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Difficulty: MediumQuadratic Functions and Graphs

The height of a diver, in meters, above the pool surface tt seconds after leaving the diving board is modeled by the function f(t)=−4.9(t−1)2+10f(t) = -4.9(t - 1)^2 + 10. If the diving board is moved so that the diver's trajectory is shifted 0.50.5 seconds later in time and the maximum height is increased by 22 meters, which of the following functions gg models the diver's new trajectory?

  1. A
    g(t)=−4.9(t−0.5)2+12g(t) = -4.9(t - 0.5)^2 + 12
  2. B
    g(t)=−4.9(t−1.5)2+8g(t) = -4.9(t - 1.5)^2 + 8
  3. g(t)=−4.9(t−1.5)2+12g(t) = -4.9(t - 1.5)^2 + 12Answer
  4. D
    g(t)=−4.9(t+0.5)2+12g(t) = -4.9(t + 0.5)^2 + 12

Answer

The function g(t)=−4.9(t−1.5)2+12g(t) = -4.9(t - 1.5)^2 + 12 models the diver's new trajectory.
The correct answer represents the translated quadratic function. Since the original function f(t)=−4.9(t−1)2+10f(t) = -4.9(t - 1)^2 + 10 has its vertex at (1,10)(1, 10), shifting the trajectory 0.50.5 seconds later in time moves the vertex horizontally to the right to t=1.5t = 1.5 seconds, replacing (t−1)(t - 1) with (t−1.5)(t - 1.5). Increasing the maximum height by 22 meters shifts the vertex vertically upward to 1212 meters, replacing the constant term 1010 with 1212. This results in the function g(t)=−4.9(t−1.5)2+12g(t) = -4.9(t - 1.5)^2 + 12.

Step-by-Step Solution

1
Identify the vertex and its meaning in the original function f(t)=−4.9(t−1)2+10f(t) = -4.9(t - 1)^2 + 10.
The vertex is (1,10)(1, 10), indicating that the maximum height of 1010 meters occurs at t=1t = 1 second.
The vertex form of a quadratic function is y=a(t−h)2+ky = a(t - h)^2 + k, where (h,k)(h, k) is the vertex representing the extreme value.
2
Apply the horizontal translation of 0.50.5 seconds later in time.
The new tt-coordinate of the vertex is 1+0.5=1.51 + 0.5 = 1.5 seconds, which changes the term (t−1)(t - 1) to (t−1.5)(t - 1.5).
A horizontal shift to the right by cc units is represented by replacing the variable tt with t−ct - c.
3
Apply the vertical translation of 22 meters upward.
The new yy-coordinate of the vertex is 10+2=1210 + 2 = 12 meters, which changes the constant term of the function to 1212.
A vertical shift upward by dd units is represented by adding dd to the function value.
4
Combine the horizontal and vertical translations to write the new equation.
g(t)=−4.9(t−1.5)2+12g(t) = -4.9(t - 1.5)^2 + 12.
Substituting the shifted vertex coordinates (1.5,12)(1.5, 12) into the vertex form while keeping the same vertical stretch/direction factor a=−4.9a = -4.9 yields the final equation.

Key Concept

Vertex form and transformations of quadratic functions
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Topics

Quadratic Functions and Graphs

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