepts would occur at x = 1 and x = −3. Also, the student did not find the

correct y-intercept. The student needs to focus on identifying the correct graph of a quadratic function

in the form h(x) = (x – u)(x – v).

Option H is incorrect  The student likely correctly found the correct zeros of the function and interpreted those values to be

the x-intercepts but did not find the correct y-intercept. The student needs to focus on identifying the

correct graph of a quadratic function in the form h(x) = (x – u)(x – v).

Option J is incorrect  The student likely found the correct y-intercept but used the values of the constants in the binomial

factors (x + 1) and (x – 3) and concluded that the x-intercepts would occur at x = 1 and x = −3. The

student needs to identify the correct graph of a quadratic function in the form

h(x) = (x – u)(x – v).

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

41 Option B is correct To determine which equation can be used to find the n

term in the geometric sequence, the student

could have used the pattern, or the sequence, to determine the initial value, the number where the

sequence begins, a

, and the common ratio by dividing each term in the sequence by the preceding

term. Examining the sequence, the student could have realized that

16 64 256 1,024 4,096

4 16 64 256 1,024

− − − − −

= = = = =

− − − − −

. Next, the student could have realized that the pattern is

= −1(4)

, a

= −16 = −1(4)

, a

= −64 = −1(4)

, a

= −256 = −1(4)

, a

= −1,024 = −1(4)

and a

= −4,096 = −1(4)

and concluded that the equation for the n

term of the sequence is

= −1(4)

. This is an efficient way to solve the problem; however, other methods could be used to

solve the problem correctly.

Option A is incorrect The student likely chose a negative coefficient, since all the term values are negative, and likely

misinterpreted that the common ratio is multiplied by the term number to generate a geometric

sequence rather than multiplying the common ratio by the preceding term. The student needs to

focus on how to write the equation for the n

term of a geometric sequence when given several terms

of the sequence.

Option C is incorrect The student likely recognized that each term value is negative and is a perfect square so used an

equation of a sequence that generates the opposite of perfect squares. The student needs to focus on

how to write the equation for the n

term of a geometric sequence when given several terms of the

sequence.

Option D is incorrect The student likely calculated the correct common ratio, 4, but recognizing that all term values are

negative used −4 as the common ratio in the equation instead of 4. The student needs to focus on

how to write the equation for the n

term of a geometric sequence when given several terms of the

sequence.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

42 13 and any equivalent

values are correct

To determine the solution to 4(q + 56.5) = 30q – 112, the student could have first distributed

(multiplied) the number immediately in front of the parentheses by the terms inside the parentheses.

This step would result in the equation 4q + 226 = 30q – 112. Next, the student could have subtracted

4q from both sides, obtaining 226 = 26q – 112. The student could then have added 112 to both sides

with the result 338 = 26q. Lastly, the student could have divided both sides of the equation by 26,

obtaining 13 = q or q = 13. This is an efficient way to solve the problem; however, other methods

could be used to solve the problem correctly.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

43 Option B is correct To determine the expression equivalent to 36m

– 100, the student could have first recognized that

36 and 100 have a greatest (largest) common factor of 4 and factored that out, resulting in

36m

– 100 = 4(9m

– 25). Next, the student could have recognized that the expression inside the

parentheses, 9m

– 25, can be rewritten as (3m)

– (5)

, which represents the difference-of-squares

pattern, where a

– b

can be written as the product of the binomial factors (a – b) and (a + b).

Applying this pattern, the student could have rewritten the expression 4(9m

– 25) = 4[(3m)

– (5)

]

as the product 4(3m – 5)(3m + 5). This is an efficient way to solve the problem; however, other

methods could be used to solve the problem correctly.

Option A is incorrect The student likely determined that two factors of 36m

are 9m and 4m, and that two factors of −100

are −20 and 5, but disregarded the value of the linear term of the quadratic equation. The student

needs to focus on understanding how to factor quadratic expressions.

Option C is incorrect The student likely recognized that 36 and 100 have a common factor of 2 and factored out the 2,

obtaining 2(18m

– 50). Then the student likely recognized that 2m and 9m are factors of 18m

and

that −5 and 10 are factors of −50 but disregarded the value of the linear term of the quadratic

equation. The student needs to focus on understanding how to factor quadratic expressions.

