select the correct answer. which statement describes the graph of the function $f(x)=\frac{x^{2}-1}{x^{2}-2x…

select the correct answer. which statement describes the graph of the function $f(x)=\frac{x^{2}-1}{x^{2}-2x + 1}$? a. there is a hole at $x=-1$. b. there is a vertical asymptote at $x=-1$. c. the y - intercept is $y=-1$. d. there is a horizontal asymptote at $y=-1$.

select the correct answer. which statement describes the graph of the function $f(x)=\frac{x^{2}-1}{x^{2}-2x + 1}$? a. there is a hole at $x=-1$. b. there is a vertical asymptote at $x=-1$. c. the y - intercept is $y=-1$. d. there is a horizontal asymptote at $y=-1$.

Answer

Explanation:

Step1: Factor the function

First, factor the numerator and denominator. $x^{2}-1=(x + 1)(x - 1)$ and $x^{2}-2x + 1=(x - 1)^{2}$. So $f(x)=\frac{(x + 1)(x - 1)}{(x - 1)^{2}}=\frac{x + 1}{x - 1},x\neq1$.

Step2: Analyze vertical asymptotes

Set the denominator of the simplified function equal to zero. For $y=\frac{x + 1}{x - 1}$, when $x-1=0$, i.e., $x = 1$, there is a vertical asymptote. The original function is not affected by $x=-1$ in terms of vertical - asymptotes.

Step3: Analyze holes

Since we canceled out the factor $(x - 1)$ in the original function, there is a hole at $x = 1$, not $x=-1$.

Step4: Find the y - intercept

Set $x = 0$ in the function $y=\frac{x + 1}{x - 1}$. Then $y=\frac{0+1}{0 - 1}=-1$.

Step5: Analyze horizontal asymptotes

For $y=\frac{x + 1}{x - 1}$, since the degrees of the numerator and denominator are the same (both degree 1), the horizontal asymptote is $y=\frac{1}{1}=1$ (the ratio of the leading coefficients).

Answer:

C. The y - intercept is $y=-1$.