2023 STAAR Algebra Math Rationales



2023 STAAR Algebra I Math Rationales

Item #

Rationale

1

Option A is correct

To determine the function that represents the relationship shown in the

table, 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 substituted the x- and y-coordinates of (–4, 10) and (–2, 7) into the

 

slope formula, resulting in     . Next, the student could

 

have substituted the x- and y-coordinates of (6, –5) and the slope,



   , into y = mx + b and solved for b:



  

  







 , or b = 4. Since          . 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 the slope of the line correctly but

subtracted 9 instead of adding when solving for b. The student needs to

focus on understanding how to write a linear function in slope-intercept

form when given a table.

Option C is incorrect

The student likely used the change in x divided by the change in y to find

the slope of the line and subtracted 4 instead of adding when solving for

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

function in slope-intercept form when given a table.

Option D is incorrect

The student likely used the change in x divided by the change in y to find

the slope of the line and correctly used the point (6, –5) to find the value

of b. The student needs to focus on understanding how to write a linear

function in slope-intercept form when given a table.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

2

decreases, 0.25

To complete the statement that describes the rate of change (constant

rate of increase or decrease) of the water level with respect to time, the

student could have chosen two points from the graph and calculated the

amount of change. The student could have used the ordered pairs (0, 9)









and (4, 8) and applied the slope formula,   , resulting in









 

     . Since the rate of change is negative, this

 

indicates that the water level is decreasing at a rate of 0.25 meter per

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

methods could be used to solve the problem correctly.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

3

Option A is correct

To determine the function that best represents the graph, the student

could have identified the solutions (x-values when y is equal to zero) of

the function as u and v and used the solutions to construct and simplify

the equation of a quadratic function using h(x) = a(x – u)(x – v), where a,

u, and v represent values. The solutions can be identified as the x-values

where the parabola (U-shaped graph) crosses the x-axis (at x = –3 and

x = 1). Letting u = –3 and v = 1, the student could have substituted those

values into the equation h(x) = a(x – u)(x – v), resulting in

h(x) = a[x – (–3)](x – 1) → h(x) = a(x + 3)(x – 1). The student could have

then multiplied the expressions (x + 3) and (x – 1), resulting in

2

h(x) = a(x – x + 3x – 3) → h(x) = a(x

2

+ 2x – 3). Next, the student could

have solved for a by substituting the coordinates of the vertex (high or

low point of the curve), (–1, –4), into the function h(x) = a(x

2

+ 2x – 3),

resulting in –4 = a((–1)

2

+ 2(–1) – 3) → –4 = a(1 – 2 – 3) → –4 = –4a →

a = 1. The student could have then substituted the value of a into the

function h(x) = a(x

2

+ 2x – 3), resulting in h(x) = 1(x

2

+ 2x – 3) →

h(x) = x

2

+ 2x – 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 made a sign error when substituting u = –3 and v = 1

into the equation h(x) = a(x – u)(x – v), resulting in h(x) = a(x – 3)(x + 1).

The student needs to focus on understanding how to identify the

solutions of a quadratic function and write the equation of the function

using those solutions.

Option C is incorrect

The student likely made a sign error when substituting u = –3 and v = 1

into the equation h(x) = a(x – u)(x – v), resulting in h(x) = a(x + 3)(x + 1).

The student needs to focus on understanding how to identify the

solutions of a quadratic function and write the equation of the function

using those solutions.

Option D is incorrect

The student likely made a sign error when substituting u = –3 and v = 1

into the equation h(x) = a(x – u)(x – v), resulting in h(x) = a(x – 3)(x – 1).

The student needs to focus on understanding how to identify the

solutions of a quadratic function and write the equation of the function

using those solutions.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

4

Option D is correct

To determine the value of x, the student could have modeled the

equation by using the formula for the area of a rectangle (A = bh where

A is the area of the rectangle, b is the length of the base of the

rectangle, and h is the height of the rectangle) and the formula for the



area of a triangle (  , where A is the area of the triangle, b is the



length of the base of the triangle, and h is the height of the rectangle)

to find the expressions on either side of the equation. Substituting

b = 5 + 2x and h = 10 into the formula for the area of a rectangle, the

student should obtain (5 + 2x)(10) or 50 + 20x. Substituting b = 30 and

h = 4x – 10 into the formula for the area of triangle, the student should



obtain    or 60x – 150. Next, the student should have set



the two expressions equal one another since the area of the rectangle

is equal to the area of the triangle, resulting in 50 + 20x = 60x – 150.

The student could have then subtracted 20x from both sides of the

equation, resulting in 50 = 40x – 150, and then added 150 to both sides,

resulting in 200 = 40x. Finally, the student could have divided both

sides of the equation by 40, resulting in 5 = x, or x = 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 subtracted 50 from 150 on the left side of the

equation instead of adding 150, resulting in 100 = 40x. Dividing by 40

on both sides, the student likely concluded that x = 2.5. The student

needs to focus on understanding the arithmetic of solving equations.

Option B is incorrect

The student likely used the incorrect formula for the area of a triangle,

resulting in 10(5 + 2x) = 2(30)(4x – 10) → 10(5 + 2x) = 60(4x – 10). Next,

the student likely distributed (multiplied) the number in front of the

parentheses to the terms inside the parentheses, resulting in

50 + 20x = 240x – 600. After subtracting 20x from both sides and adding

600 to both sides, the student likely obtained the equation 650 = 220x.

Dividing by 220 on both sides, the student likely concluded that

x ≈ 2.95, which is close to x = 3. The student needs to focus on

understanding the arithmetic of solving equations.

Option C is incorrect

The student likely set the two given expressions equal, resulting in

5 + 2x = 4x – 10. After subtracting 2x from both sides and adding 10 to

both sides, the student likely obtained the equation 15 = 2x. Dividing by

2 on both sides, the student likely concluded that x = 7.5. The student

needs to focus on understanding the arithmetic of solving equations.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

5

Option C is correct

2

To determine a factor of the given expression, 30x – 4x – 16, 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 factored out the greatest common factor

(largest factor that divides evenly into all the terms) from each term,

2 2

resulting in 2(15x – 2x – 8). Next, the student could have multiplied 15x

2

by –8, resulting in –120x . The student then could have identified two

terms that have a product of –120x

2

and a sum of –2x, which are –12x

and 10x. Then the student could have rewritten the expression in

expanded form using these two terms, resulting in

2

2(15x – 12x + 10x – 8). The student could have grouped the first two

terms and last two terms of the expression and factored out the greatest

common factor (largest factor that divides evenly into all the terms) from

each group of terms, resulting in 2[3x(5x – 4) + 2(5x – 4)]. Next, the

student could have factored out the binomial (5x – 4) from the

expression, resulting in the factored form 2(5x – 4)(3x + 2). Finally, the

student could have recognized that (5x – 4) 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

2

The student likely determined that two factors of 15x are 5x and 3x and

that two factors of –8 are 4 and –2 but disregarded the value of the

linear term of the quadratic expression, resulting in 2(5x + 4)(3x – 2). The

student needs to focus on understanding how to factor an expression of

the form ax

2

+ bx + c.

Option B is incorrect

The student likely determined that two factors of 15x

2

are 5x and 3x and

that two factors of –8 are 4 and –2 but disregarded the value of the

linear term of the quadratic expression, resulting in 2(5x + 4)(3x – 2). The

student needs to focus on understanding how to factor an expression of

the form ax

2

+ bx + c.

Option D is incorrect

The student likely determined that two factors of 15x

2

are 5x and 3x and

that two factors of –8 are 2 and –4 but disregarded the value of the

linear term of the quadratic expression, resulting in 2(5x + 2)(3x – 4). The

student needs to focus on understanding how to factor an expression of

the form ax

2

+ bx + c.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

6

Option B is correct

To determine the correlation (measure of strength of a relationship

between two variables), the student should have determined the

correlation coefficient (a value represented by r that measures the

strength of a linear association) using the linear regression feature on a

graphing calculator. The correlation coefficient that best models this

data is r ≈ –0.09. Since the correlation coefficient is negative and close to

zero, the linear association is a weak negative correlation.

