STAAR Algebra I Practice Questions (2026) — Free Sample Questions with Answers

Answer: C

1. 1. Uncovered area = backyard area − pool area: $(3x^2+40x)-(x^2-12x+150)$.
2. 2. Distribute the negative: $3x^2+40x-x^2+12x-150$.
3. 3. Combine like terms: $2x^2+52x-150$.
4. 4. Check with $x=10$: $700-130=570$, and $200+520-150=570$.

$2x^2+28x+150$: Subtracts only $x^2$. The subtraction must apply to all three terms of the pool's area.

$2x^2+52x+150$: Changes the sign of $-12x$ but not of $+150$. Subtracting 150 gives −150.

$2x^2+52x-150$: Correct. $(3x^2+40x)-(x^2-12x+150)=3x^2+40x-x^2+12x-150=2x^2+52x-150$ (A.10A).

$4x^2+28x+150$: Adds the two areas instead of subtracting the pool from the backyard.

"Not covered" means subtract. Put the whole polynomial being subtracted in parentheses and change the sign of each of its terms.

Answer: A

1. 1. In $C(h)$, h is hours and $C(h)$ is the total cost in dollars.
2. 2. $C(6)=45+12.5(6)=45+75=120$.
3. 3. So 6 hours of rental costs $120 in all.

Renting a kayak for 6 hours costs a total of $120.: Correct. The input h = 6 is the number of hours, and the output 120 is the total cost in dollars: $45+12.5(6)=120$ (A.12B).

Renting for 120 hours costs a total of $6.: Swaps the input and the output. The number inside the parentheses is the number of hours.

Each hour after the first 6 hours costs $120.: Treats the output as a rate. The rate is $12.50 per hour; 120 is a total cost.

Renting for 6 hours costs $120 plus the $45 fee.: The $45 fee is already part of the function, so $120 is the whole cost, fee included.

In function notation $f(a)=b$, a is the input and b is the output. Name both in the units of the context.

Answer: A

1. 1. Multiply a and c: $6(-20)=-120$.
2. 2. Find two integers with a product of −120 and a sum of −7: −15 and 8.
3. 3. Split the middle term: $6x^2-15x+8x-20$.
4. 4. Group: $3x(2x-5)+4(2x-5)=(3x+4)(2x-5)$.
5. 5. Check: $(3x+4)(2x-5)=6x^2-15x+8x-20=6x^2-7x-20$.

$2x-5$: Correct. $(2x-5)(3x+4)=6x^2+8x-15x-20=6x^2-7x-20$ (A.10E).

$2x+5$: Comes from $(2x+5)(3x-4)=6x^2+7x-20$. The signs are reversed, so the middle term is $+7x$.

$3x-4$: Also comes from the sign-reversed pair $(2x+5)(3x-4)$, whose middle term is $+7x$.

$x-4$: Comes from $(6x+5)(x-4)=6x^2-19x-20$. The constant −20 is right but the middle term is not.

For $a{x}^2+bx+c$ with $a\ne 1$, find two numbers whose product is ac and whose sum is b, split the middle term, and factor by grouping.

Answer: C

1. Read the graph: the Mesa line starts at about $40 and the Ridgeview line starts at about $10.
2. The lines cross at about $x=12$, $y=76$, so the costs are equal at about 12 classes.
3. To the right of the intersection, the Mesa line is below the Ridgeview line. At 20 classes the graph shows Mesa at about $100 and Ridgeview at about $120.
4. Check with rates read from the graph (Mesa rises $60 over 20 classes, Ridgeview $110): $40+3(12)=76$ and $10+5.5(12)=76$.

The costs are equal at about 12 classes; for fewer classes, Mesa Fitness costs less.: Reverses the comparison. At 0 classes the graph shows Ridgeview at $10 and Mesa at $40, so Ridgeview is cheaper below about 12 classes.

The costs are equal at about 76 classes; the cost then is about $12.: Swaps the coordinates of the intersection point, about $(12,76)$.

The costs are equal at about 12 classes; for more classes, Mesa Fitness costs less.: Correct. The lines cross near $(12,76)$, and to the right of that point the Mesa line is lower (A.3G).

The costs are never equal, because the gyms have different membership fees.: Lines with different slopes always cross; different starting values on the graph do not prevent an intersection.

The intersection of two cost lines estimates where the costs are equal; on each side of it, the lower line is the cheaper option.

Answer: B

Building more libraries in a city causes the number of traffic accidents to rise.: Treats correlation as causation; the data show only that the two quantities tend to rise together.

The two variables are associated, and city population may explain both.: Correct. Larger cities tend to have more libraries and more drivers, so population is a likely lurking variable; the association does not show causation (A.4B).

Closing some of a city's libraries would lower its number of traffic accidents.: Assumes a cause-and-effect link that correlation alone cannot establish.

The variables are not associated, because libraries cannot cause traffic accidents.: Confuses association with causation; a strong correlation is an association even when neither variable causes the other.

Correlation shows association, not causation; look for a lurking variable (such as population) that could drive both quantities.