Option D is incorrect The student likely recognized that 36 and 100 have a greatest (largest) common factor of 4 and

factored that out, resulting in 36m

– 100 = 4(9m

– 25). Next, the student likely recognized that

9 and 25 are perfect squares and applied the perfect-square trinomial pattern for factoring

– 2ab + b

= (a – b)

) instead of the difference-of-squares factoring pattern, obtaining

4(3m – 5)

. The student needs to focus on understanding how to factor quadratic expressions.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

44 Option F is correct To determine which function (relationship where each input has a single output) best models the data,

the student could have used a graphing calculator to generate the function using exponential

regression (method of determining the exponential function, r(x) = ab

, where a is the initial

[beginning] value, b is the common factor [constant rate by which successive values increase or

decrease], and x is the variable [symbol used to represent an unknown number]). The function that

best models the data is r(x) = 223.06(1.09)

This is an efficient way to solve the problem; however,

other methods could be used to solve the problem correctly.

Option G is incorrect The student likely transposed the values of a and b when writing the exponential function that models

the data. The student needs to focus on understanding how to identify the different parts of an

exponential regression equation that are generated using technology.

Option H is incorrect The student likely switched the independent values (inputs, or exponent values) with the dependent

values (outputs, or the values that change depending on the inputs) when calculating the exponential

regression, using the time in months as the dependent variable and the net revenue as the

independent variable. The student needs to focus on how to identify independent and dependent

values when using technology to find a regression equation.

Option J is incorrect The student likely switched the independent values (inputs, or exponent values) with the dependent

values (outputs, or the values that change depending on the inputs) when calculating the exponential

regression, using the time in months as the dependent variable and the net revenue as the

independent variable. Next, the student likely transposed the values of a and b when writing the

exponential function that models the data. The student needs to focus on how to identify independent

and dependent values and how to identify the different parts of an exponential regression equation

when using technology to find a regression equation.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

45 Option B is correct To determine which graph best represents the system of equations (two or more equations containing

the same set of variables [symbols used to represent unknown numbers]) and its solution, the

student can first rewrite each equation into slope-intercept form. Slope-intercept form of a linear

equation is y = mx + b, where m represents the slope (steepness of a straight line graphed on a

coordinate grid;

−

) of each line and b represents the y-intercept (value where a line crosses

the y-axis) of each line. To rewrite the first equation, 2x = 6 – y, the student could have first added y

to both sides of the equation, obtaining 2x + y = 6. Next, the student could have subtracted 2x from

both sides, resulting in the equation y = −2x + 6. Next, for the first equation, the student could have

recognized that the graph of the line will cross the y-axis (vertical axis) at the point (0, 6) and will

have a slope (steepness of a straight line graphed on a coordinate grid;

−

) of −2.

To find the slope-intercept form of the second equation, 5x – 4y = 28, the student could first have

subtracted 5x from both sides, resulting in −4y = −5x + 28. Next, the student could have divided

both sides of the equation by −4, obtaining the equation

yx=−

. Finally, the student could have

recognized that the graph of the line will cross the y-axis (vertical axis) at (0, −7) and will have a

slope (steepness of a straight line graphed on a coordinate grid;

−

) of

. This is an

efficient way to solve the problem; however, other methods could be used to solve the problem

correctly.

Option A is incorrect The student likely made sign errors when converting each equation to slope-intercept form, resulting

in identifying the slopes and y-intercepts as being the opposite signs of the correct values. The

student needs to focus on understanding how to rewrite linear equations from standard form or other

forms into slope-intercept form.

Option C is incorrect The student likely made sign errors when converting each equation to slope-intercept form, resulting

in identifying the y-intercepts as being the opposite signs of the correct values. The student needs to

focus on understanding how to rewrite linear equations from standard form or other forms into slope-

intercept form.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option D is incorrect The student likely made sign errors when converting each equation to slope-intercept form, resulting

in identifying the slopes as being the opposite signs of the correct values. The student needs to focus

on understanding how to rewrite linear equations from standard form or other forms into slope-

intercept form.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

46  Option H is correct  To determine the function equivalent to k(x) = x

+ 2x – 15, the student could have recognized the

need to find the factors (numbers or expressions that can be multiplied to get another number or

expression) of x

+ 2x – 15. The student could have determined that x

is equal to x ∙ x and written x

as the first term in each factor. The student could then have determined that the second terms in

each factor are 5 and −3 because their product (answer when multiplying) is −15 (last term in the

expression given) and their sum is 2 (coefficient of middle term in the expression given). The student

could have then written the factors as (x + 5)(x – 3). This is an efficient way to solve the problem;

however, other methods could be used to solve the problem correctly.