Option A is incorrect

The student likely interpreted the strength of the linear association

correctly but identified –(1 + r) as the correlation coefficient. The

student needs to focus on understanding how to determine the

correlation coefficient between two quantitative variables and how to

interpret this quantity as a measure of the strength of the linear

association.

Option C is incorrect

The student likely identified –(1 + r) as the correlation coefficient. The

student needs to focus on understanding how to determine the

correlation coefficient between two quantitative variables and how to

interpret this quantity as a measure of the strength of the linear

association.

Option D is incorrect

The student likely identified the correlation coefficient but interpreted a

correlation coefficient close to zero as representing a strong negative

correlation. The student needs to focus on understanding how to

determine the correlation coefficient between two quantitative

variables and how to interpret this quantity as a measure of the strength

of the linear association.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

7

greater than –4, less

than or equal to 2

To determine the range (all possible y-values) of the part of the linear

function shown, the student could have identified all the values of y for

which the graph has an x-value. The graph extends from –4 at its lowest

point to 2 at its highest point and includes y = 2 and all y-values between

y = –4 and y = 2. Therefore, the range is all real numbers greater than –4

and less than or equal to 2. This is an efficient way to solve the problem;

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

2023 STAAR Algebra I Math Rationales

Item #

Rationale

8

Option D 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

2

another number or expression) of 4x – 36x + 81. The student could

have recognized that 4x

2

and 81 represent perfect squares (numbers

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

also noticed that the square root of 4x

2

is 2x and the square root of 81

is 9. Multiplying the two square roots gives 2x • 9 = 18x. Since 36x is

2

twice 18x, the student could have correctly realized that 4x – 36x + 81

2

has the form of a perfect square trinomial a – 2ab + b

2

, which factors

as (a – b)

2

. In this case, a = 2x and b = 9, so that the factored form of

the function can be written as (2x – 9)

2

. Finally, the student could have

solved for the zero (input value, x, that produces an output value, y, of

zero) by setting the factor equal to zero and solving for x, resulting in 2x



– 9 = 0 or   . 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 4x

2

are 4x and x and

that two factors of 81 are –3 and –27 but disregarded the value of the

linear term of the quadratic equation, resulting in (4x – 3)(x – 27). The

student needs to focus on understanding how to factor an expression

representing a perfect square trinomial.

Option Bis incorrect

The student likely determined that two factors of 4x

2

are 2x and 2x and

that two factors of 81 are 3 and 27 but disregarded the value of the

linear term of the quadratic equation, resulting in (2x + 3)(2x + 27). The

student needs to focus on understanding how to factor an expression

representing a perfect square trinomial.

Option C is incorrect

The student likely incorrectly identified the perfect square trinomial

pattern as a difference of squares, resulting in (2x – 9)(2x + 9). The

student needs to focus on understanding how to factor an expression

representing a perfect square trinomial.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

9

y, 0 or 0, y

To determine the equation of the asymptote (a line that a curve

approaches), the student could have used a graphing calculator to

generate the graph of y = 16(0.75)

x

. Since the graph is an exponential

curve that extends forever to the left and the right and never crosses the

x-axis (horizontal axis), the equation of the asymptote of the graph is

y = 0. This is an efficient way to solve the problem; however, other

methods could be used to solve the problem correctly.

 

2023 STAAR Algebra I Math Rationales

Item #

Rationale

10

Option C is correct

To determine the equivalent expression, the student could have

m

)

n

applied the power of a power property ((a = a

mn

), resulting in

 

 





, or 



. 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 multiplied both the numerator (top number in a fraction)

and the denominator (bottom number in a fraction) of the exponent

 

(power that a number is raised to) by 2, resulting in 



, or 



. The

student needs to focus on understanding how to use the properties of

exponents to simplify expressions with a power raised to a power.

Option B is incorrect

The student added instead of multiplying the exponents, resulting in

 







, or 



. The student needs to focus on understanding how to use

the properties of exponents to simplify expressions with a power raised

to a power.

Option D is incorrect

The student likely added the numerator (top number in a fraction) of

 

the exponent to the power of 2, resulting in 



, or 



. The student

needs to focus on understanding how to use the properties of

exponents to simplify expressions with a power raised to a power.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

11

Option B is correct

To determine which statement is true, the student could have graphed

the quadratic function and analyzed the parabola (U-shaped graph). To

identify the domain (all possible x-values) of the function, the student

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

The student could have determined that the graph continues to expand

upward and outward indefinitely, making the domain all real numbers.

To identify the range (all possible y-values) of the function, the student

could have identified all the y-values for which the graph has an x-value.

The student could have identified- the y-coordinate of the graph’s

lowest point, (2, –8), and all the y-values greater than that y-coordinate,

which means that the range is all values greater than or equal to –8, or

n(x) ≥ –8. 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 domain of the function as the

x-coordinate of the graph’s lowest point and all the x-values greater than

that x-coordinate, resulting in all values greater than or equal to 2, or

x ≥ 2. The student needs to focus on understanding how to represent the

domain of a quadratic function.

Option C is incorrect

The student likely identified the domain of the function as the positive

x-values, resulting in all values greater than or equal to 0, or x ≥ 0. The

student needs to focus on understanding how to represent the domain

of a quadratic function.

Option D is incorrect

The student likely identified the range of the function as the

y-coordinate of the graph’s y-intercept and all the y-values less than that

y-coordinate, resulting in all values less than or equal to 12, or n(x) ≤ 12.

The student needs to focus on understanding how to represent the

range of a quadratic function.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

12

7, 10, >, 100

To determine the inequality that represents all possible combinations of

hats, x, and T-shirts, y, in an order that qualifies for free shipping, the

student should have first identified that each hat costs $7 and each

T-shirt costs $10 and represented those costs by the expressions 7x and

10y. The total cost of the hats and T-shirts in an order should be

represented by the expression 7x + 10y. Then the student should have

realized that the phrase “is over” can be represented by the inequality

symbol “>”, so the phrase “is over $100” can be represented by “ > 100”.

Therefore, all possible combinations of x and y in an order that qualifies

for free shipping should be represented by the inequality 7x + 10y > 100.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

13

Option A is correct

To determine the best estimate for the miles per gallon when the speed

is 65 miles per hour, the student could have first used a graphing

calculator to generate the function using quadratic regression (a method

of determining a quadratic function, y = ax

2

+ bx + c, where a, b, and c are

real numbers). The quadratic function that best models the data is

y = –0.00734x

2

+ 0.689x + 14.115. Next, the student could have

substituted 65 for x in the function and solved for y, resulting in

y = –0.00734(65)

2

+ 0.689(65) + 14.115 = 27.8885. Therefore, 27.9

represents the best estimate for the miles per gallon when the speed is

65 miles per hour. 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 generated the function using a linear regression (a

method of determining a linear function, y = mx + b, where m represents

the slope of the linear function and b represents the y-intercept),

resulting in y = 0.096x + 24.807. Next, the student likely substituted 65

for x in the function and solved for y, resulting in

y = 0.096(65) + 24.807 = 31.047. The student needs to focus on

understanding how to write a quadratic function that was generated

using quadratic regression.

Option C is incorrect

The student likely generated the function using a linear regression using

only the first two ordered pairs in the table, resulting in y = 0.34x + 18.1.

Next, the student likely substituted 65 for x in the function and solved for

y, resulting in y = 0.34(65) + 18.1 = 40.2. The student needs to focus on

understanding how to write a quadratic function that was generated

using quadratic regression.

Option D is incorrect

The student likely interpreted the point (40, 30.1) as the vertex (high or

low point of the curve) and chose the first y-value in the table. The

student needs to focus on understanding how to write a quadratic

function that was generated using quadratic regression.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

14

Option A is correct

To determine which exponential function models the given values, the

student could have recognized that an exponential function is of the

form p(x) = ab

x

, where a is the initial value (starting 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). From the information given, the student could have

determined that the initial population of the town was 48,000,

resulting in a = 48,000. Next, the student could have determined the

common factor, b, by dividing the population of the town after one



year by the initial population, resulting in   . Substituting





a = 48,000 and   into the exponential function p(x) = ab

x

, the







student could have obtained 







    . 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 value of b as . When



determining the initial value, a, the student likely used the value of p(x)

when x = 1 instead of when x = 0, resulting in a = 50,400. The student

needs to focus on understanding how to determine the initial value of

an exponential function from the given information.

