# How do you graph y=(x^2-4)/(x^2+3) using asymptotes, intercepts, end behavior?

Oct 26, 2016

Only a horizontal asymptote $y = 1$
Intercepts are $0 , - \frac{4}{3}$ , (2,0) and (-2,0)
minimum at $\left(0 , - \frac{4}{3}\right)$
And limit $x \to \pm \infty$ is $y \to 1$

#### Explanation:

We start by finding the Domain
Here it is $\mathbb{R}$ as the denominator is always positive
${x}^{2} + 3 > 0$
So there is no horizontal asymptote
The intercepts are $\left(0 , - \frac{4}{3}\right)$ on the y axis
and $\left(2 , 0\right) \mathmr{and} \left(- 2 , 0\right)$ on the x axis
The we look for symmetry, $f \left(- x\right) = f \left(x\right)$
So thre is symmetry about the y axis
limit $y = 1$
$x \to - \infty$

limit $y = 1$
$x \to \infty$

So $y = 1$ is a horizontal asymptote
We can calculate the first derivative $y ' = \left(2 x \left({x}^{2} + 3\right) - 2 x \left({x}^{2} - 4\right)\right)$
$= \frac{2 {x}^{3} + 6 x - 2 {x}^{3} + 8 x}{{x}^{2} + 3} = \frac{14 x}{{x}^{2} + 3} ^ 2$
So there is a minimum at $\left(0 , - \frac{4}{3}\right)$

graph{(x^2-4)/(x^2+3) [-8.89, 8.89, -4.444, 4.445]}