Question

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 $lim_{x
ightarrow2}r(x)=\frac{4}{3}$, so $r(2)=\frac{4}{3}$. b $lim_{x
ightarrow2}r(x)=\frac{4}{3}$ and r is undefined at $x = 2$ so the graph of r has a hole at $(2,\frac{4}{3})$. c $lim_{x
ightarrow1}r(x)=2$ and r is undefined at $x = 1$, so the graph of r has a hole at $(2,\frac{4}{3})$. d $lim_{x
ightarrow2^{+}}r(x)=infty$, $lim_{x
ightarrow2^{-}}r(x)=infty$ and r is undefined at $x = 2$ so the graph of r has a vertical asymptote at $x = 2$.

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, \(r(x)=\frac{(x + 2)(x - 2)}{(x - 2)(x + 1)}=\frac{x + 2}{x + 1},x
eq2\).

Step2: Find the limit as \(x\to2\)

\(\lim_{x
ightarrow2}r(x)=\lim_{x
ightarrow2}\frac{x + 2}{x + 1}\). 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 \(\lim_{x
ightarrow2}r(x)=\frac{4}{3}\) and \(r(2)\) is undefined, the graph of \(r\) has a hole at the point \((2,\frac{4}{3})\).

Answer:

B. \(\lim_{x
ightarrow2}r(x)=\frac{4}{3}\) and \(r\) is undefined at \(x = 2\) so the graph of \(r\) has a hole at \((2,\frac{4}{3})\)