Option C is incorrect

The student likely identified the initial value as the value of p(x) when

x = 1 instead of when x = 0, resulting in a = 50,400. Then the student

likely divided 1 by the population of the town when x = 0 to determine



the common factor, resulting in   . The student needs to focus



on understanding how to determine the initial value and common

factor of an exponential function from the given information.

Option D is incorrect

The student likely correctly determined that the value of a is 48,000.

When determining the value of the common factor, b, the student

likely divided 1 by the population of the town after one year, resulting



in   . The student needs to focus on understanding how to



determine the common factor of an exponential function from the

given information.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

15

Option D is correct

To determine which graph represents functions f and g, the student

should have compared the slope (steepness of a straight line when









graphed on a coordinate grid;   ) and y-intercept (value where









a line crosses the y-axis) of both lines on the grid. The graph of function f

increases at a rate of 1 (each time the x-value increases by 1 unit, the

y-value also increases by 1 unit) and has a y-intercept of 0. The graph of

function g increases at a rate of 3 (each time the x-value increases by 1

unit, the y-value increases by 3 units) and has a y-intercept of 0. The

difference between the two graphs is that the graph of function g is

increasing 3 times as fast as the graph of function f. This relationship is

best represented by the graph where each y-value of function g is 3

times the y-value of function f for the same x-value. 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 graph of a function g that has a slope of

1 instead of 3 and a y-intercept of –3 instead of 0. The student needs to

focus on understanding how transformations affect the slope and

y-intercept of the graph of a line.

Option B is incorrect

The student likely identified the graph of a function g that has a slope of

1 instead of 3 and a y-intercept of 3 instead of 0. The student needs to

focus on understanding how transformations affect the slope and

y-intercept of the graph of a line.

Option C is incorrect

The student likely identified the graph of a function g that has a slope of



instead of 3 and a y-intercept of 0. The student needs to focus on



understanding how transformations affect the slope and y-intercept of

the graph of a line.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

16

Option B is correct

To determine the quotient (answer when divided) represented by the

expression, the student could have eliminated the factors (numbers or

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

in the numerator (top number in a fraction) that are in common with

factors in the denominator (bottom number or expression in a fraction).

2

To determine the factors of the numerator, 8w – 20w – 12, the student

could have first factored out the greatest common factor (largest factor

that can be divided evenly into all the terms) from each term, resulting

2

in 4(2w – 5w – 3). Next, the student could have multiplied 2w

2

by –3,

–6w

2

. The student then could have identified two terms that

have a product (answer when multiplied) of –6w

2

and a sum (answer

when added) of –5w, which are w and –6w. Then the student could have

rewritten the expression in expanded form using these two terms,

2

resulting in 4(2w – 6w + w – 3). The student could have grouped the

first two terms and last two terms of the expression and factored out the

greatest common factor from each group of terms, resulting in 4[2w(w –

3) + 1(w – 3)]. Next, the student could have factored out the binomial (w

– 3) from the expression, resulting in the factored form 4(w – 3)(2w + 1).

Finally, the student could have recognized that (2w + 1) is a factor that

the numerator and denominator have in common and eliminated that

factor from both expressions, resulting in 4(w – 3), or 4w – 12. This is an

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

used to solve the problem correctly.

Option A is incorrect

2

The student likely determined that two factors of 8w are 4w and 2w

and that two factors of –12 are 12 and 1, but disregarded the value of

the linear term of the quadratic equation, resulting in (4w + 12)(2w + 1).

Then the student likely recognized that (2w + 1) is a factor that the

numerator and denominator have in common and eliminated that factor

from both expressions, resulting in 4w + 12. The student needs to focus

on understanding how to determine the quotient of a polynomial of

degree two divided by a polynomial of degree one.

Option B is incorrect

The student likely removed the greatest common factor from the







numerator before finding the quotient, resulting in or





. Then the student likely recognized that (2w + 1) is a factor



that the numerator and denominator have in common and eliminated

that factor from both expressions, resulting in w – 3. The student needs

to focus on understanding how to determine the quotient of a

polynomial of degree two divided by a polynomial of degree one.

Option D is incorrect

The student likely removed the greatest common factor from the







numerator before finding the quotient, resulting in . Then, the



2

student likely determined that two factors of 2w are 2w and w, and that

two factors of –3 are 3 and 1, but disregarded the value of the linear

term of the quadratic equation, resulting in (2w + 1)(w + 3). Then, the

student likely recognized that (2w + 1) is a factor that the numerator and

denominator have in common and eliminated that factor from both

resulting in

2023 STAAR Algebra I Math Rationales

expressions, resulting in w + 3. The student needs to focus on

understanding how to determine the quotient of a polynomial of degree

two divided by a polynomial of degree one.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

17

Option A is correct



To determine which ordered pair is in the solution set of     ,



the student should have recognized that the “≤” symbol indicates that

the solution set of the inequality includes the points on the boundary

line. Next, the student could have used the test point (5, –4) to

determine which half-plane is included in the solution set. Substituting



(5, –4) into     , the student could have obtained





     , which simplifies to –4 ≤ –3. Since that is a true



statement, the student could have concluded that the solution set of the

inequality is the half-plane that contains (5, –4). 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 “≤” symbol in the inequality as

meaning “greater than or equal to” and identified an ordered pair in that

solution set. The student needs to focus on understanding the meaning

of the inequality symbol.

Option C is incorrect

The student likely used the point with coordinates of (9, –1) instead of



(9, 1) as a test point and substituted it into      resulting in





 







 , which simplifies to –1 ≤ –0.6. 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.

Option D is incorrect

The student likely reversed the coordinates of (–8, 3) when using it as a



test point and substituted it into      resulting in





 







 , which simplifies to –8 ≤ –4.2. 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.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

18

Option B is correct

To determine which statement is true, the student could have used a

graphing calculator to generate the graph of p(x) = –7(4)

x

. Since the

graph is an exponential curve that extends infinitely to the left and the

right, the domain (all possible x-values) is all real numbers. No matter

which x-value is chosen, its corresponding y-value is negative; therefore,

the range (all possible y-values) is all real numbers less than 0. 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 base (value of b in an exponential

function in the form of p(x) = ab

x

) of the exponential function, b = 4, as

representing the lower boundary of the domain of the function. The

student needs to focus on understanding how to identify and express

the domain and range of a function.

Option C is incorrect

The student likely identified the set of values of the range as the domain

and included zero. The student needs to focus on understanding how to

identify and express the domain and range of a function.

Option D is incorrect

The student likely identified the set of values of the domain as the range.

The student needs to focus on understanding how to identify and

express the domain and range of a function.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

19

Option D is correct

To determine which statement is true, the student could have analyzed

the parabola (U-shaped graph) graphed on the grid. The student could

have recognized that the graph of the function has exactly two

x-intercepts (points where the curve touches the x-axis [horizontal axis]),

which are located at (–1, 0) and (3, 0). The student then should have

concluded that the zeros (input value, x, that produces an output value,

y, of zero) of the function are x = –1 and x = 3. The student could have

also recognized that the vertex (high or low point of the curve) of the

graph is located at (1, 4) and that the graph is facing downward; thus,

the maximum value of the function is 4. Next, the student could have

recognized that the axis of symmetry (an imaginary vertical line that

goes through the vertex of a parabola) is represented by the equation

x = n, where n represents the x-coordinate of the vertex. Therefore, the

equation of the axis of symmetry of the graph of the function is x = 1.

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 counted the y-intercept (value where a graph crosses

the y-axis) as a zero of the function. The student needs to focus on

understanding how to identify the key features of a quadratic function

when given a graph of the function.

