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A cubic polynomial function pp is defined by p(x)=a(x+3)(x−1)(x−k)p(x) = a(x + 3)(x - 1)(x - k), where aa and kk are constants. In the xyxy-plane, the yy-intercept of the graph of y=p(x)y = p(x) is (0,6)(0, 6). The graph of the shifted function y=p(x−2)y = p(x - 2) passes through the point (4,−30)(4, -30). What is the third xx-intercept of the graph of pp?

  1. −1-1Cevap
  2. B
    99
  3. C
    13\frac{1}{3}
  4. D
    −12-\frac{1}{2}

Cevap

−1-1
The correct answer is −1-1 because the yy-intercept (0,6)(0, 6) gives 3ak=63ak = 6, which simplifies to ak=2ak = 2. The shift y=p(x−2)y = p(x-2) passing through (4,−30)(4, -30) implies p(2)=−30p(2) = -30. Substituting x=2x=2 into the polynomial expression yields 5a(2−k)=−305a(2-k) = -30, which expands to 10a−5ak=−3010a - 5ak = -30. Substituting ak=2ak = 2 gives 10a−10=−3010a - 10 = -30, solving to a=−2a = -2. Using ak=2ak = 2, we find k=−1k = -1. The third factor is (x−k)=(x+1)(x - k) = (x + 1), which corresponds to the third xx-intercept at x=−1x = -1.

Adım Adım Çözüm

1
Use the yy-intercept of the graph of y=p(x)y = p(x) to establish a relationship between aa and kk.
p(0)=a(0+3)(0−1)(0−k)=3ak=6⇒ak=2p(0) = a(0 + 3)(0 - 1)(0 - k) = 3ak = 6 \Rightarrow ak = 2.
The yy-intercept is the point on the graph where x=0x = 0.
2
Translate the point on the shifted graph back to the original function p(x)p(x).
p(4−2)=p(2)=−30p(4 - 2) = p(2) = -30.
Since the graph of y=p(x−2)y = p(x - 2) passes through (4,−30)(4, -30), substituting x=4x = 4 yields y=−30y = -30.
3
Substitute x=2x = 2 into the expression for p(x)p(x) to set up the second equation.
p(2)=a(2+3)(2−1)(2−k)=5a(2−k)=10a−5ak=−30p(2) = a(2 + 3)(2 - 1)(2 - k) = 5a(2 - k) = 10a - 5ak = -30.
This establishes a system of equations with the relation from Step 1.
4
Solve the system of equations by substituting ak=2ak = 2 into the second equation.
10a−5(2)=−30⇒10a−10=−30⇒10a=−20⇒a=−210a - 5(2) = -30 \Rightarrow 10a - 10 = -30 \Rightarrow 10a = -20 \Rightarrow a = -2. Since ak=2ak = 2, we have −2k=2⇒k=−1-2k = 2 \Rightarrow k = -1.
Solving for the unknown constants aa and kk determines the specific polynomial expression.
5
Identify the third xx-intercept from the factored form of the polynomial.
p(x)=−2(x+3)(x−1)(x+1)p(x) = -2(x + 3)(x - 1)(x + 1). The factors correspond to roots at x=−3x = -3, x=1x = 1, and x=−1x = -1. The third xx-intercept is −1-1.
The third factor (x−k)(x - k) becomes (x+1)(x + 1) when k=−1k = -1, yielding the root and intercept at x=−1x = -1.

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