question 8 (1 point)\nfor the function\nf(x)=\begin{cases}-1, & xleq2;\\\frac{x^{2}+3x + 2}{x^{2}-1},&x >…

question 8 (1 point)\nfor the function\nf(x)=\begin{cases}-1, & xleq2;\\\frac{x^{2}+3x + 2}{x^{2}-1},&x > 2.end{cases}\nwhich of the following statements is true:\nthe function does not have any horizontal asymptotes.\nthe function has a single horizontal asymptote: (y = 1).\nthe function has two vertical asymptotes at (x=-1) and (x = 1).\nthe function has two horizontal asymptotes: (y=-1) and (y = 1).\nthe function has a single vertical asymptote: (x = 1).

question 8 (1 point)\nfor the function\nf(x)=\begin{cases}-1, & xleq2;\\\frac{x^{2}+3x + 2}{x^{2}-1},&x > 2.end{cases}\nwhich of the following statements is true:\nthe function does not have any horizontal asymptotes.\nthe function has a single horizontal asymptote: (y = 1).\nthe function has two vertical asymptotes at (x=-1) and (x = 1).\nthe function has two horizontal asymptotes: (y=-1) and (y = 1).\nthe function has a single vertical asymptote: (x = 1).

Answer

Explanation:

Step1: Analyze the function for $x\leq2$

For $x\leq2$, $f(x)= - 1$, which is a constant - function. The graph of $y = - 1$ is a horizontal line, and it is a horizontal asymptote for $x\to-\infty$ (since the function is constant for $x\leq2$).

Step2: Simplify the function for $x > 2$

For $x>2$, $f(x)=\frac{x^{2}+3x + 2}{x^{2}-1}=\frac{(x + 1)(x+2)}{(x + 1)(x - 1)}=\frac{x + 2}{x - 1}$ (for $x\neq - 1$). We find the horizontal asymptote as $x\to+\infty$ by taking the limit: $\lim_{x\to+\infty}\frac{x + 2}{x - 1}=\lim_{x\to+\infty}\frac{1+\frac{2}{x}}{1-\frac{1}{x}} = 1$.

Step3: Analyze vertical asymptotes

The function $f(x)=\frac{x + 2}{x - 1}$ (for $x>2$) has a vertical asymptote when the denominator is zero. Setting $x - 1=0$, we get $x = 1$. But we consider the domain $x>2$, so there are no vertical asymptotes in the domain of the piece - wise function.

Answer:

The function has two horizontal asymptotes: $y=-1$ and $y = 1$.