# How do you graph f(x)=(x^2+7x+12)/(-4x^2-8x+12) using holes, vertical and horizontal asymptotes, x and y intercepts?

Dec 26, 2017

hole: $x = - 3$
vertical asymptote: $x = 1$
horizontal asymptote: $y = - \frac{1}{4}$
x and y intercepts: $\left(0 , 1\right) \mathmr{and} \left(- 4 , 0\right)$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + 7 x + 12}{- 4 {x}^{2} - 8 x + 12} = \frac{\left(x + 3\right) \left(x + 4\right)}{- 4 \left[{x}^{2} + 2 x - 3\right]} = \frac{\cancel{\left(x + 3\right)} \left(x + 4\right)}{- 4 \left[\cancel{\left(x + 3\right)} \left(x - 1\right)\right]} = - \frac{\left(x + 4\right)}{4 \left(x - 1\right)}$
$\implies$
hole: $x = - 3$

From now on we will work only with our new nicer $f \left(x\right)$:
$f \left(x\right) = - \frac{\left(x + 4\right)}{4 \left(x - 1\right)}$
vertical asymptote: $4 \left(x - 1\right) = 0 \iff x = 1$
horizontal asymptotes:
${\lim}_{x \rightarrow + \infty} f \left(x\right) = - \frac{1}{4}$
${\lim}_{x \rightarrow - \infty} f \left(x\right) = - \frac{1}{4}$
$\implies$
horizontal asymptote: $y = - \frac{1}{4}$

x and y intercepts:
$f \left(0\right) = - \frac{\left(0 + 4\right)}{4 \left(0 - 1\right)} = - \frac{4}{-} 4 = 1$
$f \left(x\right) = 0 \iff - \frac{\left(x + 4\right)}{4 \left(x - 1\right)} = 0 \implies x = - 4$
$\implies$
x and y intercepts: $\left(0 , 1\right) \mathmr{and} \left(- 4 , 0\right)$

graph:
graph{(x^2+7x+12)/(-4x^2-8x+12) [-10, 10, -5, 5]}