the rational function r is given by $r(x)=\frac{x^{2}-4}{x^{2}-x - 2}$. the table gives values of $r(x)$ for…

the rational function r is given by $r(x)=\frac{x^{2}-4}{x^{2}-x - 2}$. the table gives values of $r(x)$ for selected values of x. which of the following statements is true? a $lim_{x\rightarrow2}r(x)=\frac{4}{3}$, so $r(2)=\frac{4}{3}$. b $lim_{x\rightarrow2}r(x)=\frac{4}{3}$ and r is undefined at $x = 2$ so the graph of r has a hole at $(2,\frac{4}{3})$. c $lim_{x\rightarrow1}r(x)=2$ and r is undefined at $x = 1$, so the graph of r has a hole at $(2,\frac{4}{3})$. d $lim_{x\rightarrow2^{+}}r(x)=infty$, $lim_{x\rightarrow2^{-}}r(x)=infty$ and r is undefined at $x = 2$ so the graph of r has a vertical asymptote at $x = 2$.

the rational function r is given by $r(x)=\frac{x^{2}-4}{x^{2}-x - 2}$. the table gives values of $r(x)$ for selected values of x. which of the following statements is true? a $lim_{x\rightarrow2}r(x)=\frac{4}{3}$, so $r(2)=\frac{4}{3}$. b $lim_{x\rightarrow2}r(x)=\frac{4}{3}$ and r is undefined at $x = 2$ so the graph of r has a hole at $(2,\frac{4}{3})$. c $lim_{x\rightarrow1}r(x)=2$ and r is undefined at $x = 1$, so the graph of r has a hole at $(2,\frac{4}{3})$. d $lim_{x\rightarrow2^{+}}r(x)=infty$, $lim_{x\rightarrow2^{-}}r(x)=infty$ and r is undefined at $x = 2$ so the graph of r has a vertical asymptote at $x = 2$.

Answer

Explanation:

Step1: Simplify the rational - function

First, factor the numerator and denominator of (r(x)=\frac{x^{2}-4}{x^{2}-x - 2}). The numerator (x^{2}-4=(x + 2)(x - 2)), and the denominator (x^{2}-x - 2=(x-2)(x + 1)). So, (r(x)=\frac{(x + 2)(x - 2)}{(x - 2)(x + 1)}=\frac{x + 2}{x + 1},x\neq2).

Step2: Find the limit as (x\to2)

(\lim_{x\rightarrow2}r(x)=\lim_{x\rightarrow2}\frac{x + 2}{x + 1}). Substitute (x = 2) into (\frac{x+2}{x + 1}), we get (\frac{2+2}{2+1}=\frac{4}{3}).

Step3: Analyze the function value at (x = 2)

The original function (r(x)=\frac{x^{2}-4}{x^{2}-x - 2}) is undefined at (x = 2) because when (x = 2), the denominator (x^{2}-x - 2=2^{2}-2 - 2=0). A hole in the graph of a rational - function occurs when the numerator and denominator have a common factor that cancels out, and the function is undefined at the value of (x) that makes the common factor zero. Here, since (\lim_{x\rightarrow2}r(x)=\frac{4}{3}) and (r(2)) is undefined, the graph of (r) has a hole at the point ((2,\frac{4}{3})).

Answer:

B. (\lim_{x\rightarrow2}r(x)=\frac{4}{3}) and (r) is undefined at (x = 2) so the graph of (r) has a hole at ((2,\frac{4}{3}))