Option B is incorrect

The student likely used the greatest value of the x-intercept (value

where a graph crosses the x-axis), x = 3, as the maximum value of the

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

the key features of a quadratic function when given a graph of the

function.

Option C is incorrect

The student likely switched the coordinates in the vertex of the graph of

the function, resulting in (4, 1). The student needs to focus on

understanding how to identify the key features of a quadratic function

when given a graph of the function.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

20

Option B is correct

To determine the equivalent expression, the student could have

rew

ritten



 as



   and then calculated the square root (a

value that when multiplied by itself is equal to the number under the



) of 100, to

get

 



, or 



. 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 reversed the placement of the values after

calculating the

square root, resulting in





. The student needs to

focus on understanding how to simplify square roots.

Option C is incorrect

The student likely r

ewrote 600 as

25 • 24 and then used the 24 from

under the radical as the coefficient. The student needs to focus on

understanding how to simplify square roots.

Option D is incorrect

The student likely r

ewrote 600 as

25 • 24 and then used the 25 from

under the radical as the coefficient. The student needs to focus on

understanding how to simplify square roots.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

21

Option A is correct

To determine which statement is true, the student could have

recognized that the equation of a vertical line can be written as x = a,

where a is the value where the line intersects (crosses) the x-axis

(horizontal axis). Therefore, the equation of the line is x = –4. To

determine the slope (steepness of a straight line when graphed on a









coordinate grid, represented by   ), the student should have









chosen two points on the line and substituted the corresponding values

in the equation for the slope of a line. Using (–4, 0) and (–4, 2),

 

   , which is “undefined” because division by zero is not

 

possible. 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 the variable (symbol used to represent an

unknown number) y for the equation because the line is parallel to the

y-axis (vertical axis), and then used the x-value of –4 as the slope. The

student needs to focus on understanding how to write the equation of

a vertical line and on understanding that the slope of all vertical lines is

undefined.

Option C is incorrect

The student likely used the correct variable, x, for the equation because

the line is perpendicular to the x-axis, but used the x-value of –4 as the

slope. The student needs to focus on understanding that the slope of

all vertical lines is undefined.

Option D is incorrect

The student likely found the correct slope of the line but used the

variable y for the equation because the line is parallel to the y-axis. The

student needs to focus on understanding how to write the equation of

a vertical line.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

22

Option C is correct

To determine the zero (input value, x, that produces an output value, y,

of zero) of linear function f, the student could have identified the

x-value where the line crosses the x-axis (horizontal axis), which is 2.

Therefore, the zero of f is 2. 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 interpreted the zero of the function as representing

the opposite value of the slope (steepness of a straight line when









graphed on a coordinate grid, represented by   ), which is –3.









The student needs to focus on understanding how to identify key

features of linear functions.

Option B is incorrect

The student likely interpreted the zero of the function as representing

the slope, which is 3. The student needs to focus on understanding how

to identify key features of linear functions.

Option D is incorrect

The student likely interpreted the zero of the function as representing

the y-value where the line crosses the y-axis (vertical axis), which is –6.

The student needs to focus on understanding how to identify key

features of linear functions.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

23

Option B 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 –6x + 10y = 42. The student could

have added this to the second equation, 6x – y = –15, to get the result

9y = 27. The student then could have divided both sides of the resulting

equation by 9, obtaining y = 3. Next, to find the corresponding value of

x, the student could have substituted y = 3 into the second equation,

resulting in 6x – 3 = –15. Adding 3 to both sides of that equation results

in 6x = –12. Finally, the student could have divided both sides of the

equation by 6, resulting in x = –2. Since x = –2 and y = 3, the ordered

pair that is a solution to the system of equations is (–2, 3). 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 multiplied the second equation by 5, obtaining

30x – 5y = –75 but made a sign error when adding to the first equation,

resulting in –27x = –54. The student then likely divided both sides of

the resulting equation by –27, obtaining x = 2. Next, the student likely

substituted x = 2 into the second equation of the system, resulting in

6(2) – y = –15 or 12 – y = –15. Subtracting 12 from both sides of the

equation results in –y = –27. Finally, the student likely divided both

sides of the equation by –1, resulting in y = 27. Since x = 2 and y = 27,

the student likely determined that the solution to the system of

equations is (2, 27). The student needs to focus on understanding how

to complete all the steps to calculate the solution to a system of

equations.

Option C is incorrect

The student likely used the elimination method incorrectly by adding

the two equations without eliminating a variable, resulting in

3x + 4y = 6. The student likely then recognized that (2, 0) is a solution to

the resulting equation, since 3(2) + 4(0) = 6. The student needs to focus

on understanding how to complete all the steps to calculate the

solution to a system of equations.

Option D is incorrect

The student likely multiplied the first equation by 2, obtaining

–6x + 10y = 42, but made a sign error when adding to the second

equation, resulting in –9y = 27. The student then likely divided both

sides of the resulting equation by –9, obtaining y = –3. Next, the

student likely substituted y = –3 into the second equation, resulting in

6x + 3 = –15. Subtracting 3 from both sides of the equation results in

6x = –18. Finally, the student likely divided both sides of the equation

by 6, resulting in x = –3. Since x = –3 and y = –3, the student determined

that the solution to the system of equations is (–3, –3). The student

needs to focus on understanding how to complete all the steps to

calculate the solution to a system of equations.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

24

Option A is correct

2

To determine which expressions are equivalent to 12x – 48x + 48, 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 factored out the greatest common factor

(largest factor that divides evenly into all the terms), 12, from each

2

term, resulting in 12(x – 4x + 4). Next, the student could have

recognized that x

2

is equal to x times x and written x as the first term in

each factor. The student could then have determined that the second

terms in the factors are –2 and –2 because their product (answer when

multiplied) is 4 and their sum (answer when added) is –4. The student

could have then written the factors as 12(x – 2)(x – 2), or 12(x – 2)

2

.

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

could be used to solve the problem correctly.

Option E is correct

2

To determine which expressions are equivalent to 12x – 48x + 48, the

student could have found the factors of the expression. The student

could have first factored out the greatest common factor, 12, from

2

each term, resulting in 12(x – 4x + 4). 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 made a sign error when factoring out the greatest

common factor from each term, resulting in –12(x

2

+ 4x + 4). The

student needs to focus on understanding how to factor an expression

of the form ax

2

+ bx + c.

Option C is incorrect

2

The student likely determined that two factors of x are x and x and

that two factors of 4 are –4 and –1 but disregarded the value of the

linear term of the quadratic equation, resulting in 12(x – 4)(x – 1). The

student needs to focus on understanding how to factor an expression

of the form ax

2

+ bx + c.

Option D is incorrect

The student likely made a sign error when factoring out the greatest

common factor from each term, resulting in –12(x

2

+ 4x + 4). Next, the

student likely determined that the first term in each factor is x and that

the second term in each factor is 2, resulting in –12(x + 2)(x + 2), or

–12(x + 2)

2

. The student needs to focus on understanding how to factor

an expression of the form ax

2

+ bx + c.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

25

Option D is correct

To determine the quadratic function in vertex form (y = a(x – h)

2

+ k,

where (h, k) is the vertex [high or low point of the curve] and a is the

coefficient of the quadratic term), the student could have identified the

vertex of the function as (h, k) and used the additional given point to

create the equation of the quadratic function. Letting h = 1 and k = 46,

the student could have substituted those values into the function

y = a(x – h)

2

+ k, resulting in y = a(x – 1)

2

+ 46. Next, the student could

have solved for a by substituting the coordinates of the additional point,

(3, 10), into the function y = a(x – 1)

2

+ 46, resulting in 10 = a(3 – 1)

2

+ 46

→ 10 = 4a + 46 → –36 = 4a → –9 = a. The student could have then

substituted the value of a into the function y = a(x – 1)

2

+ 46, resulting in

y = –9(x – 1)

2

+ 46. 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 subtracted 10 from 46 instead of subtracting 46 from

10 when solving for a, resulting in 36 = 4a or 9 = a. Then the student

likely identified the coordinates of the given point as the values of h and

k, resulting in y = 9(x – 3)

2

+ 10. The student needs to focus on

understanding how to write quadratic functions in vertex form.

