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Difficulty: HardQuadratic Functions and Graphs

In the xyxy-plane, the graph of the quadratic function f(x)=−2x2+12x−10f(x) = -2x^2 + 12x - 10 is translated 44 units to the left and kk units up, where kk is a constant, to produce the graph of a new quadratic function gg. If the graph of gg passes through the origin (0,0)(0,0), what is the value of kk?

  1. A
    90
  2. B
    10
  3. -6Answer
  4. D
    3

Answer

The correct answer is −6-6.
The correct answer is −6-6. By completing the square on the original quadratic function, we rewrite f(x)=−2(x2−6x)−10f(x) = -2(x^2 - 6x) - 10 as f(x)=−2(x−3)2+8f(x) = -2(x-3)^2 + 8. A translation of 44 units to the left is represented by replacing xx with x+4x+4, and a translation of kk units up is represented by adding kk, giving g(x)=f(x+4)+k=−2(x+1)2+8+kg(x) = f(x+4) + k = -2(x+1)^2 + 8 + k. Since the graph of gg passes through the origin, we substitute (0,0)(0,0) into the equation: 0=−2(0+1)2+8+k0 = -2(0+1)^2 + 8 + k, which simplifies to 0=6+k0 = 6 + k, yielding k=−6k = -6.

Step-by-Step Solution

1
Rewrite the function f(x)f(x) in vertex form by completing the square.
f(x)=−2(x−3)2+8f(x) = -2(x-3)^2 + 8
Converting the standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c to vertex form f(x)=a(x−h)2+kf(x) = a(x-h)^2 + k helps easily identify the vertex of the parabola. First, factor out −2-2 from the variable terms: f(x)=−2(x2−6x)−10f(x) = -2(x^2 - 6x) - 10. To complete the square inside the parentheses, add and subtract 99 (since (−62)2=9(\frac{-6}{2})^2 = 9): f(x)=−2(x2−6x+9−9)−10f(x) = -2(x^2 - 6x + 9 - 9) - 10. This simplifies to f(x)=−2((x−3)2−9)−10=−2(x−3)2+18−10=−2(x−3)2+8f(x) = -2((x-3)^2 - 9) - 10 = -2(x-3)^2 + 18 - 10 = -2(x-3)^2 + 8.
2
Determine the equation of the translated function g(x)g(x).
g(x)=−2(x+1)2+8+kg(x) = -2(x+1)^2 + 8 + k
Translating a function 44 units to the left is represented by replacing xx with x+4x + 4. Translating a function kk units up is represented by adding kk to the entire function. Therefore, g(x)=f(x+4)+k=−2((x+4)−3)2+8+k=−2(x+1)2+8+kg(x) = f(x+4) + k = -2((x+4)-3)^2 + 8 + k = -2(x+1)^2 + 8 + k.
3
Substitute the point (0,0)(0,0) into g(x)g(x) to solve for kk.
k=−6k = -6
Since the graph of gg passes through the origin (0,0)(0,0), we have g(0)=0g(0) = 0. Substituting x=0x=0 yields 0=−2(0+1)2+8+k⇒0=−2(1)+8+k⇒0=6+k⇒k=−60 = -2(0+1)^2 + 8 + k \Rightarrow 0 = -2(1) + 8 + k \Rightarrow 0 = 6 + k \Rightarrow k = -6.

Key Concept

Quadratic transformations and translations in the coordinate plane using vertex form.
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Topics

Quadratic Functions and Graphs

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