How do you find vertical, horizontal and oblique asymptotes for  f(x) = (x^2 - 9) / (x - 4)?

Jan 19, 2017

The vertical asymptote is $x = 4$
The oblique asymptote is $y = x + 4$
No horizontal asymptote.

Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{4\right\}$

As you cannot divide by $0$, $x \ne 4$

The vertical asymptote is $x = 4$

As the degree of the numerator is $>$ than the denominator, there is an oblique asymptote.

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{2}$$\textcolor{w h i t e}{a a a a a a}$$- 9$$\textcolor{w h i t e}{a a a a}$∣$\textcolor{red}{x - 4}$

$\textcolor{w h i t e}{a a a a}$${x}^{2} - 4 x$$\textcolor{w h i t e}{a a a a a a a a}$∣$\textcolor{b l u e}{x + 4}$

$\textcolor{w h i t e}{a a a a a}$$0 + 4 x - 9$

$\textcolor{w h i t e}{a a a a a a a}$$+ 4 x - 16$

$\textcolor{w h i t e}{a a a a a a a a a}$$+ 0 + 7$

Therefore,

$f \left(x\right) = \frac{{x}^{2} - 9}{x - 4} = \textcolor{b l u e}{x + 4} + \frac{7}{x - 4}$

${\lim}_{x \to - \infty} \left(f \left(x\right) - \left(x + 4\right)\right) = {\lim}_{x \to - \infty} \frac{7}{x} = {0}^{-}$

${\lim}_{x \to + \infty} \left(f \left(x\right) - \left(x + 4\right)\right) = {\lim}_{x \to + \infty} \frac{7}{x} = {0}^{+}$

The oblique asymptote is $y = x + 4$

graph{(y-(x^2-9)/(x-4))(y-x-4)=0 [-52, 52, -26.03, 26]}