Option B is incorrect

The student likely subtracted 10 from 46 instead of subtracting 46 from

10 when solving for a, resulting in 36 = 4a or 9 = a. The student likely

then substituted the value of a into the function y = a(x – 1)

2

+ 46,

resulting in y = 9(x – 1)

2

+ 46. The student needs to focus on

understanding how to write quadratic functions in vertex form.

Option C is incorrect

The student likely identified the coordinates of the given point as the

values of h and k, resulting in y = –9(x – 3)

2

+ 10. The student needs to

focus on understanding how to write quadratic functions in vertex form.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

26

Option B is correct

To determine which function (relationship where each input has a single

output) best models the data in the table, the student could have used a

graphing calculator to generate the function using linear regression (a

method of determining a linear function, y = mx + b, where m represents

the slope and b represents the y-intercept). The function that best

models the data is f(x) = –41x + 337, or f(x) = 337 – 41x. 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 identified the slope of the linear function

and used the first y-coordinate from the table, y = 296, as the value of b,

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.

Option C is incorrect

The student likely generated the function using only the last two

ordered pairs from the table, (6, 89) and (7, 51). Substituting the two

ordered pairs into the formula for the slope of a line, the student likely



obtained    . Next, the student likely substituted the



coordinates (7, 51) and the slope m = –38 into y = mx + b and solved for

b, resulting in 51 = –38(7) + b → 51 = –266 + b → 317 = b. Since b = 317

and m = –38, the student obtained the equation f(x) = –38x + 317, or

f(x) = 317 – 38x. 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.

Option D is incorrect

The student likely used only the last two ordered pairs from the table,

(6, 89) and (7, 51), to calculate the slope of the line and then used the

first y-coordinate from the table as the value of b, 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.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

27

Option D is correct

To determine what 1.029 represents in the function w(t) = 270(1.029)

t

,

the student should have recognized that in an exponential function

w(t) = ab

t

, a represents the initial value (starting value), b is the

common factor (constant rate by which successive values increase or

decrease), and t is the variable (symbol used to represent an unknown

number). In this situation, the variable t represents the number of

years. In w(t) = 270(1.029)

t

, the student should have recognized that

the initial number of whales is 270 since a = 270. The student should

have also recognized that the number of whales are increasing at a rate

of 2.9% per year since b = 1.029 and that 1.029 represents the growth

factor of the number of whales since 1.029 > 1.

Option A is incorrect

The student likely interpreted b = 1.029 as the initial number of whales

in the North Atlantic Ocean, instead of recognizing that b = 1.029 is a

growth factor since b > 1 and that a = 270 is the initial value. The

student needs to focus on interpreting the meaning of the values of a

and b of an exponential function in the form w(t) = ab

t

.

Option B is incorrect

The student likely interpreted b = 1.029 as the decay factor of the

number of whales in the North Atlantic Ocean, which indicates a

decrease of 2.9% per year, instead of recognizing that b = 1.029 is a

growth factor since b > 1. The student needs to focus on interpreting

the meaning of the values of a and b of an exponential function in the

form w(t) = ab

t

.

Option C is incorrect

The student likely interpreted b = 1.029 as the number of whales in the

North Atlantic Ocean at the end of the first year, instead of recognizing

that b = 1.029 is a growth factor since b > 1 and that the value of w(t)

when t = 1 is the number of whales at the end of the first year. The

student needs to focus on interpreting the meaning of the values of a

and b of an exponential function in the form w(t) = ab

t

.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

28

Option C is correct

To determine the slope (steepness of a straight line graphed on a

coordinate grid) of the line, the student could have used the given









ordered pairs and applied the slope formula,    Substituting









the x-values and y-values of (−2, –2) and (4, 2) into the slope formula,

  

the student could have calculated     . 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 as the change in the y-values

divided by the change in the x-values but made a sign error. The student

needs to focus on understanding how to use the formula for the slope of

a line when given two ordered pairs.

Option B is incorrect

The student likely calculated the slope as the change in the x-values









  

divided by the change in the y-values,      . The









  

student needs to focus on understanding how to use the formula for the

slope of a line when given two ordered pairs.

Option D is incorrect

The student likely calculated the slope as the change in the x-values

divided by the change in the y-values and made a sign error. The student

needs to focus on understanding how to use the formula for the slope of

a line when given two ordered pairs.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

29

Option B is correct

To determine the solution to the equation 5(2w + 4) = 4(2w + 9), the

student could first have distributed (multiplied) the number in front of

the parentheses to the terms inside the parentheses, resulting in

10w + 20 = 8w + 36. Next, the student could have subtracted 8w from

both sides of the equation, resulting in 2w + 20 = 36. The student then

could have subtracted 20 from both sides of the equation, resulting in

2w = 16. Finally, the student could have divided both sides of the

equation by 2, resulting in w = 8. 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 only to the first terms in the parentheses,

resulting in 10w + 4 = 8w + 9. Then, subtracting 8w and subtracting 4

from both sides of the equation, the student likely obtained 2w = 5.

Finally, dividing both sides of the equation by 2, the student likely



found that   . The student needs to focus on understanding how to



apply the distributive property when solving equations.

Option C is incorrect

The student likely added instead of subtracting when moving terms

across the equal sign, resulting in 10w + 20 = 8w + 36 → 18w = 56.

Dividing both sides of the equation by 18, the student likely found that

 

   . The student needs to focus on understanding the

 

arithmetic of solving equations.

Option D is incorrect

The student likely used addition instead of multiplication when

distributing and distributed only to the first terms in the parentheses,

resulting in 7w + 4 = 6w + 9. Then, subtracting 6w and subtracting 4

from both sides of the equation, the student likely obtained w = 5. The

student needs to focus on understanding how to apply the distributive

property when solving equations.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

30

Option D is correct

To determine the best estimate of the price of a discounted ticket for

the baseball game, the student could have graphed the first equation

and the second equation on the same coordinate plane and estimated

the coordinates of the point where the two lines intersect (cross) given

the graph of the first equation. The student then could have estimated

that the two lines intersect at (13.95, 11.15). Since y represents the price

of a discounted ticket in dollars, the student could have concluded that

the price of a discounted ticket is $11.15. 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 estimated the price of a standard ticket instead of

estimating the price of a discounted ticket, resulting in $13.95. The

student needs to focus on interpreting the point of intersection of two

intersecting lines.

Option B is incorrect

The student likely overestimated the price of a standard ticket instead of

estimating a discounted ticket, resulting in $14.55. The student needs to

focus on estimating the solution to a system of equations using the

graphing method.

Option C is incorrect

The student likely divided the total amount of ticket sales by the sum

(answer when added) of the number of standard tickets sold and the

number of discounted tickets sold, resulting in

 

  . The student needs to focus on estimating the

 

solution to a system of equations using the graphing method.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

31

Option A is correct



To determine the first four terms of the sequence 







 



  



,



where f(1) = 27, the student could have substituted n = 2, n = 3, and

n = 4 into the function to determine the second, third, and fourth terms

of the sequence, respectively. Since f(1) = 27, the student should have

concluded that the first term of the sequence is 27. Substituting n = 2

into the function, the student could have obtained

  









 



  



 















  so the second term of the

  

sequence is 9. Substituting n = 3 into the function, the student could

  

have obtained 







 



  



 















  so the third

  

term of the sequence is 3. Last, substituting n = 4 into the function, the

student could have obtained

  









 



  



 















  so the fourth term of the

  

sequence is 1. The first four terms of the sequence are 27, 9, 3, 1. 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 multiplied by 3 instead of multiplying by when



evaluating the function for n = 2, n = 3, and n = 4, resulting in the terms

27, 81, 243, 729. The student needs to focus on understanding how to

identify terms of a geometric sequence when the sequence is given in

function form using a recursive process.

