Question

1. given ( f(x)=\frac{2 x - 4}{x^{2}-x - 6} )

a. state vertical asymptote(s).
b. horizontal asymptote.
c. intercepts
d. use limit notation to describe end behavior.
e. sketch

Explanation:

Step1: Factor the numerator and denominator

• Numerator: \(2x - 4=2(x - 2)\)
• Denominator: \(x^{2}-x - 6=(x - 3)(x + 2)\)

So, \(f(x)=\frac{2(x - 2)}{(x - 3)(x + 2)}\)

Step2: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is non - zero).
Set \((x - 3)(x + 2)=0\), then \(x = 3\) and \(x=-2\)

Step3: Find horizontal asymptote

For rational functions \(\frac{f(x)}{g(x)}\) where \(f(x)=a_nx^n+\cdots\) and \(g(x)=b_mx^m+\cdots\)
Here \(n = 1\) (degree of numerator) and \(m = 2\) (degree of denominator). Since \(n\lt m\), \(y = 0\)

Step4: Find \(x\) - intercepts

Set numerator equal to zero: \(2(x - 2)=0\Rightarrow x = 2\)
\(x\) - intercept is \((2,0)\)
Set \(x = 0\) for \(y\) - intercept: \(y=\frac{2(0 - 2)}{(0 - 3)(0 + 2)}=\frac{-4}{-6}=\frac{2}{3}\)
\(y\) - intercept is \((0,\frac{2}{3})\)

Step5: End - behavior using limit notation

\(\lim_{x
ightarrow\infty}\frac{2x-4}{x^{2}-x - 6}=\lim_{x
ightarrow\infty}\frac{\frac{2x}{x^{2}}-\frac{4}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{x}{x^{2}}-\frac{6}{x^{2}}}=\lim_{x
ightarrow\infty}\frac{\frac{2}{x}-\frac{4}{x^{2}}}{1-\frac{1}{x}-\frac{6}{x^{2}}}=0\)
\(\lim_{x
ightarrow-\infty}\frac{2x-4}{x^{2}-x - 6}=\lim_{x
ightarrow-\infty}\frac{\frac{2x}{x^{2}}-\frac{4}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{x}{x^{2}}-\frac{6}{x^{2}}}=\lim_{x
ightarrow-\infty}\frac{\frac{2}{x}-\frac{4}{x^{2}}}{1-\frac{1}{x}-\frac{6}{x^{2}}}=0\)

Answer:

A. Vertical asymptotes: \(x = 3\) and \(x=-2\)
B. Horizontal asymptote: \(y = 0\)
C. \(x\) - intercept: \((2,0)\); \(y\) - intercept: \((0,\frac{2}{3})\)
D. \(\lim_{x
ightarrow\infty}f(x)=0\) and \(\lim_{x
ightarrow-\infty}f(x)=0\)