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Difficulty: MediumQuadratic Functions and Graphs

In the xyxy-plane, the graph of the quadratic function ff is a parabola with vertex (4,−3)(4, -3). If the graph passes through the point (1,15)(1, 15), what is the value of f(2)f(2)?

Answer: 5

Answer

The value of f(2)f(2) is 55.
By using the vertex form of a quadratic function, f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k with vertex (4,−3)(4, -3), the function can be written as f(x)=a(x−4)2−3f(x) = a(x - 4)^2 - 3. Substituting the point (1,15)(1, 15) gives 15=a(1−4)2−315 = a(1 - 4)^2 - 3, which simplifies to 18=9a18 = 9a, leading to a=2a = 2. Substituting a=2a = 2 back into the function gives f(x)=2(x−4)2−3f(x) = 2(x - 4)^2 - 3. Finally, evaluating at x=2x = 2 gives f(2)=2(2−4)2−3=5f(2) = 2(2 - 4)^2 - 3 = 5.

Step-by-Step Solution

1
Write the quadratic function in vertex form and substitute the vertex (4,−3)(4, -3).
f(x)=a(x−4)2−3f(x) = a(x - 4)^2 - 3
The vertex form of a quadratic function is f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.
2
Substitute the point (1,15)(1, 15) into the equation and solve for the constant aa.
a=2a = 2
Since the graph passes through (1,15)(1, 15), substituting x=1x = 1 and f(x)=15f(x) = 15 allows us to solve for aa.
3
Substitute x=2x = 2 into the completed function f(x)=2(x−4)2−3f(x) = 2(x - 4)^2 - 3 to find f(2)f(2).
f(2)=5f(2) = 5
Evaluating the function at x=2x = 2 yields the required value.

Key Concept

Determining a quadratic function's equation from its vertex and a point, then evaluating it.
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Topics

Quadratic Functions and Graphs

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