Question

State the vertical asymptote of the rational function. f(x) =((x-9)(x+7))/(x^2-4)

Answer 1

$D:x^2-4\not=0\\ D:x^2\not=4\\ D:x\not=-2 \wedge x\not =2\\\\ \displaystyle \lim_{x\to-2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\ \dfrac{(-2-9)(-2+7)}{(-2^-)^2-4}=\dfrac{-11\cdot5}{4^+-4}=\dfrac{-55}{0^+}=-55\cdot\infty=-\infty\\ \dfrac{(-2-9)(-2+7)}{(-2^+)^2-4}=\dfrac{-11\cdot5}{4^--4}=\dfrac{-55}{0^-}=-55\cdot(-\infty)=\infty$

$\displaystyle \lim_{x\to2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\ \dfrac{(2-9)(2+7)}{(2^-)^2-4}=\dfrac{-7\cdot9}{4^--4}=\dfrac{-63}{0^-}=-63\cdot(-\infty)=\infty\\ \dfrac{(2-9)(2+7)}{(2^+)^2-4}=\dfrac{-7\cdot9}{4^+-4}=\dfrac{-63}{0^+}=-63\cdot\infty=-\infty$

So, the vertical asymptotes are $x=\pm 2$

Answer 2

Answer:

There is a vertical asymptote for the rational function at x = −7. Set the denominator equal to 0 and solve for x.

x + 7 = 0 → x = −7

Step-by-step explanation: