11. given ( f(x)=\frac{2 x - 4}{x^{2}-x - 6} )\na. state vertical asymptote(s).\nb. horizontal…

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