Algebra 1 Quadratic Worksheet Answer Key

Contents

This algebra 1 quadratic worksheet answer key solves all 24 problems of the graphing quadratic functions worksheet step by step.

Every solution below shows the formula, the substitution and the final answer in bold. Compare each step with your own work, and re-check any vertex by substituting its x-coordinate back into the function.

Algebra 1Common Core: F-IF.C.7a · F-IF.C.8a · F-IF.B.4 · A-SSE.B.3b

Algebra 1 Quadratic Worksheet Answer Key: problems 1–5

Problem 1: Name the coefficients

Compare with $ax^{2}+bx+c$: $a=3$, $b=-5$, $c=2$.
The missing $x$-term means $b=0$: $a=-1$, $b=0$, $c=4$.

Problem 2: Opens up or down

$a=-2 < 0$, so it opens down and the vertex is a maximum. The y-intercept is $f(0)=-4$, the point $(0, -4)$.
$a=\frac{1}{2} > 0$, so it opens up and the vertex is a minimum. The y-intercept is $(0, -3)$.

Problem 3: Axis of symmetry

$x=-\frac{8}{2(1)}=-4$. $x=-4$
$x=-\frac{-12}{2(2)}=3$. $x=3$
$x=-\frac{5}{2(-1)}=\frac{5}{2}$. $x=\frac{5}{2}$

Problem 4: Vertex from standard form

$x=-\frac{-6}{2}=3$ and $y=9-18+5=-4$. Vertex: $(3, -4)$.
$x=-\frac{8}{2(-2)}=2$ and $y=-2(4)+16-3=5$. Vertex: $(2, 5)$.

Problem 5: Read vertex form

Compare with $a(x-h)^{2}+k$: $h=3$, $k=1$. Vertex $(3, 1)$, axis $x=3$. Since $a=2 > 0$, the minimum value is 1.
Write $x+2$ as $x-(-2)$: $h=-2$, $k=5$. Vertex $(-2, 5)$, axis $x=-2$. Since $a=-1 < 0$, the maximum value is 5.

Algebra 1 Quadratic Worksheet Answer Key: problems 6–10

Problem 6: Zeros from factored form

Set each factor to 0: $x+2=0$ gives $x=-2$, and $x-6=0$ gives $x=6$. The x-intercepts are $(-2, 0)$ and $(6, 0)$.

The axis is halfway between them: $x=\frac{-2+6}{2}=2$.

At $x=2$: $y=(4)(-4)=-16$. Vertex: $(2, -16)$.

Problem 7: Complete a table

Substitute each value. For example, $x=-1$: $1+4+1=6$, and $x=2$: $4-8+1=-3$.

x	-1	0	1	2	3	4	5
y	6	1	-2	-3	-2	1	6

The lowest value is $-3$ at $x=2$, and the values repeat on both sides of it. Vertex: $(2, -3)$. Axis: $x=2$.

Problem 8: Read a graph

Reading the graph:

Vertex: $(1, -4)$, the lowest point.
Axis of symmetry: $x=1$.
x-intercepts: $(-1, 0)$ and $(3, 0)$.
y-intercept: $(0, -3)$.
The graph never goes below $-4$, so the range is $y\ge -4$.

Features of $y=x^{2}-2x-3$.

Problem 9: Effect of a

Test the point with $x=2$: $y=x^{2}$ gives 4, $y=2x^{2}$ gives 8, and $y=\frac{1}{2}x^{2}$ gives 2.

Curve C passes through $(2, 4)$, curve B through $(1, 2)$ (so $y=2x^{2}$), and curve A through $(2, 2)$. So A is $y=\frac{1}{2}x^{2}$, B is $y=2x^{2}$, C is $y=x^{2}$.

The graph of $y=-2x^{2}$ has the same width as B but opens down, because $a$ is negative.

Problem 10: Intercepts from standard form

At $x=0$, $y=6$: the y-intercept is $(0, 6)$.

Set $y=0$ and factor: $x^{2}-5x+6=(x-2)(x-3)=0$, so $x=2$ or $x=3$. The x-intercepts are $(2, 0)$ and $(3, 0)$.

Algebra 1 Quadratic Worksheet Answer Key: problems 11–15

Problem 11: Maximum or minimum value

Since $a=-1 < 0$, the parabola opens down, so there is a maximum. The axis is $x=-\frac{6}{2(-1)}=3$, and $f(3)=-9+18-5=4$. The maximum value is 4, at $x=3$.

Zeros: $-(x^{2}-6x+5)=-(x-1)(x-5)=0$, so $x=1$ and $x=5$.

Problem 12: Factored to standard form

Multiply: $x^{2}+5x-3x-15=$ $x^{2}+2x-15$.

The axis is $x=-\frac{2}{2}=-1$, which is halfway between the zeros 3 and $-5$. At $x=-1$: $y=(-4)(4)=-16$. Vertex: $(-1, -16)$.

