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Difficulty: HardFunction Notation and Transformations

In the xyxy-plane, the graph of the function gg is obtained by applying a sequence of transformations to the graph of the function f(x)=∣x+2∣−3f(x) = |x + 2| - 3. Specifically, the graph of gg is a vertical stretch and translation of the graph of ff, such that g(x)=af(x−h)+kg(x) = a f(x - h) + k for some constants aa, hh, and kk. The vertex of the graph of gg is located at (1,5)(1, 5), and the graph of gg passes through the point (0,−1)(0, -1). What is the value of g(3)g(3)?

  1. A
    -5
  2. -7Answer
  3. C
    17
  4. D
    -43

Answer

The value of g(3)g(3) is −7-7.
The correct value of −7-7 is obtained by first identifying the vertex of f(x)=∣x+2∣−3f(x) = |x + 2| - 3 at (−2,−3)(-2, -3). Comparing this to the vertex of g(x)g(x) at (1,5)(1, 5) yields the horizontal shift parameter h=3h = 3 and the equation −3a+k=5-3a + k = 5. Using the point (0,−1)(0, -1) yields the second equation −2a+k=−1-2a + k = -1. Solving this system gives a=−6a = -6 and k=−13k = -13. Substituting these into the formula for g(3)g(3) yields −7-7.

Step-by-Step Solution

1
Identify the vertex of the function f(x)=∣x+2∣−3f(x) = |x + 2| - 3.
The vertex of f(x)f(x) is at (−2,−3)(-2, -3).
The vertex of an absolute value function of the form y=∣x−x0∣+y0y = |x - x_0| + y_0 is located at (x0,y0)(x_0, y_0).
2
Relate the vertex of f(x)f(x) to the vertex of g(x)=af(x−h)+kg(x) = a f(x - h) + k.
h=3h = 3 and −3a+k=5-3a + k = 5.
The horizontal shift hh moves the vertex from x=−2x = -2 to x=1x = 1, so −2+h=1  ⟹  h=3-2 + h = 1 \implies h = 3. The vertical stretch and translation transform the yy-coordinate of the vertex from −3-3 to 55, so a(−3)+k=5a(-3) + k = 5.
3
Use the given point (0,−1)(0, -1) to set up a second equation.
−2a+k=−1-2a + k = -1.
Since the graph of g(x)g(x) passes through (0,−1)(0, -1), we evaluate g(0)=af(0−3)+k=−1g(0) = a f(0 - 3) + k = -1. Evaluating f(−3)=∣−3+2∣−3=−2f(-3) = |-3 + 2| - 3 = -2 yields the equation −2a+k=−1-2a + k = -1.
4
Solve the system of equations for aa and kk.
a=−6a = -6 and k=−13k = -13.
Subtracting −3a+k=5-3a + k = 5 from −2a+k=−1-2a + k = -1 gives a=−6a = -6. Substituting a=−6a = -6 back into either equation yields k=−13k = -13.
5
Evaluate g(3)g(3) using the completed function formula g(x)=−6f(x−3)−13g(x) = -6 f(x - 3) - 13.
g(3)=−7g(3) = -7.
We substitute x=3x = 3 into the equation to get g(3)=−6f(0)−13g(3) = -6 f(0) - 13. Evaluating f(0)=∣0+2∣−3=−1f(0) = |0 + 2| - 3 = -1 gives g(3)=−6(−1)−13=6−13=−7g(3) = -6(-1) - 13 = 6 - 13 = -7.

Key Concept

Function transformations including horizontal translations, vertical translations, and vertical scaling.
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Topics

Function Notation and Transformations

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