# How do you find all the asymptotes for function f(x) = (2x^2+3x+8)/(x+3)?

May 14, 2018

Vertical asymptote $x = 3$ and slanting or oblique asymptote $y = 2 x + 3$

#### Explanation:

There are three types of asymptotes.

Vertical asymptotes are indicated by denominator. Here we have $\left(x + 3\right)$ in denominator, so vertical asymptote is given by $x + 3 = 0$ or $x = - 3$. Observe that in case $x + 3$ is a factor of numerator, it will cancel out and we will not have an asymptote, rather we will havea hole at $x = - 3$. But here $x + 3$ is not a factor of numerator, we do have a vertical asymptote as

${\lim}_{x \to - 3} \frac{2 {x}^{2} + 3 x + 8}{x + 3} = \infty$

Horizontal asymptotes in such cases (algebraic expressions) are there when degree of numerator is equal to that of denominator. Here it is not so we do not a horizontal asymptote. Stll assume numerator as only $3 x + 8$ and then

${\lim}_{x \to \infty} \frac{3 x + 8}{x + 3} = {\lim}_{x \to \infty} \frac{3 + \frac{8}{x}}{1 + \frac{3}{x}} = \frac{3}{1} = 3$ and we would have $y = 3$ as horizontal asymptote.

Here we have $\frac{2 {x}^{2} + 3 x + 8}{x + 3}$ and we have

${\lim}_{x \to \infty} \frac{2 {x}^{2} + 3 x + 8}{x + 3} = {\lim}_{x \to \infty} \frac{2 x + 3 + \frac{8}{x}}{1 + \frac{3}{x}}$

= $2 x + 3$

Hence we have a slanting or oblique asymptote $y = 2 x + 3$

It is apparent that we have this only when degree of numerator is one more than that of denominator.