Option C is incorrect



The student likely identified as the first term of the sequence and then



added 27 to the numerator (top number in a fraction) for each

   

additional term, resulting in the terms , , , . The student needs to

   

focus on understanding how to identify terms of a geometric sequence

when the sequence is given in function form using a recursive process.

Option D is incorrect



The student likely identified as the first term of the sequence and then



multiplied the denominator (bottom number in a fraction) by 27 for

   

each additional term, resulting in the terms , , , . The

   

student needs to focus on understanding how to identify terms of a

geometric sequence when the sequence is given in function form using a

recursive process.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

32

Option B is correct

To determine the range (all possible y-values) of g, the student could

have plotted the six ordered pairs presented in the table on a coordinate

grid and analyzed the shape of the graph. The student could have

     

plotted points at   , (1, 4),   , (2, 4),   , and (3, –2) on a

     

coordinate grid and determined that the points of the graph represent a

parabola (U-shaped graph). Since the graph of the parabola opens



downward, the maximum value should be identified as because the



 

vertex (high point of the curve)    is the same distance horizontally

 

from the points (1, 4) and (2, 4), which means the vertex must be

halfway between the two points. Therefore, the range of function g is all



real numbers less than or equal to . 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 x-coordinate, instead of the

y-coordinate, of the vertex as the maximum value of the range. The

student needs to focus on understanding how to represent the range of

a quadratic function from a table of values.

Option C is incorrect

The student likely identified the x-coordinate, instead of the

y-coordinate, of the vertex as the boundary for the range and did not

recognize that the parabola opens downward. The student needs to

focus on understanding how to represent the range of a quadratic

function from a table of values.

Option D is incorrect

The student likely identified the y-coordinate of the vertex as the

boundary for the range but did not recognize that the parabola opens

downward. The student needs to focus on understanding how to

represent the range of a quadratic function from a table of values.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

33

Option C is correct

To determine which expression is equivalent to (2a + 5)(3a – 2), the

student could have multiplied each term in the factor (2a + 5) by each

term in the factor (3a – 2) and then combined like terms (terms that

contain the same variables raised to the same powers or constant

terms). The multiplication steps are 2a(3a – 2) + 5(3a – 2), resulting in

2

6a – 4a + 15a – 10. The student could have combined like terms to

obtain 6a

2

+ 11a – 10. 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 multiplied each term in the factor (2a + 5) by each

term in the factor (3a – 2) but determined the product of 2a and 3a to

be 6a instead of 6a

2

. The student likely multiplied 2a(3a – 2) + 5(3a – 2)

to get a result of 6a – 4a + 15a – 10, and then combined like terms,

resulting in 17a – 10. The student needs to focus on understanding how

to find the product of two binomials.

Option B is incorrect

The student likely multiplied only the first terms and the last terms in

2

the two factors, resulting in 2a(3a) + 5(–2), or 6a – 10. The student

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

binomials.

Option D is incorrect

The student likely did not apply the negative sign when distributing

(2a + 5) to the factor (3a – 2), resulting in 6a

2

+ 4a + 15a + 10. The

student likely then combined like terms to obtain 6a

2

+ 19a + 10. The

student needs to focus on understanding how to find the product of

two binomials.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

34

Option D is correct

To determine the linear function that models the total cost, t, for a

single order of c cartridges, the student could have used the slope-

intercept form of a linear equation (t = mc + b, where m represents the

slope of the line and b represents the value of the t-intercept). The

student could have identified the slope and y-intercept from the

situation. Since the printer ink cost is a constant rate, $18.99 per

cartridge, the student could have recognized that this represents the

slope of the function, so m = 18.99. Since the shipping fee is a flat rate,

$7.95, no matter the number of cartridges purchased in a single order,

the student could have recognized that this represents the initial value

(when c = 0) or t-intercept of the function, so b = 7.95. Therefore, the

linear function that models the total cost for a single order of cartridges

is t = 18.99c + 7.95. 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 reversed the variables (symbols used to represent an

unknown number) in the function and added the cost per cartridge and

the shipping fee, resulting in c = 26.94t. The student needs to focus on

understanding how to write linear functions given a description of a

situation.

Option B is incorrect

The student likely reversed the variables in the function, resulting in

c = 18.99t + 7.95. The student needs to focus on understanding how to

write linear functions given a description of a situation.

Option C is incorrect

The student likely added the cost per cartridge and the shipping fee,

resulting in t = 26.94c. The student needs to focus on understanding

how to write linear functions given a description of a situation.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

35

4, 4

To determine the correct value of the exponent for each term, the



student could have applied the negative exponent property, 









,

to the y in the denominator (bottom of a fraction), resulting in

6 2 –1

x y

3

÷ x y . Next, the student could have applied the quotient of powers





 



property, , to the factors containing x and y, resulting in





6 – 2 3 – (–1) 4 4

x y = x y . This is an efficient way to solve the problem; however,

other methods could be used to solve the problem correctly.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

36

Option C is correct

To determine the function that models the number of students

participating in sports after x years, the student could have used an

exponential function of the form f(x) = ab

x

, where a is the initial value

(starting 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). From the given information,

the student could have determined that the initial number of students

who participated in sports was 317, so a = 317. The student should have

recognized that since the number of students who participate in sports

increases, this situation represents exponential growth, with a growth

factor of b = 1 + 0.04, or b = 1.04. Substituting a = 317 and b = 1.04 into

the exponential function f(x) = ab

x

, the student could have obtained

f(x) = 317(1.04)

x

. 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 correct initial value, a = 317, but used

the growth rate, 4%, as the growth factor, resulting in f(x) = 317(4)

x

. The

student needs to focus on understanding how to determine the

common factor of an exponential function from the given information.

Option B is incorrect

The student likely identified the growth rate, 4%, as a rate of change

(constant rate of increase or decrease) and represented the situation

with a linear function instead of an exponential function, resulting in

f(x) = 4x + 317. The student needs to focus on understanding how to

determine the initial value and common factor of an exponential

function from the given information.

Option D is incorrect

The student likely identified the growth factor, b = 1.04, as a constant

rate of change and represented the situation with a linear function

instead of an exponential function, resulting in f(x) = 1.04x + 317. The

student needs to focus on understanding how to determine the initial

value and common factor of an exponential function from the given

information.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

37

Option D is correct

To determine which system of equations (two or more equations

containing the same set of variables [symbols used to represent

unknown numbers]) represents the graph of the two lines, the student

could write each equation in slope-intercept form (y = mx + b, where m

is the slope and b is the y-intercept) and then convert each equation to

standard form (ax + by = c where a, b, and c are integers).

To find the equation for line a, the student could have used the first

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









 

  , resulting in     . Next, the student









 

could have substituted one of the ordered pairs from the table, (4, –9),

and the slope, m = –6, into y = mx + b and solved for b, resulting in

–9 = –6(4) + b → –9 = –24 + b → 15 = b. Since b = 15 and m = –6, the

equation for line a in slope-intercept form is y = –6x + 15. To convert

the equation from slope-intercept form to standard form, the student

could have added 6x to both sides of the equation, resulting in

6x + y = 15.

To find the equation for line b, the student could have used the first

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









 

  , resulting in     . Next, the student









 

could have substituted one of the ordered pairs from the table,

(1, –12), and the slope, m = –3, into y = mx + b and solved for b,

resulting in –12 = –3(1) + b → –12 = –3 + b → –9 = b. Since b = –9 and

m = –3, the equation for line b in slope-intercept form is y = –3x – 9. To

convert the equation from slope-intercept form to standard form, the

student could have added 3x to both sides of the equation, resulting in

3x + y = –9.

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 slopes as positive instead of negative

and multiplied the y-term and constant by the coefficient of the x-term

in each equation, resulting in 6y = x + 90 for line a and 3y = x – 27 for

line b. The student likely then converted the equations from slope-

intercept form to standard form by subtracting the y-term and the

constant value from both sides of the equations, obtaining x – 6y = –90

for line a and x – 3y = 27 for line b. The student needs to focus on

understanding how to write a linear function in standard form when

given a table.

