How do you find the vertical, horizontal or slant asymptotes for g(x)=(x+7)/(x^2-4)?

Dec 31, 2016

The vertical asymptotes are $x = - 2$ and $x = 2$
No slant asymptote.
The horizontal asymptote is $y = 0$

Explanation:

We need

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

Let's factorise the denominator

${x}^{2} - 4 = \left(x + 2\right) \left(x - 2\right)$

Then,

$g \left(x\right) = \frac{x + 7}{{x}^{2} - 4} = \frac{x + 7}{\left(x + 2\right) \left(x - 2\right)}$

The domain of $g \left(x\right)$ is ${D}_{g} \left(x\right) = \mathbb{R} - \left\{- 2 , 2\right\}$

As we cannot divide by $0$, $x \ne - 2$ and $x \ne 2$

The vertical asymptotes are $x = - 2$ and $x = 2$

The degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

To find the horizontal asymptotes, we calculate $\lim g \left(x\right)$ as $x \to \pm \infty$

${\lim}_{x \to - \infty} g \left(x\right) = {\lim}_{x \to - \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{1}{x} = {0}^{-}$

${\lim}_{x \to + \infty} g \left(x\right) = {\lim}_{x \to + \infty} \frac{x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{1}{x} = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{(y-(x+7)/(x^2-4))(y)=0 [-11.25, 11.26, -5.62, 5.62]}