Answer: A

1. The boundary passes through $(0,-2)$ and $(3,0)$: slope $\frac{2}{3}$, so $y=\frac{2}{3}x-2$, which is $2x-3y=6$.
2. The line is dashed, so the symbol is $<$ or $>$.
3. The region below the line is shaded. Test $(0,-5)$: $2(0)-3(-5)=15$, and $15>6$.
4. So the inequality is $2x-3y>6$. Check with the unshaded point $(0,0)$: $0>6$ is false, as expected.

$2x-3y>6$: Correct. The dashed boundary through $(0,-2)$ and $(3,0)$ is $2x-3y=6$, and the shaded test point $(0,-5)$ gives $15>6$ (A.3D).

$2x-3y<6$: Describes the side above the line; $(0,0)$ gives $0<6$, but $(0,0)$ is not shaded. The sign flips when $-3y>-2x+6$ is solved for $y$.

$2x-3y\ge 6$: Uses $\ge$, which needs a solid boundary; the boundary is dashed, so points on the line are not included.

$3x-2y>6$: Swaps the coefficients; $3x-2y=6$ has intercepts $(2,0)$ and $(0,-3)$, not $(3,0)$ and $(0,-2)$.

In standard form, decide the symbol with a test point instead of "above means >": a negative y-coefficient reverses the relationship.

Answer: A

1. Subtract 85: $-6.75n\ge -65$.
2. Divide by $-6.75$ and reverse: $n\le 9.63\dots$
3. The greatest whole number is 9.
4. Check: 9 sessions leave $85-60.75=24.25\ge 20$; 10 sessions leave $17.50<20$.

9: Correct. $6.75n\le 65$ gives $n\le 9.63\dots$; sessions are whole numbers, so the greatest is 9, leaving $85-60.75=\$24.25$ (A.5B).

10: Rounds 9.63 up; 10 sessions cost $67.50 and leave only $17.50.

12: Ignores the $20 he wants to keep: $\frac{85}{6.75}\approx 12.6$.

15: Adds the $20 to the balance instead of subtracting it: $\frac{105}{6.75}\approx 15.6$.

Solve the inequality, then apply the context: a count must be a whole number, so round down to stay within a maximum.

Answer: C

1. Let $a$ = adult tickets and $s$ = student tickets: $a+s=380$ and $12a+7s=3,860$.
2. Multiply the first equation by 7: $7a+7s=2,660$.
3. Subtract from the second equation: $5a=1,200$, so $a=240$ and $s=140$.
4. Check: $12(240)+7(140)=2,880+980=3,860$.

140: This is the number of student tickets; it answers the wrong question.

190: Assumes the tickets were split evenly; $12(190)+7(190)=3,610$, not $3,860.

240: Correct. With $a+s=380$ and $12a+7s=3,860$, substituting $s=380-a$ gives $5a=1,200$, so $a=240$ (A.5C).

280: Finds $12(380)-3,860=700$ but divides by the student price, 7, instead of the price difference, 5. That gives 100 student tickets and $380-100=280$ adult tickets.

Write one equation for the count and one for the money, then solve by substitution or elimination and check both equations.

Answer: A

1. 1. Line 1 passes through (0, 1) and (1, 2): slope 1, intercept 1, so $y=x+1$.
2. 2. Line 2 passes through (0, 4) and (1, 2): slope −2, intercept 4, so $y=-2x+4$.
3. 3. Check the intersection (1, 2): $1+1=2$ and $-2(1)+4=2$.

$y=x+1$ and $y=-2x+4$: Correct. One line crosses the y-axis at 1 and rises 1 for each 1 to the right. The other crosses at 4 and falls 2 for each 1 to the right. They meet at (1, 2) (A.2I).

$y=x+1$ and $y=-\frac{1}{2}x+4$: Uses run over rise for the second line. From (0, 4) to (1, 2), it falls 2 units for 1 unit of run, so its slope is −2.

$y=x-1$ and $y=-2x+4$: Uses the x-intercept of the first line, −1, as its y-intercept. That line crosses the y-axis at 1.

$y=-x+1$ and $y=2x+4$: Reverses both slope signs. The first line rises and the second falls from left to right.

Write each line from its y-intercept and slope, then check that the intersection point satisfies both equations.

Answer: B

1. 1. The initial height (at $t=0$) is 30 cm.
2. 2. Rate of change: $\frac{21-30}{4-0}=\frac{-9}{4}=-2.25$ cm per hour.
3. 3. The equation is $h=30-2.25t$.
4. 4. Check: $30-2.25(4)=30-9=21$.

$h=21-2.25t$: Uses the height after 4 hours as the starting height. At $t=0$ the candle is 30 cm tall.

$h=30-2.25t$: Correct. The candle loses $30-21=9$ cm in 4 hours, which is $9÷4=2.25$ cm per hour, starting from 30 cm (A.2C).

$h=30-9t$: Uses the total change of 9 cm as the hourly rate. That loss took 4 hours.

$h=30-0.44t$: Divides the hours by the change in height ($4÷9\approx 0.44$). The rate is centimeters per hour, $9÷4$.

