Question

state the vertical asymptote(s) and determine the end behavior of the rational function ( f(x)=\frac{-2}{x - 5} ).
equation(s) of vertical asymptote(s):
end behavior:
as ( x
ightarrow-infty,f(x)
ightarrow )
as ( x
ightarrow+infty,f(x)
ightarrow )

Explanation:

Step1: Find vertical asymptote

For a rational function \(y = \frac{N(x)}{D(x)}\), vertical asymptotes occur where \(D(x)=0\) (assuming \(N(x)
eq0\) at that point).
For \(f(x)=\frac{-2}{x - 5}\), set \(x-5 = 0\).
\(x=5\)

Step2: Determine end - behavior

As \(x\to\pm\infty\), we consider the limit of \(f(x)=\frac{-2}{x - 5}\).
We know that \(\lim_{x\to\pm\infty}\frac{-2}{x - 5}\).
Since \(\lim_{x\to\pm\infty}\frac{1}{x}=0\), for \(y=\frac{-2}{x - 5}\), we can rewrite it as \(y=\frac{-2}{x(1-\frac{5}{x})}\)
\(\lim_{x\to-\infty}\frac{-2}{x - 5}=0\) (because as \(x\to-\infty\), \(\frac{-2}{x-5}\approx\frac{-2}{x}\) and \(\lim_{x\to-\infty}\frac{-2}{x}=0\))
\(\lim_{x\to+\infty}\frac{-2}{x - 5}=0\) (because as \(x\to+\infty\), \(\frac{-2}{x-5}\approx\frac{-2}{x}\) and \(\lim_{x\to+\infty}\frac{-2}{x}=0\))

Answer:

Equation(s) of vertical asymptote(s): \(x = 5\)
As \(x\to-\infty,f(x)\to0\)
As \(x\to+\infty,f(x)\to0\)