# SAT Quadratic Functions Worksheet PDF with Answers | 1600.now

Source: https://1600.now/sat-nonlinear-functions-worksheet

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## SAT Quadratic Functions Worksheet

For a quadratic in vertex form, $f(x)=a(x-h)^2+k$, the vertex is $(h,k)$. When $a>0$, the minimum output is $k$; when $a<0$, the maximum output is $k$. That output differs from the input $h$ where it occurs.

Written by [Luke Finigan](https://1600.now/about)

These 12 quadratic-function questions cover vertices, zeros, intercepts, and functions built from points. Print the student worksheet or download the PDF with explanations.

12 questions · about 25 minutes · Advanced Math › Nonlinear functions

[Download 4-page PDF](https://1600.now/downloads/worksheets/sat-nonlinear-functions-worksheet-student-letter.pdf)

[Share to Google Classroom](https://classroom.google.com/share?url=https%3A%2F%2F1600.now%2Fsat-nonlinear-functions-worksheet) · This worksheet is free to print and share under [CC BY-NC-ND 4.0](https://creativecommons.org/licenses/by-nc-nd/4.0/). Tutors may also use it in paid sessions. If you post it online, please link to this page instead of re-uploading the PDF. These are original questions, not College Board items.

### Use the form that exposes the feature you need

Vertex form exposes the turning point. Factored form $a(x-r)(x-s)$ exposes the zeros $r$ and $s$. Standard form $ax^2+bx+c$ exposes the $y$-intercept $c$. Choose the form that answers the question with the least rearranging.

Zeros or a vertex may leave the leading coefficient unknown. Use an additional point to find $a$. For a vertex $(2,3)$ and point $(4,11)$, substitute into $f(x)=a(x-2)^2+3$ to get $11=4a+3$, so $a=2$.

Check symmetry around $x=h$, or around $x=-\dfrac b{2a}$ in standard form. Inputs equally far from that axis have equal outputs. Keep input, output, and the ordered pair distinct when selecting the final answer.

Try it yourself: For $f(x)=-2(x-4)^2+9$, what is the maximum value? · A $4$ B $9$ C $-2$

### Questions

1. **Question 1**

   For $f(x) = (x - 3)^2 + 5$, what is the minimum value of $f(x)$?

   A. 5 B. 8 C. 3 D. 9

   **A.** A square is nonnegative. Its smallest value is 0, attained at $x = 3$. Therefore the minimum output is 5, not the x-coordinate 3.
2. **Question 2**

   What are the zeros of $f(x) = (x + 2)(x - 7)$?

   A. −2 and −7 B. 2 and −7 C. −2 and 7 D. 2 and 7

   **C.** Set each factor equal to zero: $x + 2 = 0$ gives -2, and $x - 7 = 0$ gives 7.
3. **Question 3**

   What is the $y$-intercept of the graph $y=2(x-4)^{2}-3$?

   A. (0, −3) B. (4, −3) C. (0, 13) D. (0, 29)

   **D.** At the $y$-intercept $x=0$. Then $y=2(16)-3=29$, so the point is (0, 29). The vertex is (4, −3).
4. **Question 4**

   The height $h$, in meters, of a modeled object $t$ seconds after launch is $h(t)=-5t^{2}+20t+1$. What is its maximum height?

   A. 16 meters B. 21 meters C. 2 meters D. 20 meters

   **B.** The vertex occurs at $t=-\dfrac{20}{2(-5)}=2$. Evaluate $h(2)=-20+40+1=21$ meters. The time 2 is not the requested height.
5. **Question 5**

   A quadratic has zeros 1 and 5 and passes through (0, 10). Which equation represents it?

   A. $y=2(x-1)(x-5)$ B. $y=2(x+1)(x+5)$ C. $y=10(x-1)(x-5)$ D. $y=(x-1)(x-5)$

   **A.** The zeros imply $y=a(x-1)(x-5)$. At $x=0$, $10=5a$, so $a=2$.
6. **Question 6**

   If $f(x)=x^{2}-8x+11$, what is the minimum value of $f$?

   A. −4 B. 11 C. −5 D. −8

   **C.** Complete the square: $x^{2}-8x+11=(x-4)^{2}-5$. The minimum output is −5, attained at $x=4$.
7. **Question 7**

   The function $f(x)=-2(x+1)^{2}+8$ has its maximum at which value of $x$?

   A. −2 B. −1 C. 1 D. 8

   **B.** The negative coefficient makes the parabola open downward. Its vertex is (−1, 8), so the maximizing input is $x=-1$.
8. **Question 8**

   For $f(x)=x^{2}-6x+8$, what is the smaller zero?

   A. −4 B. −2 C. 2 D. 4

   **C.** Factor $f(x)$ as $(x-2)(x-4)$. The zeros are 2 and 4, and the smaller is 2.
9. **Question 9**

   A quadratic function has vertex (2, 3) and passes through (4, 11). What is $f(0)$?

