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Difficulty: MediumPolynomial Factors and Graphs

The graph of the polynomial function ff in the xyxy-plane is tangent to the xx-axis at (1,0)(1, 0) and crosses the xx-axis only at (−2,0)(-2, 0) and (4,0)(4, 0). If the yy-intercept of the graph of ff is (0,8)(0, 8), what is the value of f(−1)f(-1)?

  1. A
    -20
  2. B
    10
  3. 20Answer
  4. D
    60

Answer

20
The graph of the polynomial function is tangent to the xx-axis at (1,0)(1, 0) and crosses it at (−2,0)(-2, 0) and (4,0)(4, 0). This means x=1x = 1 is a root with multiplicity 2, while x=−2x = -2 and x=4x = 4 are roots with multiplicity 1. Thus, we can write the function in the form f(x)=a(x+2)(x−1)2(x−4)f(x) = a(x + 2)(x - 1)^2(x - 4). Substituting the yy-intercept (0,8)(0, 8) gives 8=a(0+2)(0−1)2(0−4)  ⟹  8=−8a  ⟹  a=−18 = a(0 + 2)(0 - 1)^2(0 - 4) \implies 8 = -8a \implies a = -1. Substituting a=−1a = -1 back into the formula yields f(x)=−(x+2)(x−1)2(x−4)f(x) = -(x + 2)(x - 1)^2(x - 4). Evaluating the function at x=−1x = -1 gives f(−1)=−(−1+2)(−1−1)2(−1−4)=−(1)(4)(−5)=20f(-1) = -(-1 + 2)(-1 - 1)^2(-1 - 4) = -(1)(4)(-5) = 20. Therefore, the correct value is 20.

Step-by-Step Solution

1
Determine the factored form of the polynomial from its x-intercepts and multiplicities.
f(x)=a(x+2)(x−1)2(x−4)f(x) = a(x + 2)(x - 1)^2(x - 4)
Since the graph is tangent to the xx-axis at (1,0)(1, 0), the root x=1x = 1 has a multiplicity of 2, corresponding to the factor (x−1)2(x - 1)^2. Since it crosses the xx-axis at (−2,0)(-2, 0) and (4,0)(4, 0), these roots have a multiplicity of 1, corresponding to the factors (x+2)(x + 2) and (x−4)(x - 4).
2
Use the y-intercept to find the constant coefficient aa.
a=−1a = -1
The yy-intercept is (0,8)(0, 8), so substituting x=0x = 0 into the equation gives 8=a(0+2)(0−1)2(0−4)  ⟹  8=−8a  ⟹  a=−18 = a(0 + 2)(0 - 1)^2(0 - 4) \implies 8 = -8a \implies a = -1.
3
Substitute the value of aa back into the function and evaluate f(−1)f(-1).
f(−1)=20f(-1) = 20
Substituting a=−1a = -1 gives f(x)=−(x+2)(x−1)2(x−4)f(x) = -(x + 2)(x - 1)^2(x - 4). Evaluating at x=−1x = -1 yields f(−1)=−(−1+2)(−1−1)2(−1−4)=−(1)(−2)2(−5)=−(1)(4)(−5)=20f(-1) = -(-1 + 2)(-1 - 1)^2(-1 - 4) = -(1)(-2)^2(-5) = -(1)(4)(-5) = 20.

Key Concept

Determining a polynomial function's equation from its graphical features (intercepts and tangencies) and evaluating it at a point.
Estimated Time:1m 30s
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Topics

Polynomial Factors and Graphs

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