# How do you graph f(x)=(x+3)/((x+1)(x-3)) using holes, vertical and horizontal asymptotes, x and y intercepts?

Mar 19, 2017

see explanation.

#### Explanation:

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

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve " (x+1)(x-3)=0rArrx=-1" and } x = 3$

$\Rightarrow x = - 1 \text{ and " x=3" are the asymptotes}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

f(x)=(x/x^2+3/x^2)/(x^2/x^2-(2x)/x^2-3/x^2)=(1/x+3/x^2)/(1-2/x-3/x^2

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here hence there are no holes.

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

$x = 0 \to y = \frac{3}{- 3} = - 1 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to x + 3 = 0 \to x = - 3 \leftarrow \textcolor{red}{\text{ x-intercept}}$
graph{(x+3)/(x^2-2x-3) [-10, 10, -5, 5]}