Option F is incorrect  The student likely determined that two factors of x

are x and x, and that two factors of −15 are

15 and −1 but disregarded the value of the linear term (middle term, or the term with a degree of 1)

of the quadratic equation. The student needs to focus on understanding how to factor a quadratic

equation of the form f(x) = x

+ bx + c.

Option G is incorrect  The student likely determined that two factors of x

are x and x, and that two factors of −15 are

1 and −15 but disregarded the value of the linear term (middle term, or the term with a degree of 1)

of the quadratic equation. The student needs to focus on understanding how to factor a quadratic

equation of the form f(x) = x

+ bx + c.

Option J is incorrect  The student likely determined that two factors of x

are x and x, and that two factors of −15 are

3 and −5 but disregarded the value of the linear term (middle term, or the term with a degree of 1) of

the quadratic equation. The student needs to focus on understanding how to factor a quadratic

equation of the form f(x) = x

+ bx + c.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

47 Option C is correct  To determine which graph best represents part of a quadratic function with a domain (all possible

x-values) of all real numbers less than −4, the student could have identified a graph with a vertex

(highest or lowest point of the curve) containing an x-coordinate of −4 (which represents the greatest

x-value) and a partial parabola that continues up forever (as represented by the arrow) to the left.

This is an efficient way to solve the problem; however, other methods could be used to solve the

problem correctly.

Option A is incorrect  The student likely identified a graph with a domain of all real numbers less than 0, using the y-value

of the y-intercept, (0, −4), instead of the x-value. The student needs to focus on understanding how

to identify the domain of a quadratic function from a graph.

Option B is incorrect  The student likely identified a graph with a domain of all real numbers greater than −4, confusing

“greater than” with “less than.” The student needs to focus on understanding how to identify the

domain of a quadratic function from a graph.

Option D is incorrect  The student likely identified a graph with a domain of all real numbers greater than 0, using the

y-value of the y-intercept, (0, −4), instead of the x-value, and confusing “greater than” with “less

than.” The student needs to focus on understanding how to identify the domain of a quadratic function

from a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

48 Option G is correct  To determine the slope (steepness of a straight line when graphed on a coordinate grid) when given

two points, the student could have used the given ordered pairs and applied the slope formula

−

. Substituting the values of (−3, 1) and (5, 8) into the slope formula, the student could

have calculated

( )

8 1 7

−

−−

. This is an efficient way to solve the problem; however, other

methods could be used to solve the problem correctly.

Option F is incorrect  The student likely misapplied the slope formula, adding the x-values and the y-values, obtaining

( )

8 1 9

+−

. The student needs to focus on understanding how to use the formula for the slope

of a line when given two ordered pairs.

Option H is incorrect  The student likely misapplied the slope formula, adding the x-values and the y-values and making a

sign error when calculating, obtaining

( )

8 1 9 9

= = = −

−

+−

. The student needs to focus on

understanding how to use the formula for the slope of a line when given two ordered pairs.

Option J is incorrect 

The student likely made a sign error when calculating the slope, resulting in

( )

8 1 7 7

−

= = = −

−

−−

instead of

m =

. The student needs to focus on understanding how to use the formula for the slope

of a line when given two ordered pairs.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

49 Option A is correct  To determine the distance the mail carrier traveled on the morning route, the student could set up

and solve a system of equations [two or more equations containing the same set of variables

(symbols used to represent unknown numbers)]. If x represents the number of miles the mail carrier

traveled on the morning route and y represents the number of miles the mail carrier traveled on the

afternoon route, the student could have set up the two equations; 16x + 12y = 141 (16 times the

number of miles in the morning route + 12 times the number of miles in the afternoon route = 141 miles)

and 10x + 15y = 123.75 (10 times the number of miles in the morning route + 15 times the number

of miles in the afternoon route = 123.75 miles). Next, the student could have solved the system of

equations using the elimination method, multiplying the first equation by 5 and the second equation

by −4, resulting in the equations 80x + 60y = 705 and −40x – 60y = −495. Next, the student could

have added the two equations together to eliminate the terms containing y, resulting in 40x = 210.

