# How do you graph f(x)=(x^2+3x+2)/(-3x-12) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jul 9, 2018

Below

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + 3 x + 2}{- 3 x - 12}$

Using long division, we can rearrange the polynomial
$f \left(x\right) = \frac{\left(- 3 x - 12\right) \left(- \frac{1}{3} x - \frac{1}{3}\right) - 2}{- 3 x - 12}$

$f \left(x\right) = - \frac{1}{3} x - \frac{1}{3} - \frac{2}{- 3 x - 12}$

This means that your oblique asymptote is $y = - \frac{1}{3} x - \frac{1}{3}$ which can be found by figuring out what happens when x approaches 0 and your horizontal asymptote is $x = - 4$ after solving $- 3 x - 12 = 0$ since your denominator cannot equal to 0.

To figure out your x and y intercepts, we let $y = 0$ and $x = 0$ respectively.

When $x = 0$, $y = - \frac{1}{6}$ so your y-intercept is $\left(0 , - \frac{1}{6}\right)$
When $y = 0$, $x = - 2$ and $x = - 1$ so your x-intercepts are $\left(- 2 , 0\right)$ and $\left(- 1 , 0\right)$

graph{(x^2+3x+2)/(-3x-12) [-10, 10, -5, 5]}
Above is the graph