Option B is incorrect

The student likely multiplied the y-term and constant by the coefficient

of the x-term in each equation, resulting in 6y = x + 90 for line a and

3y = x – 27 for line b. The student likely then converted the equations

from slope-intercept form to standard form by adding the x- and

y-terms on one side of the equations, obtaining x + 6y = 90 for line a

2023 STAAR Algebra I Math Rationales

and x + 3y = –27 for line b. The student needs to focus on

understanding how to write a linear function in standard form when

given a table.

Option C is incorrect

The student likely identified the y-intercepts but identified the slopes

as positive instead of negative before converting the equations into

standard form, resulting in y = 6x + 15 for line a and y = 3x – 9 for line b.

The student likely then converted the equations from slope-intercept

form to standard form by subtracting the y-term and the constant

value from both sides of the equation, obtaining 6x – y = –15 for line a

and 3x – y = 9 for line b. The student needs to focus on understanding

how to write a linear function in standard form when given a table.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

38

3, –9

To determine the coordinates of the vertex of the graph of g, the

2

student could have identified f(x) = x as the quadratic parent function

and recognized that the coordinates of the vertex of the graph of f are

(0, 0). The student could have recognized that f(x – 3) represents the

vertex of the graph of f shifted 3 units to the right. Then the student

could have recognized that –9 in g(x) = f(x – 3) – 9 represents the vertex

of the graph of f shifted down 9 units. Thus, the coordinates of the

vertex of the graph of g are (3, –9). This is an efficient way to solve the

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

correctly.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

39

Option C is correct

To determine which expression is equivalent to 24gh – 12g

2

+ 18g, the

student could have determined that 6g is the greatest common factor

(largest factor [numbers multiplied together to produce another

2

number] that the numbers share) of 24gh, 12g , and 18g. Because

6g(4h) = 24gh, 6g(2g) = 12g

2

, and 6g(3) = 18g, the student could have

factored out 6g from the expression, resulting in 6g(4h – 2g + 3). 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 6g is the greatest common factor but

did not factor out the 6 from the last two terms, resulting in

6g(4h – 12g + 18). The student needs to focus on understanding how to

identify the greatest common factor of an expression.

Option B is incorrect

The student likely recognized that 12g is the greatest common factor

for the first two terms but did not factor out the 12 from the last two

terms, resulting in 12g(2h – 12g + 18). The student needs to focus on

understanding how to identify the greatest common factor of an

expression.

Option D is incorrect

The student likely recognized that 12g is the greatest common factor

for the first two terms but divided the last term by 6g, resulting in

12g(2h – g + 3). The student needs to focus on understanding how to

identify the greatest common factor of an expression.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

40

Option D is correct

To determine the cost per click on Website 1, the student could set up

and solved a system of equations (two or more equations containing

the same set of variables [symbols used to represent unknown

numbers]). If x represents the cost per click on Website 1 and y

represents the cost per click on Website 2, the student could have set

up two equations:

15x + 29y = 94.15 (15 clicks on Website 1 and 29 clicks on Website 2 on

Monday for a total cost of $94.14) and

25x + 29y = 121.15 (25 clicks on Website 1 and 29 clicks on Website 2

on Tuesday for a total cost of $121.15).

Next, the student could have solved the system of equations using the

elimination method. Subtracting the first equation from the second

equation to eliminate the term containing y, the student could have

obtained 10x = 27. Dividing both sides of the equation by 10, the

student could have obtained the result x = 2.7. Since x represents the

cost per click on Website 1, the student could have concluded that the

cost is $2.70. 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 set up and solved the system of equations correctly

but switched the values of x and y, concluding that the cost per click on

Website 1 was $1.85 instead of $2.70. The student needs to focus on

understanding what value each variable represents in terms of the

situation when solving a system of equations.

Option B is incorrect

The student likely found the sum of the number of clicks on Website 1,

15 + 25 = 40, and the sum of the total costs, 94.15 + 121.15 = 215.3,



and then divided the sums, obtaining  . Last, the student



likely rounded 5.3825 to 5.38. 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 found the sum of the numbers of clicks on Website 1

and Website 2, 15 + 25 + 29 + 29 = 98, and the sum of the total costs,

94.15 + 121.15 = 215.3, and then divided the sums, obtaining



 . Last, the student likely rounded 2.1969 to 2.20. The



student needs to focus on understanding how to write a system of

equations from a verbal description.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

41

Option A is correct

To determine the solutions to k(x) = 0, the student could have used the







quadratic formula (  , where a, b, and c are the



coefficients of the quadratic equation ax

2

+ bx + c = 0) to solve for x.

The student could have recognized that a = 1, b = 32, and c = 248, since

the quadratic equation is x

2

+ 32x + 248 = 0. Next, the student could

have substituted a = 1, b = 32, and c = 248 into the quadratic formula

and evaluated, resulting in









 



 





  →   →   →  

  







→     



. Therefore, the solutions to k(x) = 0 are  



  



 and     



.

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 the quadratic formula to solve for x but started

the formula with b instead of –b in the numerator (top number or

expression in a fraction), resulting in









 



 





  →   →   →  

  







→     



.



The student needs to focus on understanding how to apply the

quadratic formula when finding solutions to a quadratic equation.

Option C is incorrect

The student likely used the quadratic formula to solve for x but started

the formula with a instead of 2a in the denominator (bottom number











or expression in a fraction), resulting in   →  







 



 





→   →   →     



.

  

The student needs to focus on understanding how to apply the

quadratic formula when finding solutions to a quadratic equation.

Option D is incorrect

The student likely used the quadratic formula to solve for x but started

the formula with b instead of –b in the numerator and with a instead of

2a in the denominator, resulting in









 



 





  →   →   →  

  







→     



.



The student needs to focus on understanding how to apply the

quadratic formula when finding solutions to a quadratic equation.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

42

Option A is correct

To determine which statement is true about the graph that represents

the relationship between the value of a device in dollars and the

number of years since the device was purchased, the student could

have modeled the exponential function in the form y = ab

x

, where a is

the initial value (starting value), b is the common factor (rate by which

successive values increase or decrease), and x is the variable (symbol

used to represent an unknown number) and then used a graphing

calculator to generate the graph of the function. From the given

information, the student could have determined that the initial value of

the electronic device was $650, so a = 650. The student should have

recognized that since the value of the electronic device decreases by

30%, this situation represents exponential decay, with a decay factor of

b = 1 – 0.3, or b = 0.7. Substituting a = 650 and b = 0.7 into the

exponential function y = ab

x

, the student could have obtained

y = 650(0.7)

x

. Next, the student could have used a graphing calculator

to generate the graph of y = 650(0.7)

x

. Since the graph is an exponential

curve that extends infinitely to the left and the right and never crosses

the x-axis (horizontal axis), the equation of asymptote (a line that a

curve approaches) of the graph is y = 0. Next, the student could have

recognized that the y-intercept (value where a graph crosses the y-axis)

of the graph is 650. 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 identified the initial value as the value of y when

x = 1 instead of when x = 0, resulting in a = 455. Substituting a = 455

and b = 0.7 into the exponential function y = ab

x

, the student likely

obtained y = 455(0.7)

x

. Next, the student likely recognized that the

y-intercept of the graph is 455. The student needs to focus on

understanding how to identify the key features of exponential

functions.

Option C is incorrect

The student likely confused horizontal with vertical when describing

the asymptote of the graph of the function. The student needs to focus

on understanding how to identify the key features of exponential

functions.

