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Difficulty: MediumPolynomial Factors and Graphs

The graph of a third-degree polynomial function ff in the xyxy-plane has xx-intercepts at (2,0)(2, 0), (−1,0)(-1, 0), and (k,0)(k, 0). The yy-intercept of the graph of ff is (0,12)(0, 12). If f(3)=−48f(3) = -48, what is the value of kk?

  1. -3Answer
  2. B
    -1
  3. C
    3
  4. D
    6

Answer

-3
The correct answer is −3-3. By writing the cubic polynomial in factored form as f(x)=a(x−2)(x+1)(x−k)f(x) = a(x - 2)(x + 1)(x - k), we can use the yy-intercept (0,12)(0, 12) to determine that f(0)=2ak=12f(0) = 2ak = 12, which simplifies to ak=6ak = 6. Next, using the point (3,−48)(3, -48), we find that f(3)=4a(3−k)=−48f(3) = 4a(3 - k) = -48, which simplifies to 3a−ak=−123a - ak = -12. Substituting ak=6ak = 6 into this equation gives 3a−6=−123a - 6 = -12, which simplifies to 3a=−63a = -6, or a=−2a = -2. Since ak=6ak = 6 and a=−2a = -2, solving for kk yields k=−3k = -3.

Step-by-Step Solution

1
Write the general factored form of the cubic polynomial using its xx-intercepts.
f(x)=a(x−2)(x+1)(x−k)f(x) = a(x - 2)(x + 1)(x - k)
Since the graph of ff has xx-intercepts at (2,0)(2, 0), (−1,0)(-1, 0), and (k,0)(k, 0), the factors of f(x)f(x) must be (x−2)(x - 2), (x+1)(x + 1), and (x−k)(x - k), multiplied by a constant vertical stretch factor aa.
2
Use the yy-intercept (0,12)(0, 12) to find an equation relating aa and kk.
f(0)=a(0−2)(0+1)(0−k)=2ak=12  ⟹  ak=6f(0) = a(0 - 2)(0 + 1)(0 - k) = 2ak = 12 \implies ak = 6
Plugging x=0x = 0 and f(0)=12f(0) = 12 into the factored form allows us to find the product of aa and kk.
3
Use the given function value f(3)=−48f(3) = -48 to set up a second equation.
f(3)=a(3−2)(3+1)(3−k)=4a(3−k)=−48  ⟹  a(3−k)=−12  ⟹  3a−ak=−12f(3) = a(3 - 2)(3 + 1)(3 - k) = 4a(3 - k) = -48 \implies a(3 - k) = -12 \implies 3a - ak = -12
Substituting x=3x = 3 and f(3)=−48f(3) = -48 into the factored form provides a system of equations to solve for the individual constants.
4
Substitute ak=6ak = 6 into the equation from Step 3 to solve for aa.
3a−6=−12  ⟹  3a=−6  ⟹  a=−23a - 6 = -12 \implies 3a = -6 \implies a = -2
By replacing akak with 66 in the equation 3a−ak=−123a - ak = -12, we can isolate and solve for aa.
5
Solve for kk using the values of aa and akak.
Since a=−2a = -2 and ak=6ak = 6, we divide both sides of −2k=6-2k = 6 by −2-2 to get k=−3k = -3.
This isolates the unknown value kk.

Key Concept

Polynomial Factors and Graphs
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Topics

Polynomial Factors and Graphs

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