Quadratic Functions Worksheet Answer Key – Algebra 1

Problem 1 Read the Vertex Form, Opening Up

a = 2 > 0, so it opens up. In a(x − h)² + k the vertex is (3, 5), the axis is x = 3 and the minimum value is 5.

Opens up; vertex (3, 5); axis x = 3; minimum 5

Problem 2 Read the Vertex Form, Opening Down

a = −1 < 0, so it opens down. Write x + 4 as x − (−4): the vertex is (−4, −2). The maximum value is −2.

Opens down; vertex (−4, −2); maximum −2

Problem 3 Axis from Standard Form

Mirror line: x = 18 ÷ 2 = 9. Then y = 81 − 162 + 50 = −31.

Mirror line x = 9; turning point (9, −31)

Problem 4 Complete a Table of Values

x = 0: 5. x = 1: 1 − 6 + 5 = 0. x = 2: 4 − 12 + 5 = −3. x = 3: 9 − 18 + 5 = −4. x = 4: −3. x = 5: 0. x = 6: 5.

The lowest value is at x = 3, and the values mirror around it.

Vertex (3, −4); y-intercept 5

Problem 5 Read the Factored Form

Zeros: 2 and −6. The axis is halfway: (2 + (−6)) ÷ 2 = −2. At x = −2: (−4)(4) = −16.

Zeros 2 and −6; axis x = −2; vertex (−2, −16)

Problem 6 Read a Parabola Graph

The highest point is (2, 5), so the vertex is (2, 5) and the axis is x = 2. The graph crosses the y-axis at (0, 1).

Vertex (2, 5); axis x = 2; maximum 5; y-intercept 1

Problem 7 Vertex Form to Standard Form

(x − 1)² = x² − 2x + 1. Multiply by 3: 3x² − 6x + 3. Subtract 4.

y = 3x² − 6x − 1

Problem 8 Vertex of a Downward Parabola

x = −12 ÷ (2 · (−2)) = 3. y = −2(9) + 36 − 7 = 11.

Vertex (3, 11); maximum value 11

Problem 9 Complete the Square, Easy Case

Half of 10 is 5: y = (x² + 10x + 25) − 25 + 16 = (x + 5)² − 9. Vertex (−5, −9).

Zeros: x² + 10x + 16 = (x + 2)(x + 8), so x = −2 or x = −8. They are symmetric around x = −5.

y = (x + 5)² − 9; vertex (−5, −9); zeros −2 and −8

Problem 10 Completing the Square, No Zeros

Half of −8 is −4: y = (x² − 8x + 16) − 16 + 19 = (x − 4)² + 3.

The vertex (4, 3) is the minimum and lies above the x-axis, and the parabola opens up, so y ≥ 3 for all x.

y = (x − 4)² + 3; no x-intercepts

Problem 11 Completing the Square, a = 2

Factor 2 from the first two terms: 2(x² − 6x) + 13. Add 9 inside: 2(x² − 6x + 9) − 2 · 9 + 13.

= 2(x − 3)² − 18 + 13 = 2(x − 3)² − 5.

y = 2(x − 3)² − 5; vertex (3, −5)

Problem 12 Graph from Zeros and Vertex

Factor: (x − 3)(x + 1), so the zeros are 3 and −1. Axis x = 1, and y = 1 − 2 − 3 = −4. The y-intercept is −3.

Zeros −1 and 3; vertex (1, −4); y-intercept −3

Problem 13 Vertex from Factored Form

Zeros: −1 and 5. Axis: (−1 + 5) ÷ 2 = 2. At x = 2: −2(3)(−3) = 18.

Since a = −2 < 0, the parabola opens down.

Vertex (2, 18); maximum value 18

Problem 14 Compare Three Parabolas

A larger |a| makes the parabola narrower. |3| > |1| > |−0.5|.

The sign of a sets the direction: only −0.5x² opens down.

Narrowest: y = 3x²; opens down and is widest: y = −0.5x²

Problem 15 Ball Launched from a Roof

Starting height: h(0) = 64. Maximum at t = −48 ÷ (2 · (−16)) = 1.5, so h(1.5) = −36 + 72 + 64 = 100.

Ground: −16t² + 48t + 64 = −16(t − 4)(t + 1) = 0, so t = 4 (reject −1).

64 ft; 100 ft at 1.5 s; lands at 4 s

Problem 16 Largest Rectangular Pen

Zeros: w = 0 and w = 48, so the vertex is at w = 24. A(24) = 24 · 24 = 576.

The best pen is a square, 24 ft by 24 ft.

Width 24 ft; maximum area 576 square feet

Problem 17 Best Ticket Price

Zeros: p = 0 and 250 − 5p = 0, so p = 50. The vertex is halfway at p = 25.

R(25) = 25(250 − 125) = 25 × 125 = 3125.

Price $25; maximum revenue $3,125; revenue is 0 at $0 and $50

Problem 18 Equation from Vertex and a Point

Start with y = a(x + 1)² + 6. Substitute (2, −3): −3 = a(3)² + 6, so 9a = −9 and a = −1.

Standard: −(x² + 2x + 1) + 6 = −x² − 2x + 5.

y = −(x + 1)² + 6 = −x² − 2x + 5

Problem 19 Equation from Zeros and a Point

Start with y = a(x + 2)(x − 4). At (0, −16): −16 = a(2)(−4) = −8a, so a = 2.

Check the standard form 2x² − 4x − 16: at x = 0 it equals −16 ✓.

y = 2(x + 2)(x − 4)

Problem 20 Compare Two Quadratic Functions

f(x) = (x − 3)² + 2, so the minimum is 2 at x = 3.

The table of g is symmetric around x = 1, where the lowest value is 1. So the minimum of g is 1 at x = 1.

g has the smaller minimum (1 at x = 1, against 2 at x = 3 for f)