Option D is incorrect

The student likely interpreted the rate of decrease as the horizontal

asymptote. The student needs to focus on understanding how to

identify the key features of exponential functions.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

43

solid boundary line

with a slope of –1 and

y-intercept 2; shaded

half plane that does

not include test point

(0, 0)

To determine the solution set for the linear inequality y ≥ –x + 2, the

student should have recognized that the graph of the solution set of the

inequality would have a boundary line that is solid because the “≥”

symbol indicates that the solution set includes the boundary line. The

student should have recognized that the y-intercept of the inequality is 2

and the slope is –1 and graphed the 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 y ≥ –x + 2, the student could have

obtained 0 ≥ 0 + 2, or 0 ≥ 2. Since that is a false statement, the student

could have then concluded that the solution set of the inequality is the

half plane that does not contain (0, 0) and shaded that region. This is an

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

used to solve the problem correctly.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

44

Option C is correct

To determine the equation that best represents the line that passes

through point P and is parallel (lines that do not intersect [cross] and

are always the same distance from each other) to line m, the student

could have first chosen two points from the graph of line m and

calculated the slope (steepness of a straight line when graphed on a









coordinate grid, represented by   ). The student could have









substituted the x- and y-coordinates of (0, –3) and (2, 0) from the graph

 

of line m into the slope formula, resulting in    . Since

 

parallel lines have the same slope, the student should have concluded



that the line passing through point P has a slope of . Next, the student



could have substituted the x- and y-coordinates of point P, (–2, 2), and

 

  into y = mx + b and solved for b, resulting in  







 , or

 



b = 5. Since b = 5 and   , the equation of the line that passes





through point P and is parallel to line m 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 correct slope for both lines but

switched the x- and y-coordinates of point P when substituting into



y = mx + b, resulting in  







 , or b = –5. Since b = –5 and



 

  , the student likely obtained the equation     . The

 

student needs to focus on understanding how to determine the

y-intercept of a line that is parallel to a given line.

Option B is incorrect

The student likely used the slope of the line that passes through point P



and is perpendicular to line m, which is  . Next, the student likely





substituted the x- and y-coordinates of point P, (–2, 2), and   





into y = mx + b and solved for b, resulting in   







 , or



  

  . Since   and    , the student likely obtained the

  

 

equation      . The student needs to focus on understanding

 

how to determine the slope of a line that is parallel to a given line.

Option C is incorrect

The student likely used the slope of the line that passes through point P

and is perpendicular to line m and switched the x- and y-coordinates of

point P when substituting into y = mx + b, resulting in

   

  







  or    . Since    and    , the student

   

 

likely obtained the equation      . The student needs to focus

 

on understanding how to determine the slope and y-intercept of a line

that is parallel to a given line.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

45

Option A is correct

To determine which equation is equivalent to S = Ph + 2B when solved

for B, the student could have first subtracted Ph from both sides of the

equation, resulting in S – Ph = 2B. Next, the student could have divided



both sides of the equation by 2, resulting in   . 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 divided both sides of the equation by Ph and then



multiplied both sides of the equation by 2, resulting in   . The



student needs to focus on understanding the arithmetic of solving

literal equations.

Option C is incorrect

The student likely multiplied both sides of the equation by Ph instead



of subtracting, resulting in   . The student needs to focus on



understanding the arithmetic of solving literal equations.

Option D is incorrect

The student likely subtracted 2 from both sides of the equation and



then divided both sides of the equation by Ph, resulting in    The



student needs to focus on understanding the arithmetic of solving

literal equations.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

46

Option B is correct

To determine which graph could represent the part of a linear function

(a function representing a straight line) given x > 5 and y < 2, the

student could have identified the graph containing an open circle at the

point with coordinates (5, 2) and a partial line that continues down and

to the right infinitely (as represented by the arrow) since y < 2 and

x > 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 interpreted x > 5 as meaning x is less than 5 and y < 2

as meaning y is greater than 2, resulting in a graph that contains an

open circle at the point with coordinates (5, 2) and a partial line that

continues up and to the left infinitely. The student needs to focus on

understanding how to identify the domain and range of a linear

function from a graph.

Option C is incorrect

The student likely interpreted x > 5 as meaning x is less than 5, resulting

in a graph that contains an open circle at the point with coordinates

(5, 2) and a partial line that continues down and to the left infinitely.

The student needs to focus on understanding how to identify the

domain and range of a linear function from a graph.

Option D is incorrect

The student likely interpreted y < 2 as meaning y is greater than 2,

resulting in a graph that contains an open circle at the point with

coordinates (5, 2) and a partial line that continues up and to the right

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

the domain and range of a linear function from a graph.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

47

Option B is correct

To determine the value of g(20), the student should have substituted

20 for x in the function (relationship where each input value has a

single output value) and then simplified the function, resulting in

g(20) = 6(2(20) + 7) = 6(40 + 7) = 6(47) = 282.

Option A is incorrect

The student likely added 6 to 2x before evaluating the function,

obtaining 8x + 7. The student then likely substituted x = 20 into the

function, resulting in g(20) = 8(20) + 7 = 160 + 7 = 167. The student

needs to focus on understanding how to apply the order of operations

when simplifying a numeric expression.

Option C is incorrect

The student likely added 20 and 7 before multiplying 20 by 2, resulting

in 6(2(20 + 7)) = 6(2(27)) = 6(54) = 324. 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 distributed (multiplied) the factor 6 to 2x but not to

7, obtaining 12x + 7. The student then likely substituted x = 20 into the

function, resulting in g(20) = 12(20) + 7 = 240 + 7 = 247. The student

needs to focus on understanding how to apply the order of operations

when simplifying a numeric expression.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

48

Option C is correct

To determine the x-intercept (value where a graph crosses the x-axis)

and y-intercept (value where a graph crosses the y-axis) of the line, the

student could have determined that the graph of the linear function

intersects (crosses) the x-axis when x = –2, so the x-intercept is (–2, 0).

Next, the student could have determined that the graph of the linear

function intersects the y-axis when y = 8, so the y-intercept is (0, 8).

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 switched the intercepts. The student needs to focus

on understanding how to identify the intercepts of a linear function

when given a graph.

Option B is incorrect

The student likely chose the correct value for the x-intercept but used

the slope (steepness of a straight line when graphed on a coordinate









grid, represented by   ) of the line as the y-intercept. The









student needs to focus on understanding how to identify the intercepts

of a linear function when given a graph.

Option D is incorrect

The student likely chose the correct value for the y-intercept but used

the slope of the line as the x-intercept. The student needs to focus on

understanding how to identify the intercepts of a linear function when

given a graph.

2023 STAAR Algebra I Math Rationales

Item #

Rationale

49

Option C is correct

To determine the values that best represent the zeros (input values, x,

that produce an output value, y, of zero) of the function, the student

could have identified the x-values where the graph crosses the x-axis

(horizontal axis), which are –1 and 3. Therefore, the zeros of the

function are x = –1 and x = 3. This is an efficient way to solve the

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

correctly.

Option D is correct

To determine the values that best represent the zeros of the function,

the student could have identified the x-values where the graph crosses

the x-axis, which are –1 and 3. Therefore, the zeros of the function are

x = –1 and x = 3. 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 x-coordinate of the vertex (high or low

point of the curve) as a zero of the function, resulting in x = 1. The

student needs to focus on understanding how to identify the key

features of a quadratic function when given a graph of the function.

Option B is incorrect

The student likely identified the y-coordinate of the vertex as a zero of

the function, resulting in x = –4. The student needs to focus on

understanding how to identify the key features of a quadratic function

when given a graph of the function.

Option E is incorrect

The student likely identified the x-coordinate of the y-intercept (value

where a graph crosses the y-axis) as a zero of the function, resulting in

x = 0. The student needs to focus on understanding how to identify the

key features of a quadratic function when given a graph of the function.

Option F is incorrect

The student likely identified the y-coordinate of the y-intercept as a

zero of the function, resulting in x = –3. The student needs to focus on

understanding how to identify the key features of a quadratic function

when given a graph of the function.



2023 STAAR Algebra I Math Rationales

Item #

Rationale

50

Option C is correct

To determine the rate of change (constant rate of increase or decrease)

of height off the ground with respect to the number of steps, the

student could have chosen two points from the table and calculated

the rate of change. The student could have used the first two sets of









values in the table and applied the slope formula,   , resulting









 

in     . Therefore, the rate of change is 7 inches per

 

step. 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 change in y but did not divide by the

change in x, resulting in 14 inches per step. 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 B is incorrect

The student likely calculated the change in x over the change in y,



resulting in inch per step. 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 D is incorrect

The student likely calculated the change in x over the change in y and



used 1 as the change in x, resulting in inch per step. 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.