Question

Difficulty: EasyQuadratic Functions and Graphs

A quadratic function ff is defined by f(x)=2(x+3)28f(x) = 2(x + 3)^2 - 8. What are the coordinates of the vertex of the graph of y=f(x)y = f(x) in the xyxy-plane?

  1. (3,8)(-3, -8)Answer
  2. B
    (3,8)(3, -8)
  3. C
    (3,8)(-3, 8)
  4. D
    (3,8)(3, 8)

Answer

The correct answer is the coordinate pair (3,8)(-3, -8).
The vertex form of a quadratic function is f(x)=a(xh)2+kf(x) = a(x - h)^2 + k, where the vertex of the graph of the function in the xyxy-plane is (h,k)(h, k). For the function f(x)=2(x+3)28f(x) = 2(x + 3)^2 - 8, the expression inside the parentheses can be rewritten as (x(3))(x - (-3)), showing that h=3h = -3. The constant term outside the parentheses is 8-8, showing that k=8k = -8. Therefore, the vertex of the graph of the function is (3,8)(-3, -8).

Step-by-Step Solution

1
Identify the standard vertex form of a quadratic function.
The vertex form is y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
Comparing the given equation to this template allows us to read the coordinates directly.
2
Rewrite the given equation to align with the negative sign inside the parentheses of the vertex form.
The term (x+3)2(x + 3)^2 can be rewritten as (x(3))2(x - (-3))^2, and the subtraction of 88 can be written as adding 8-8. This yields f(x)=2(x(3))2+(8)f(x) = 2(x - (-3))^2 + (-8).
Matching the template exactly prevents sign errors when identifying the values of hh and kk.
3
Extract the values of hh and kk.
h=3h = -3 and k=8k = -8, which means the vertex is located at (3,8)(-3, -8).
The vertex (h,k)(h, k) coordinates represent the minimum point of this upward-opening parabola.

Key Concept

Identifying the vertex coordinates directly from the vertex form of a quadratic function.

Alternative Method

An alternative method is to expand the function into the standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c. Expanding f(x)=2(x+3)28f(x) = 2(x + 3)^2 - 8 gives f(x)=2(x2+6x+9)8f(x) = 2(x^2 + 6x + 9) - 8, which simplifies to f(x)=2x2+12x+10f(x) = 2x^2 + 12x + 10. The xx-coordinate of the vertex can be found using the formula x=b2a=122(2)=3x = -\frac{b}{2a} = -\frac{12}{2(2)} = -3. Substituting x=3x = -3 back into the function gives f(3)=2(3+3)28=8f(-3) = 2(-3 + 3)^2 - 8 = -8. This confirms the vertex is at (3,8)(-3, -8).
Estimated Time:45s
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