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Difficulty: MediumPolynomial Factors and Graphs

The graph of a third-degree polynomial function gg has xx-intercepts at (−2,0)(-2, 0) and (3,0)(3, 0). At x=−2x = -2, the graph is tangent to the xx-axis, and the graph passes through the point (1,−12)(1, -12). What is the yy-intercept of the graph of gg?

  1. A
    (0,−12)(0, -12)
  2. (0,−8)(0, -8)Answer
  3. C
    (0,−18)(0, -18)
  4. D
    (0,−36)(0, -36)

Answer

The yy-intercept of the graph is (0,−8)(0, -8).
The polynomial has a double root at x=−2x = -2 because it is tangent to the xx-axis at (−2,0)(-2, 0), and a single root at x=3x = 3 because it crosses the xx-axis at (3,0)(3, 0). This allows us to write the cubic polynomial as g(x)=a(x+2)2(x−3)g(x) = a(x+2)^2(x-3). Substituting the point (1,−12)(1, -12) gives −12=a(1+2)2(1−3)  ⟹  −12=−18a  ⟹  a=23-12 = a(1+2)^2(1-3) \implies -12 = -18a \implies a = \frac{2}{3}. To find the yy-intercept, we evaluate the function at x=0x = 0: g(0)=23(0+2)2(0−3)=23(4)(−3)=−8g(0) = \frac{2}{3}(0+2)^2(0-3) = \frac{2}{3}(4)(-3) = -8, which corresponds to the point (0,−8)(0, -8).

Step-by-Step Solution

1
Set up the general form of the cubic function using the given intercepts.
g(x)=a(x+2)2(x−3)g(x) = a(x+2)^2(x-3)
An xx-intercept at (−2,0)(-2, 0) where the graph is tangent to the xx-axis represents a root of multiplicity 2, which corresponds to the factor (x+2)2(x+2)^2. An xx-intercept at (3,0)(3, 0) represents a root of multiplicity 1, corresponding to the factor (x−3)(x-3).
2
Use the point (1,−12)(1, -12) to solve for the constant coefficient aa.
a=23a = \frac{2}{3}
Substitute x=1x = 1 and g(1)=−12g(1) = -12 into the equation: −12=a(1+2)2(1−3)  ⟹  −12=a(9)(−2)  ⟹  −12=−18a  ⟹  a=23-12 = a(1+2)^2(1-3) \implies -12 = a(9)(-2) \implies -12 = -18a \implies a = \frac{2}{3}.
3
Calculate the yy-intercept of the function by evaluating g(0)g(0).
g(0)=−8g(0) = -8, which corresponds to the point (0,−8)(0, -8)
The yy-intercept of a graph occurs where x=0x = 0. Substituting x=0x = 0 into the equation g(x)=23(x+2)2(x−3)g(x) = \frac{2}{3}(x+2)^2(x-3) yields g(0)=23(0+2)2(0−3)=23(4)(−3)=−8g(0) = \frac{2}{3}(0+2)^2(0-3) = \frac{2}{3}(4)(-3) = -8.

Key Concept

Identifying polynomial factors and equations from graphical features like intercepts and tangency.
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Topics

Polynomial Factors and Graphs

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