Dividing by 40, the student obtained the result x = 5.25. Since x represents the number of miles the

mail carrier traveled on the morning route, the student could have concluded that the distance of the

morning route in miles is 5.25 miles. This is an efficient way to solve the problem; however, other

methods could be used to solve the problem correctly.

Option B is incorrect  The student likely subtracted the number of times the mail carrier delivered mail on the morning route

this month, 10, from the number of times the mail carried delivered mail on the morning route last

month, 16, and concluded that the difference represents the distance of the morning route in miles.

The student needs to focus on understanding how to write a system of equations from a verbal

description.

Option C is incorrect  The student likely set up and solved the system of equations correctly, but switched the values of

x and y, concluding that the distance of the morning route was 4.75 miles instead of 5.25. The

student needs to focus on understanding what value each variable represents in terms of the situation

when solving a system of equations.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

Option D is incorrect  The student likely found the sum of the total distances traveled both months, 141 + 123.75 = 264.75,

and the sum of the total number of routes the mail carrier delivered, 16 + 12 + 10 + 15 = 53, and

then divided the sums, obtaining

264.75

4.995



. Lastly, the student likely rounded 4.99 to 5. The

student needs to focus on understanding how to write a system of equations from a verbal

description.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

50 Option J is correct To determine which function best represents the graph of q, the student could have first identified

p(x) = x

as the quadratic parent function and used the function q(x) = af(x) + d to analyze the

transformation. Next, the student could have recognized that the graph of p was reflected over the

x-axis and translated up 2 units to create the graph of q. The student could then have determined

that a reflection over the x-axis indicates that the coefficient of the quadratic term, a, is −1, and that

a vertical translation up 2 units indicates that the value of d is 2. This is an efficient way to solve the

problem; however, other methods could be used to solve the problem correctly.

Option F is incorrect The student likely correctly identified the reflection of the graph over the x-axis but did not recognize

replacing x with x – 2 in a quadratic function would indicate that the graph was translated 2 units

right instead of up 2 units. The student needs to focus on how the direction of the transformation

affects the function.

Option G is incorrect The student likely correctly identified the reflection of the graph over the x-axis but did not recognize

that replacing x with x + 2 in a quadratic function would indicate that the graph was translated 2 units

left instead of up 2 units. The student needs to focus on how the direction of the transformation

affects the function.

Option H is incorrect The student likely correctly identified the reflection of the parent graph over the x-axis but interpreted

a vertical translation 2 units up as 2 being subtracted from the quadratic term instead of added. The

student needs to focus on how the direction of the transformation affects the function.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

51 Option C is correct To determine the solution to the equation 2(40 – 5y) = 10y + 5(1 – y), the student could first have

distributed (multiplied) the number in front of the parentheses by the terms inside of the parentheses,

resulting in 80 – 10y = 10y + 5 – 5y. Next, the student could have combined like terms (terms that

contain the same variables raised to the same powers) on the right side of the equation, obtaining

80 – 10y = 5y + 5. The student could then have added 10y to both sides of the equation, resulting in

the equation 80 = 15y + 5, and then subtracted 5 from both sides with the result 75 = 15y. Finally,

the student could have divided both sides of the equation by 15, with the result that 5 = y, or y = 5.

This is an efficient way to solve the problem; however, other methods could be used to solve the

problem correctly.

Option A is incorrect The student likely distributed 2 to only the first term in the parentheses, resulting in

80 – 5y = 10y + 5 – 5y. After combining like terms, the student likely obtained 80 – 5y = 5y + 5.

Then, adding 5y and subtracting 5 from both sides, the student likely obtained the result 75 = 10y.

Finally, dividing both sides by 10, the student found that y = 7.5. The student needs to focus on

understanding how to apply the distributive property when solving equations.

Option B is incorrect The student likely made a sign error when applying the distributive property and identified all the

terms as positive after distributing (multiplying) the numbers immediately in front of the parentheses

to the terms inside the parentheses, resulting in 80 + 10y = 10y + 5 + 5y. When combining like

terms, the student likely obtained 80 + 10y = 15y + 5. After subtracting 10y and subtracting 5 from

both sides, the student likely obtained the equation 75 = 5y. Dividing by 5 on both sides, the student

concluded that y = 15. The student needs to focus on understanding how to apply the distributive

property when solving equations.