   A. 3 B. 7 C. 11 D. 19

   **C.** Use $f(x)=a(x-2)^{2}+3$. Since $11=4a+3$, $a=2$. Then $f(0)=2(4)+3=11$.
10. **Question 10**

    If $g(x)=(x-5)^{2}-9$, what is the positive difference between its two zeros?

    A. 3 B. 5 C. 6 D. 9

    **C.** Set $(x-5)^{2}=9$. The roots are $x=2$ and $x=8$, whose positive difference is 6.
11. **Question 11**

    A rectangle has perimeter 40 meters. If its width is $x$ meters, its area is $\mathrm{A}(x)=x(20-x)$. What is its greatest possible area?

    A. 40 square meters B. 80 square meters C. 100 square meters D. 400 square meters

    **C.** $\mathrm{A}(x)=-x^{2}+20x=-(x-10)^{2}+100$. The maximum is 100 square meters when both sides are 10 meters.
12. **Question 12**

    If $f(x)=3x^{2}+bx+7$ and the axis of symmetry is $x=2$, what is $b$?

    A. −12 B. −6 C. 6 D. 12

    **A.** For $ax^{2}+bx+c$, the axis is $x=-\dfrac{b}{2a}$. Set $2=-\dfrac{b}{6}$ to obtain $b=-12$.

0/12 correct

### Check your quadratics answers

Each row is a wrong answer choice from one of the sheet's problems. Find the one you picked, then run the check in the last column.

| What your answer looked like | What actually happened | Fix it next time |
| --- | --- | --- |
| 3 for the minimum of $f(x) = (x - 3)^2 + 5$ (problem 1) | 3 is where the minimum happens, the $x$-coordinate of the vertex. The minimum value is the output there. | $(x - 3)^2$ is smallest at 0, so the minimum is $0 + 5 = 5$. Problem 4 is the same trap: the height peaks at $t = 2$, and the height is $h(2) = 21$ meters. |
| 2 and $-7$ for the zeros of $f(x) = (x + 2)(x - 7)$ (problem 2) | Both signs are reversed. $x + 2 = 0$ gives $x = -2$. | Set each factor equal to 0 and solve: $-2$ and 7. Check: $f(7) = 9 \cdot 0 = 0$, but $f(2) = 4(-5) = -20$. |
| $(0, -3)$ or $(0, 13)$ for the y-intercept of $y = 2(x - 4)^2 - 3$ (problem 3) | $-3$ is the $y$-value of the vertex, not the y-intercept. 13 comes from $16 - 3$, which drops the 2. | Set $x = 0$ and keep every factor: $2(0 - 4)^2 - 3 = 2(16) - 3 = 29$. |
| $y = (x - 1)(x - 5)$ in problem 5 | Zeros at 1 and 5 fix the factors but not the leading coefficient. This choice assumes $a = 1$. | Use the third point. At $x = 0$, $a(-1)(-5) = 5a = 10$, so $a = 2$. Check the choice you picked: $(0 - 1)(0 - 5) = 5$, not 10. |
| 11 for the minimum of $f(x) = x^2 - 8x + 11$ (problem 6) | 11 is $f(0)$, the y-intercept. The minimum is at the vertex, $x = 4$. | Complete the square: $f(x) = (x - 4)^2 - 5$, so the minimum is $-5$. Check: $f(4) = 16 - 32 + 11 = -5$. |
| $-6$ in problem 12 | You used $x = -\dfrac{b}{a}$. The axis of symmetry is $x = -\dfrac{b}{2a}$. | $2 = -\dfrac{b}{2(3)}$ gives $b = -12$. Check: $3x^2 - 12x + 7 = 3(x - 2)^2 - 5$. |

#### Worked example: building the function from its vertex

Problem 9 describes a quadratic with vertex $(2, 3)$ that passes through $(4, 11)$, and asks for $f(0)$. The vertex fills in everything except $a$: $f(x) = a(x - 2)^2 + 3$. Substitute the other point: $11 = a(4 - 2)^2 + 3$, so $4a = 8$ and $a = 2$.

Now $f(0) = 2(0 - 2)^2 + 3 = 8 + 3 = 11$. Symmetry gives a quick check: 0 and 4 are both 2 units from the axis $x = 2$, so $f(0)$ must equal $f(4)$, which is 11. The choice 7 skips finding $a$ and uses $a = 1$, and the choice 3 is the $y$-value of the vertex.

### Apply the check to fresh questions

On a fresh quadratic question, label the requested feature before calculating. Check your result using the intercept, a known point, or symmetry, then continue with nonlinear-functions practice.

[Practice nonlinear functions questions](https://1600.now/bank/math/skill/Nonlinear%20functions)

[Review the nonlinear functions method](https://1600.now/sat-skill/nonlinear-functions)

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