Question

follow the steps for graphing a rational function to graph the function ( r(x)=\frac{x + 2}{x(x + 9)} )
(type an ordered pair using integers or fractions.)
d. there are no holes in the graph.
determine the behavior of the graph on either side of any vertical asymptotes, if any exist. select the correct choice and,
if necessary, fill in the answer box(es) to complete your choice.
a. it approaches ( infty ) on one side of the asymptote(s) at ( x=square ) and ( -infty ) on the other. it approaches either ( infty ) or
( -infty ) on both sides of the asymptote(s) at ( x=square ).
(type integers or simplified fractions. use a comma to separate answers as needed. type each answer
only once.)
b. it approaches either ( infty ) or ( -infty ) on both sides of the asymptote(s) at ( x=square ).
(type integers or simplified fractions. use a comma to separate answers as needed. type each answer
only once.)
c. it approaches ( infty ) on one side of the asymptote(s) at ( x=square ) and ( -infty ) on the other.
(type integers or simplified fractions. use a comma to separate answers as needed. type each answer
only once.)
d. the function has no vertical asymptote.

Explanation:

Step1: Find the vertical asymptotes

For a rational function \(R(x)=\frac{N(x)}{D(x)}\), vertical asymptotes occur at the values of \(x\) that make \(D(x) = 0\) (when \(N(x)
eq0\) at those values).
Set \(D(x)=x(x + 9)=0\). Solving \(x(x + 9)=0\) gives \(x=0\) or \(x=-9\).
Check if \(N(x)=x + 2\) is non - zero at \(x = 0\) and \(x=-9\).
When \(x = 0\), \(N(0)=0 + 2=2
eq0\). When \(x=-9\), \(N(-9)=-9 + 2=-7
eq0\).

Step2: Analyze the behavior near \(x = 0\)

Let's consider the one - sided limits as \(x\to0^{+}\) and \(x\to0^{-}\).
\(\lim_{x\to0^{+}}\frac{x + 2}{x(x + 9)}=\lim_{x\to0^{+}}\frac{x+2}{x}\cdot\frac{1}{x + 9}\). As \(x\to0^{+}\), \(\frac{x + 2}{x}\to+\infty\) (since \(x+2\to2\) and \(x\to0^{+}\)) and \(\frac{1}{x + 9}\to\frac{1}{9}\). So \(\lim_{x\to0^{+}}\frac{x + 2}{x(x + 9)}=+\infty\).
\(\lim_{x\to0^{-}}\frac{x + 2}{x(x + 9)}=\lim_{x\to0^{-}}\frac{x + 2}{x}\cdot\frac{1}{x + 9}\). As \(x\to0^{-}\), \(\frac{x + 2}{x}\to-\infty\) (since \(x + 2\to2\) and \(x\to0^{-}\)) and \(\frac{1}{x + 9}\to\frac{1}{9}\). So \(\lim_{x\to0^{-}}\frac{x + 2}{x(x + 9)}=-\infty\).

Step3: Analyze the behavior near \(x=-9\)

Let's consider the one - sided limits as \(x\to-9^{+}\) and \(x\to-9^{-}\).
\(\lim_{x\to-9^{+}}\frac{x + 2}{x(x + 9)}=\lim_{x\to-9^{+}}\frac{x + 2}{x}\cdot\frac{1}{x + 9}\). As \(x\to-9^{+}\), \(x+9\to0^{+}\), \(x\to-9\), and \(x + 2\to-7\). So \(\frac{x + 2}{x}\to\frac{-7}{-9}=\frac{7}{9}\) and \(\frac{1}{x + 9}\to+\infty\). Then \(\lim_{x\to-9^{+}}\frac{x + 2}{x(x + 9)}=+\infty\).
\(\lim_{x\to-9^{-}}\frac{x + 2}{x(x + 9)}=\lim_{x\to-9^{-}}\frac{x + 2}{x}\cdot\frac{1}{x + 9}\). As \(x\to-9^{-}\), \(x + 9\to0^{-}\), \(x\to-9\), and \(x + 2\to-7\). So \(\frac{x + 2}{x}\to\frac{-7}{-9}=\frac{7}{9}\) and \(\frac{1}{x + 9}\to-\infty\). Then \(\lim_{x\to-9^{-}}\frac{x + 2}{x(x + 9)}=-\infty\).

Answer:

C. It approaches \(\infty\) on one side of the asymptote(s) at \(x = 0\) and \(-\infty\) on the other. It approaches \(\infty\) on one side of the asymptote(s) at \(x=-9\) and \(-\infty\) on the other.