# How do you find vertical, horizontal and oblique asymptotes for (2x^2 + 5x) /( x - 2)?

May 16, 2018

Vertical asymptote: $x = 2$
No horizontal asymptote
Oblique asymptote: $y = 2 x + 9$

#### Explanation:

Vertical asymptotes: we must look for points in which the function is not defined. In this case, the point is $x = 2$, since it causes the denominator to vanish:

$f \left(2\right) = \setminus \frac{2 \cdot {2}^{2} + 5 \cdot 2}{2 - 2} = \setminus \frac{18}{0}$

which is undefined.

Horizontal asymptotes: we can look for horizontal asymptotes, since the domain is unbounded. If they exist, they are given by ${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right)$:

$\setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{2 {x}^{2} + 5 x}{x - 2} = \setminus \frac{x \left(2 x + 5\right)}{x \left(1 - \setminus \frac{2}{x}\right)} = \setminus \frac{2 x + 5}{1 - \setminus \frac{2}{x}} \setminus \to \setminus \frac{\setminus \pm \setminus \infty}{1} = \setminus \pm \setminus \infty$

Oblique asymptotes: we can look for oblique asymptotes, since the function diverges. Here's the theory: if $y = f \left(x\right)$ approaches some line $y = m x + q$ as $x \setminus \to \setminus \pm \setminus \infty$, then we have

$f \left(x\right) - \left(m x + q\right) \setminus \to 0 \setminus \iff f \left(x\right) \setminus \to m x + q$

If we divide both sides by $x$, we have

$f \frac{x}{x} \setminus \to \frac{m x}{x} + \frac{q}{x} \setminus \to m$

since $\frac{q}{x} \setminus \to 0$ because $x \setminus \to \setminus \pm \setminus \infty$

So, if $f \frac{x}{x}$ has a finite limit, that would be the slope of the line it is approaching. Once $m$ is known, we simply have

$f \left(x\right) \setminus \to m x + q \setminus \iff f \left(x\right) - m x \setminus \to q$.

Let's do it!

$m = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \frac{x}{x} = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{2 {x}^{2} + 5 x}{{x}^{2} - 2 x} = 2$

$q = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right) - m x = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{2 {x}^{2} + 5 x}{x - 2} - 2 x$
$= \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{9 x}{x - 2} = 9$