Questions: Question 13 13. Let h(x)=(x-2)(x+1)^2/(x+1)(x-2)^2, Which of the following statements about the graph of h is correct? (A) The graph of h has one hole and one vertical asymptote. (B) The graph of h has two holes and no vertical asymptotes. (C) The graph of h has no holes and two vertical asymptotes. (D) The graph of h has one hole and two vertical asymptotes.

Solution Steps

To determine the correct statement about the graph of the function \( h(x) = \frac{(x-2)(x+1)^{2}}{(x+1)(x-2)^{2}} \), we need to analyze the points where the function is undefined and where it can be simplified. A hole occurs when a factor cancels out in the numerator and denominator, while a vertical asymptote occurs when a factor remains in the denominator after simplification.

1. Identify the factors in the numerator and denominator.
2. Simplify the function by canceling common factors.
3. Determine the points where the function is undefined and classify them as holes or vertical asymptotes based on the simplification.

The given function is \( h(x) = \frac{(x-2)(x+1)^{2}}{(x+1)(x-2)^{2}} \). The factors in the numerator are \((x-2)\) and \((x+1)^2\), and the factors in the denominator are \((x+1)\) and \((x-2)^2\).

Simplifying the function involves canceling out the common factors in the numerator and the denominator. The factor \((x-2)\) appears once in the numerator and twice in the denominator, so one \((x-2)\) can be canceled from both, leaving:

\[ h(x) = \frac{(x+1)^{2}}{(x-2)(x+1)} = \frac{x+1}{x-2} \]

The function \( h(x) = \frac{x+1}{x-2} \) is undefined at \( x = 2 \) because the denominator becomes zero. Since the factor \((x-2)\) was canceled during simplification, this point is a hole. The function is also undefined at \( x = -1 \) in the original form, but after simplification, this point is no longer a discontinuity.

After simplification, the function \( h(x) = \frac{x+1}{x-2} \) has a vertical asymptote at \( x = 2 \) because the denominator is zero and the factor does not cancel out.

Final Answer