Option D is incorrect The student likely solved the equation for y and did not get a solution of either 5, 7.5, or 15. The

student needs to focus on understanding how to apply the distributive property and on the arithmetic

of solving equations.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

52 Option F is correct To determine which graph best represents the situation in which the initial value of a home is

$200,000 and the value of the home increases at the rate of 6% per year, the student first could have

recognized that the graph will represent an exponential function in the form y = ab

, where a is the

y-intercept (value where the graph crosses the y-axis), b is the common factor (constant rate by

which successive values increase or decrease), and x is the variable (symbol used to represent an

unknown number). Since it is given that the initial value of the house is $200,000, the student could

have recognized that the value of a is 200,000. Since the value of the home increases at a rate of 6%

per year, the student could have understood that the common factor, b, will be 1 + 0.06, or b = 1.06.

Substituting these values into the exponential function y = ab

, the student could have obtained the

result y = 200,000(1.06)

, where x represents the time in years. The student then could have

calculated the value of the function when x = 5, resulting in y = 200,000(1.06)

. Finally, the student

could have concluded that the point located at approximately (5, 267,645) lies on the graph of the

exponential function. This is an efficient way to solve the problem; however, other methods could be

used to solve the problem correctly.

Option G is incorrect The student likely miscalculated the value of the base, b, as b = 1.6 instead of 1.06, and used an

initial value of a = 100,000. The student needs to focus on understanding how to identify the graph of

an exponential function.

Option H is incorrect The student likely miscalculated the value of the base, b, as b = 1.6 instead of 1.06, but used the

correct initial value of a = 200,000. The student needs to focus on understanding how to identify the

graph of an exponential function.

Option J is incorrect The student likely correctly calculated the value of the base, b, as b = 1.06, but used an incorrect

initial value of a = 100,000. The student needs to focus on understanding how to identify the graph of

an exponential function.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

53 Option C is correct To determine the range (all possible y-values) of the part of the discrete linear function shown, the

student could have identified all the y-values of the points that are plotted. The ordered pairs on the

graph are (0, 96), (1, 88), (2, 80), (3, 72), (4, 64), (5, 56), and (6, 48). The student could have

realized that the set of y-values {96, 88, 80, 72, 64, 56, 48} represents the range of the function for

this situation. This is an efficient way to solve the problem; however, other methods could be used to

solve the problem correctly.

Option A is incorrect The student likely identified the set of seven y-values on the scale of the y-axis, beginning with 96

and decreasing in increments of 12, {96, 84, 72, 60, 48, 36, 24}, as representing the range of the

function. The student needs to focus on understanding how to identify and express the domain and

range of a function from a graph.

Option B is incorrect The student likely identified the set of sums of the values in the domain and the number of balls given

to each player, {8, 9, 10, 11, 12, 13, 14}, as the range. The student needs to focus on understanding

how to identify and express the domain and range of a function from a graph.

Option D is incorrect The student likely identified the set of values of the domain, {0, 1, 2, 3, 4, 5, 6}, as the range. The

student needs to focus on understanding how to identify and express the domain and range of a

function from a graph.

Texas Education Agency

Student Assessment Division

May 2022

2022 STAAR Algebra I Math Rationales

Item # Rationale

54 Option G is correct To determine which graph best represents linear function (a relationship where each input has a

single output) k, the student could have recognized that the zero of a linear function is located at the

x-intercept of the graph. The student could then have identified the graph of a line that appears to

have an x-intercept of −2 and a y-intercept of 6. The student could have determined that the line

intersects (crosses) the x-axis at (−2, 0) and the y-axis at (0, 6), representing an x-intercept of −2

and a y-intercept of 6. This is an efficient way to solve the problem; however, other methods could be

used to solve the problem correctly.

Option F is incorrect The student likely correctly identified the y-intercept of 6 to be located at the point (0, 6), but likely

confused the slope (steepness of a straight line graphed on a coordinate grid;

−

) of the line

with the x-intercept of the line, choosing a line with a slope of −2. The student needs to focus on

understanding how to identify the zero and the y-intercept of a linear function.

Option H is incorrect The student likely reversed the values of the zero of the function and the y-intercept of the function.

The student needs to focus on understanding how to identify the zero and the y-intercept of a linear

function.

Option J is incorrect The student likely reversed the values of the zero of the function and the y-intercept of the function

and then used −6, the opposite of 6, as the zero of the function. The student needs to focus on

understanding how to identify the zero and the y-intercept of a linear function.