Back to Student Worksheet
Math Grade 9-12 Answer Key

Quadratic Functions and Parabolas

Graphing, analyzing, and solving quadratic relationships

Answer Key
Name:
Date:
Score: / 15

Quadratic Functions and Parabolas

Graphing, analyzing, and solving quadratic relationships

Math - Grade 9-12

Instructions: Read each problem carefully. Show your work and explain your reasoning when needed.
  1. 1

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

    Use the vertex formula x = -b / 2a or complete the square.

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

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

    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.
  3. 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).

    Start with factored form using the given x-intercepts.

    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.
  4. 4

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

    The solutions are x = -3 and x = -4. Factoring gives (x + 3)(x + 4) = 0, so each factor can be set equal to zero.
  5. 5

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

    The y-intercept occurs where x = 0.

    The y-intercept is 7, or the point (0, 7). Substituting x = 0 into the function gives y = 7.
  6. 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.

    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.
  7. 7

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

    Complete the square by adding and subtracting the same value.

    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.
  8. 8

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

    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.
  9. 9

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

    Use the difference of squares.

    The zeros are x = 3 and x = -3. Setting y = 0 gives x^2 - 9 = 0, which factors as (x - 3)(x + 3) = 0.
  10. 10

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

    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.
  11. 11

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

    Substitute carefully into x = [-b plus or minus square root of (b^2 - 4ac)] / 2a.

    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.
  12. 12

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

    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.
  13. 13

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

    Use the form y = (x - h)^2 + k to identify shifts.

    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.
  14. 14

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

    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.
  15. 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.

    Use area = length times width and reject any negative dimension.

    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.
LivePhysics.com Math - Grade 9-12 - Answer Key

Related Mission Packs