Question

Difficulty: MediumQuadratic Functions and Graphs

The graph of the quadratic function f(x)=2(x3)2+af(x) = -2(x - 3)^2 + a in the xyxy-plane has a yy-intercept at (0,10)(0, -10), where aa is a constant. What is the maximum value of f(x)f(x)?

  1. A
    28-28
  2. B
    10-10
  3. C
    33
  4. 88Answer

Answer

The maximum value of the function is 88.
The correct answer is 88. Since the yy-intercept of the graph is (0,10)(0, -10), we substitute x=0x = 0 into the function: f(0)=2(03)2+a=10f(0) = -2(0 - 3)^2 + a = -10. Simplifying this yields 2(9)+a=10-2(9) + a = -10, which becomes 18+a=10-18 + a = -10. Adding 1818 to both sides gives a=8a = 8. The equation of the function is therefore f(x)=2(x3)2+8f(x) = -2(x - 3)^2 + 8. Since this is in vertex form, the vertex is (3,8)(3, 8). Because the coefficient of the squared term is negative, the parabola opens downward, and the maximum value of the function is the yy-coordinate of the vertex, which is 88.

Step-by-Step Solution

1
Use the given yy-intercept to set up an equation for the constant aa.
f(0)=10    2(03)2+a=10f(0) = -10 \implies -2(0 - 3)^2 + a = -10
The yy-intercept occurs where the input xx is equal to 00.
2
Simplify the expression and solve for aa.
2(3)2+a=10    2(9)+a=10    18+a=10    a=8-2(-3)^2 + a = -10 \implies -2(9) + a = -10 \implies -18 + a = -10 \implies a = 8
Squaring 3-3 yields 99, and multiplying by 2-2 gives 18-18. Adding 1818 to both sides isolates aa.
3
Identify the vertex of the quadratic function and determine the maximum value.
The function is f(x)=2(x3)2+8f(x) = -2(x - 3)^2 + 8. The vertex of this parabola is (3,8)(3, 8). Since the leading coefficient 2-2 is negative, the parabola opens downward, meaning the yy-coordinate of the vertex, 88, is the maximum value of f(x)f(x).
A quadratic function in vertex form f(x)=a(xh)2+kf(x) = a(x - h)^2 + k has its vertex at (h,k)(h, k). If a<0a < 0, the maximum value is kk.

Key Concept

Identifying the vertex and maximum value of a quadratic function from its vertex form and yy-intercept.
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