Question

follow the steps for graphing a rational function to graph the function ( r(x)=\frac{x + 2}{x(x + 9)} ).

a. there is a hole in the graph at the point
(type an ordered pair using integers or fractions.)
b. 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=) and ( -infty ) on the other. it approaches either ( infty ) or
( -infty ) on both sides of the asymptote(s) at ( x=)
(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=)
(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=) 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 domain

The function \(R(x)=\frac{x + 2}{x(x + 9)}\) is undefined when \(x(x+9)=0\), i.e., \(x = 0\) or \(x=-9\). Since there is no common factor in the numerator and denominator (the numerator is \(x + 2\) and the denominator is \(x(x + 9)\)), there are no holes.

Step2: Analyze the vertical asymptotes

For vertical asymptote \(x=a\), we consider the limit as \(x\to a^{+}\) and \(x\to a^{-}\).

• For \(x = 0\):

\(\lim_{x\to0^{+}}\frac{x + 2}{x(x + 9)}=\lim_{x\to0^{+}}\frac{x+2}{x}\cdot\frac{1}{x + 9}\). As \(x\to0^{+}\), \(\frac{x + 2}{x}\to+\infty\) 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 \(x\to0^{-}\), \(\frac{x + 2}{x}\to-\infty\) and \(\frac{1}{x + 9}\to\frac{1}{9}\), so \(\lim_{x\to0^{-}}\frac{x + 2}{x(x + 9)}=-\infty\).

• For \(x=-9\):

\(\lim_{x\to - 9^{+}}\frac{x + 2}{x(x + 9)}=\lim_{x\to - 9^{+}}\frac{x + 2}{x}\cdot\frac{1}{x + 9}\). As \(x\to-9^{+}\), \(x+9\to0^{+}\), \(\frac{x + 2}{x}\to\frac{-7}{-9}=\frac{7}{9}\), so \(\lim_{x\to - 9^{+}}\frac{x + 2}{x(x + 9)}=+\infty\).
\(\lim_{x\to - 9^{-}}\frac{x + 2}{x(x + 9)}=\lim_{x\to - 9^{-}}\frac{x + 2}{x}\cdot\frac{1}{x + 9}\). As \(x\to-9^{-}\), \(x + 9\to0^{-}\), \(\frac{x + 2}{x}\to\frac{-7}{-9}=\frac{7}{9}\), so \(\lim_{x\to - 9^{-}}\frac{x + 2}{x(x + 9)}=-\infty\).

Answer:

For the hole - related part: B. There are no holes in the graph.
For the vertical asymptote - related part: C. It approaches \(\infty\) on one side of the asymptote(s) at \(x = 0\) and \(-\infty\) on the other, and it approaches \(\infty\) on one side of the asymptote(s) at \(x=-9\) and \(-\infty\) on the other.