# How do you graph f(x)=(-3x^2-12x-9)/(x^2+5x+4) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jan 22, 2017

see explanation.

#### Explanation:

First, factorise and simplify f(x)

$f \left(x\right) = \frac{- 3 \cancel{\left(x + 1\right)} \left(x + 3\right)}{\cancel{\left(x + 1\right)} \left(x + 4\right)} = \frac{- 3 \left(x + 3\right)}{x + 4}$

with exclusion x ≠ - 1 which indicates a hole at x = - 1

The graph of $f \left(x\right) = \frac{- 3 \left(x + 3\right)}{x + 4}$ is the same as

$\frac{- 3 {x}^{2} - 12 x - 9}{{x}^{2} + 5 x + 4}$ but without the hole.

$\textcolor{b l u e}{\text{Asymptotes}}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : $x + 4 = 0 \Rightarrow x = - 4 \text{ is the asymptote}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

divide terms on numerator/denominator by x

$f \left(x\right) = \frac{\frac{- 3 x}{x} - \frac{9}{x}}{\frac{x}{x} + \frac{4}{x}} = \frac{- 3 - \frac{9}{x}}{1 + \frac{4}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{- 3 - 0}{1 + 0}$

$\Rightarrow y = - 3 \text{ is the asymptote}$

$\textcolor{b l u e}{\text{Approaches to asymptotes}}$

$\text{horizontal asymptote } y = - 3$

as $x \to + \infty , f \left(x\right) \to y = - 3 \text{ from above}$

as $x \to - \infty , f \left(x\right) \to y = - 3 \text{ from below}$

$\text{vertical asymptote } x = - 4$

${\lim}_{x \to - {4}^{-}} f \left(x\right) \to - \infty$

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

$\textcolor{b l u e}{\text{Intercepts}}$

$x = 0 \to y = - \frac{9}{4} \leftarrow \text{ y-intercept}$

$y = 0 \to x + 3 = 0 \to x = - 3 \leftarrow \text{ x-intercept}$
graph{(-3(x+3))/(x+4) [-10, 10, -5, 5]}