Problem 13: Complete the square

Half of 8 is 4, and $4^{2}=16$. Add and subtract 16:

$y=(x^{2}+8x+16)-16+7=(x+4)^{2}-9$.

Vertex form: $y=(x+4)^{2}-9$. Vertex: $(-4, -9)$.

Problem 14: No x-intercepts

Half of $-10$ is $-5$, and $(-5)^{2}=25$. So $y=(x^{2}-10x+25)-25+30=(x-5)^{2}+5$.

The vertex is $(5, 5)$, so the minimum value is 5. The parabola opens up and its lowest point is above the x-axis, so it never reaches $y=0$.

Problem 15: Leading coefficient 2

Factor 2 from the first two terms: $y=2(x^{2}-6x)+10$. Add and subtract 9 inside: $y=2(x^{2}-6x+9)-18+10=2(x-3)^{2}-8$.

Vertex form: $y=2(x-3)^{2}-8$, vertex $(3, -8)$.

Zeros: $2(x-3)^{2}=8$, so $(x-3)^{2}=4$ and $x-3=\pm 2$. The zeros are $x=1$ and $x=5$.

Algebra 1 Quadratic Worksheet Answer Key: problems 16–20

Problem 16: Increasing and decreasing

The vertex is $(2, 4)$, the peak of the graph.

Increasing: $x < 2$.
Decreasing: $x > 2$.
$f(x) > 0$: the graph is above the x-axis between the zeros 0 and 4, so $0 < x < 4$.

The graph of $f(x)=-x^{2}+4x$, with the vertex and intercepts marked.

Problem 17: Equation from a graph

Start with $y=a(x-1)^{2}+4$. Substitute $(3, 0)$: $0=a(2)^{2}+4$, so $4a=-4$ and $a=-1$.

Vertex form: $y=-(x-1)^{2}+4$. Expand: $-(x^{2}-2x+1)+4=$ $-x^{2}+2x+3$.

A parabola with vertex $(1, 4)$ that passes through $(3, 0)$.

Problem 18: Compare two functions

For $f$: the axis is $x=3$ and $f(3)=9-18+11=2$. The minimum value of $f$ is 2.

For $g$: the values 5, 2, 1, 2, 5 are symmetric around $x=2$, so the lowest value is $g(2)=1$.

Since $1 < 2$, function $g$ has the lower minimum.

Problem 19: Three forms, one parabola

Expand the vertex form: $(x-3)^{2}-4=x^{2}-6x+9-4=x^{2}-6x+5$.

Expand the factored form: $(x-1)(x-5)=x^{2}-5x-x+5=x^{2}-6x+5$.

All three match. Vertex form shows the vertex $(3, -4)$; factored form shows the zeros 1 and 5; standard form shows the y-intercept 5.

Problem 20: Equation from zeros and a point

Factored form: $y=a(x-2)(x-8)$. Substitute $(5, -9)$: $-9=a(3)(-3)=-9a$, so $a=1$.

Factored: $y=(x-2)(x-8)$. Standard: $y=x^{2}-10x+16$.

The axis is halfway between 2 and 8, $x=5$, and $y=-9$ there. Vertex: $(5, -9)$.

Algebra 1 Quadratic Worksheet Answer Key: problems 21–24

Problem 21: Thrown ball

$h(0)=6$. 6 feet.
$t=-\frac{32}{2(-16)}=1$, and $h(1)=-16+32+6=22$. At 1 second, at 22 feet.
The graph is symmetric about $t=1$, and 0 is one second before it, so the other time is $t=2$. Check: $h(2)=-64+64+6=6$. At 2 seconds.

Height of a ball (feet) against time (seconds).

Problem 22: Garden fence

The third side is $40-2x$, so $A(x)=x(40-2x)=-2x^{2}+40x$.
The axis is $x=-\frac{40}{2(-2)}=10$. $x=10$ feet.
$A(10)=10(40-20)=200$. The garden is 10 ft by 20 ft, with area 200 square feet.

Problem 23: T-shirt sale

$1200-60x+100x-5x^{2}=$ $-5x^{2}+40x+1200$.
$x=-\frac{40}{2(-5)}=4$, so the price is $12+4=$ $\$16$.
Shirts sold: $100-5(4)=80$. Revenue: $16\times 80=1280$. 80 shirts, $\$1{,}280$. Check with the formula: $-80+160+1200=1280$.

Problem 24: Fountain arc

$x=-\frac{2}{2(-1/4)}=4$ and $y=-4+8=4$. 4 feet, at $x=4$.
Solve $y=0$: $-\frac{1}{4}x(x-8)=0$, so $x=0$ or $x=8$. The water lands 8 feet from the nozzle.
$y(2)=-1+4=3$, so 3 feet. By symmetry about $x=4$, the mirror point is $x=6$; check $y(6)=-9+12=3$. $x=6$.

Path of a water stream: $x$ is horizontal distance and $y$ is height, both in feet.

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Graphing Quadratic Functions: Algebra 1 Math Worksheet Answer Key – PDF Download