Question

the function ( f(x)=\frac{x + 5}{x^{2}-9x + 18} ) is a rational function.
a. find all vertical asymptotes
b. create a rough sketch of the graph near each asymptote
a. find all vertical asymptotes select the correct choice and, if necessary, fill in the answer box(es) to complete your choice.
○ a. the function has two vertical asymptotes. the leftmost asymptote is and the rightmost asymptote is
(type equations. use integers or fractions for any numbers in the equations.)
○ b. the function has one vertical asymptote,
(type an equation. use integers or fractions for any numbers in the equation.)
○ c. the function has three vertical asymptotes. the leftmost asymptote is the middle asymptote is and the rightmost asymptote is

Explanation:

Step1: Factor the denominator

Factor \(x^{2}-9x + 18\). We look for two numbers that multiply to \(18\) and add up to \(-9\). These numbers are \(-6\) and \(-3\). So, \(x^{2}-9x + 18=(x - 3)(x - 6)\)

Step2: Find the vertical asymptotes

For a rational function \(y=\frac{f(x)}{g(x)}\), vertical asymptotes occur at the values of \(x\) that make \(g(x)=0\) (provided that \(f(x)
eq0\) at those values).
Set \((x - 3)(x - 6)=0\). Using the zero - product property, \(x-3 = 0\) gives \(x = 3\) and \(x - 6=0\) gives \(x = 6\).
When \(x = 3\), \(f(3)=\frac{3 + 5}{(3)^{2}-9(3)+18}=\frac{8}{0}\) (undefined) and \(f(3)
eq0\).
When \(x = 6\), \(f(6)=\frac{6 + 5}{(6)^{2}-9(6)+18}=\frac{11}{0}\) (undefined) and \(f(6)
eq0\).

Answer:

A. The function has two vertical asymptotes. The leftmost asymptote is \(x = 3\) and the rightmost asymptote is \(x = 6\)