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Nonlinear Functions

SAT Math· difficulty 4/5

Which polynomial could correspond to a graph with x-intercepts at −2-2, 11, 33 and y-intercept at −6-6?

  • A

    f(x)=(x−2)(x+1)(x+3)f(x) = (x - 2)(x + 1)(x + 3)

  • B

    f(x)=(x+2)(x−1)(x−3)f(x) = (x + 2)(x - 1)(x - 3)

  • C

    f(x)=−(x+2)(x−1)(x−3)f(x) = -(x + 2)(x - 1)(x - 3)

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  • D

    f(x)=−(x−2)(x+1)(x+3)f(x) = -(x - 2)(x + 1)(x + 3)

Explanation

Need zeros at −2,1,3-2, 1, 3. Form: f(x)=a(x+2)(x−1)(x−3)f(x) = a(x+2)(x-1)(x-3). f(0)=a(2)(−1)(−3)=6a=−6⇒a=−1f(0) = a(2)(-1)(-3) = 6a = -6 \Rightarrow a = -1. So f(x)=−(x+2)(x−1)(x−3)f(x) = -(x+2)(x-1)(x-3).

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