In a verbal description, the starting amount is the y-intercept and the change per unit of time is the slope. Divide the change in output by the change in input.

Answer: D

$p=n+2$ and $3p+4n=15$: Reverses the relationship. This says a pen costs $2 more than a notebook.

$n=p+2$ and $4p+3n=15$: Swaps the quantities. Fatima bought 3 pens and 4 notebooks, so p is multiplied by 3.

$n=2p$ and $3p+4n=15$: Reads "$2 more than" as "2 times as much."

$n=p+2$ and $3p+4n=15$: Correct. "A notebook costs $2 more than a pen" gives $n=p+2$, and 3 pens plus 4 notebooks cost $15 (A.2I). The solution $p=1$, $n=3$ checks: $3+12=15$.

Translate each sentence into its own equation: one for the relationship between the unknowns and one for the total. Match each count with the right price.

Answer: D

1. 1. Vertex: $t=\frac{-b}{2a}=\frac{-64}{2(-16)}=2$, and $h(2)=-64+128+80=144$.
2. 2. Landing: $-16t^2+64t+80=-16(t^2-4t-5)=-16(t-5)(t+1)=0$, so $t=5$.
3. 3. For $0\le t\le 5$, the height goes from 80 up to 144 and then down to 0.
4. 4. The reasonable range is $0\le h\le 144$.

$80\le h\le 144$: Starts the range at the launch height. The ball falls to the ground, so its height goes down to 0.

$0\le h\le 80$: Uses the launch height as the greatest height. The ball rises to 144 feet at $t=2$.

$h\le 144$: This is the range for every real input. Heights below 0 feet are not reasonable here.

$0\le h\le 144$: Correct. The vertex is at $t=\frac{-64}{-32}=2$, with $h(2)=144$. The ball lands when $h=0$ (at $t=5$), so the heights run from 0 to 144 feet (A.6A).

A reasonable range in context runs from the least to the greatest output the situation allows. For a thrown object, that is ground level up to the height at the vertex.

Answer: B

1. 1. Let w be the width. Then the length is $w+4$, and $w(w+4)=96$.
2. 2. $w^2+4w-96=0$.
3. 3. Factor: $(w+12)(w-8)=0$, so $w=-12$ or $w=8$.
4. 4. Reject −12. Check: $8\times 12=96$.

−12: This solves $w^2+4w-96=0$, but a width cannot be negative.

8: Correct. $w(w+4)=96$ gives $w^2+4w-96=(w+12)(w-8)=0$. The width must be positive, so $w=8$ feet (A.8A).

12: This is the length, $8+4=12$ feet. It can also come from reading the factor $w+12$ as $w=12$.

24: Divides the area by 4. The 4 is how much longer the length is, not a side length.

Write the area equation, solve it, and keep only the solution that makes sense in the context.

Answer: C

1. 1. Identify $a=2$ and $b=-12$.
2. 2. Use the axis of symmetry formula: $x=\frac{-b}{2a}=\frac{12}{4}=3$.
3. 3. Check symmetry: $f(2)=8-24+7=-9$ and $f(4)=32-48+7=-9$, equal values on either side of $x=3$.

$x=-11$: This is the y-coordinate of the vertex, $f(3)=-11$. The axis of symmetry is a vertical line through the vertex's x-coordinate.

$x=-3$: Drops the negative in $-b$. With $b=-12$, $-b=12$, so $x=\frac{12}{4}=3$.

$x=3$: Correct. $x=\frac{-b}{2a}=\frac{-(-12)}{2(2)}=\frac{12}{4}=3$ (A.7A).

$x=6$: Divides $-b$ by a instead of by $2a$: $\frac{12}{2}=6$.

The axis of symmetry of $f(x)=a{x}^2+bx+c$ is the vertical line $x=\frac{-b}{2a}$. It is written as an equation in x.

Answer: C

1. Ratio: $\frac{18}{12}=\frac{27}{18}=\frac{40.5}{27}=1.5$, so $b=1.5$.
2. Work back to $x=0$: $f(0)=\frac{12}{1.5}=8$.
3. $f(x)=8(1.5{)}^x$.
4. Check: $f(3)=8(1.5{)}^3=8(3.375)=27$.

$f(x)=12(1.5{)}^x$: Takes $a$ from $x=1$; then $f(1)=18$, not 12. The initial value $a$ is $f(0)$.

$f(x)=8(0.5{)}^x$: Uses the percent increase, 0.5, as the base; growth of 50% means $b=1+0.5=1.5$.

$f(x)=8(1.5{)}^x$: Correct. The ratio is $\frac{18}{12}=1.5$, and $a=f(0)=\frac{12}{1.5}=8$, so $f(x)=8(1.5{)}^x$ (A.9C).

$f(x)=8{\left(\frac{2}{3}\right)}^x$: Inverts the ratio ($\frac{12}{18}$); the outputs increase, so $b>1$.

Find b as the ratio of consecutive outputs (x-steps of 1); find a = f(0) by working backward one step if the table does not start at x = 0.