Quadratic Functions and Parabolas Worksheet

Question 1

Find the vertex of the quadratic function y = x^2 - 6x + 5.

Accepted Answer

The vertex is (3, -4). Completing the square gives y = (x - 3)^2 - 4, so the vertex is at x = 3 and y = -4.

Question 2

Determine whether the parabola y = -2x^2 + 8x - 1 opens upward or downward. Then state the axis of symmetry.

Accepted Answer

The parabola opens downward because the coefficient of x^2 is negative. The axis of symmetry is x = 2 because x = -b / 2a = -8 / (2 times -2) = 2.

Question 3

Write the quadratic function in standard form that has x-intercepts at x = 1 and x = 5 and passes through the point (2, -3).

Accepted Answer

The function is y = x^2 - 6x + 5. Using the intercepts, the function has the form y = a(x - 1)(x - 5). Substituting (2, -3) gives -3 = a(1)(-3), so a = 1. Expanding gives y = x^2 - 6x + 5.

Question 4

Solve the equation x^2 + 7x + 12 = 0 by factoring.

Accepted Answer

The solutions are x = -3 and x = -4. Factoring gives (x + 3)(x + 4) = 0, so each factor can be set equal to zero.

Question 5

Find the y-intercept of the function y = 3x^2 - 2x + 7.

Accepted Answer

The y-intercept is 7, or the point (0, 7). Substituting x = 0 into the function gives y = 7.

Question 6

A ball is thrown upward, and its height is modeled by h(t) = -16t^2 + 48t + 5. Find the maximum height of the ball.

Accepted Answer

The maximum height is 41 feet. The time of the vertex is t = -b / 2a = -48 / (2 times -16) = 1.5. Substituting t = 1.5 gives h(1.5) = -16(1.5)^2 + 48(1.5) + 5 = 41.

Question 7

Convert y = x^2 + 4x - 1 into vertex form.

Accepted Answer

The vertex form is y = (x + 2)^2 - 5. Completing the square shows x^2 + 4x - 1 = (x^2 + 4x + 4) - 4 - 1 = (x + 2)^2 - 5.

Question 8

State the domain and range of the function y = (x - 1)^2 + 6.

Accepted Answer

The domain is all real numbers. The range is y is greater than or equal to 6 because the parabola opens upward and has a minimum value of 6 at its vertex.

Question 9

Find the zeros of y = x^2 - 9.

Accepted Answer

The zeros are x = 3 and x = -3. Setting y = 0 gives x^2 - 9 = 0, which factors as (x - 3)(x + 3) = 0.

Question 10

A quadratic function has vertex (4, -2) and opens upward. Write one possible equation in vertex form.

Accepted Answer

One possible equation is y = (x - 4)^2 - 2. This matches the vertex form y = a(x - h)^2 + k with h = 4, k = -2, and a positive value of 1.

Question 11

Solve x^2 - 4x - 5 = 0 using the quadratic formula.

Accepted Answer

The solutions are x = 5 and x = -1. Using the quadratic formula with a = 1, b = -4, and c = -5 gives x = [4 plus or minus square root of 36] / 2, which simplifies to x = [4 plus or minus 6] / 2.

Question 12

For the function y = -x^2 + 6x - 8, find the vertex and the maximum value.

Accepted Answer

The vertex is (3, 1), and the maximum value is 1. Using x = -b / 2a gives x = -6 / (2 times -1) = 3. Substituting x = 3 gives y = -9 + 18 - 8 = 1.

Question 13

Graphing question: Describe how the graph of y = (x + 1)^2 - 3 is related to the graph of y = x^2.

Accepted Answer

The graph is shifted 1 unit left and 3 units down from y = x^2. The shape stays the same because the coefficient of the squared term is still 1.

Question 14

Find the axis of symmetry and x-intercepts of y = x^2 - 2x - 8.

Accepted Answer

The axis of symmetry is x = 1. The x-intercepts are x = 4 and x = -2, so the intercept points are (4, 0) and (-2, 0). Factoring gives (x - 4)(x + 2) = 0.

Question 15

A rectangular garden has a length that is 3 feet more than its width. Its area is 54 square feet. Write and solve a quadratic equation to find the dimensions.

Accepted Answer

Let the width be w. Then the length is w + 3, and the area equation is w(w + 3) = 54, or w^2 + 3w - 54 = 0. Factoring gives (w + 9)(w - 6) = 0, so w = 6 or w = -9. The negative value is not reasonable, so the width is 6 feet and the length is 9 feet.

Quadratic Functions and Parabolas

Quadratic Functions and Parabolas

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