Question

15 Points!

Answer 1

$f(x)=\dfrac{x-5}{x^2-1}\\\\\text{the horizontal asymptotes}\\\\x^2-1=0\to x^2=1\to x=\pm\sqrt1\to x=-1\ \wedge\ x=1\\\\\text{the vertical asymptotes}\\\\y=\lim\limits_{x\to\pm\infty}\dfrac{x-5}{x^2-1}=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x}{x^2}+\frac{5}{x^2}}{\frac{x^2}{x^2}-\frac{1}{x^2}}=\dfrac{0+0}{1-0}=0$

Answer: B) x = 1, x = -1, y = 0