# How do you graph f(x)=x^2/(x+5)?

Jun 7, 2015

$f \left(x\right) = {x}^{2} / \left(x + 5\right) = \frac{\left({x}^{2} + 5 x\right) - \left(5 x + 25\right) + 25}{x + 5}$

$= \frac{\left(x - 5\right) \left(x + 5\right) + 25}{x + 5}$

$= x - 5 + \frac{25}{x + 5}$

For large positive or negative $x$ this will be asymptotic to $x - 5$

$f \left(x\right)$ has a simple pole at $x = - 5$ with $f \left(- 5 - \epsilon\right)$ being large and negative, and $f \left(- 5 + \epsilon\right)$ is large and positive for small $\epsilon > 0$

$f \left(0\right) = 0$ so the curve passes through $\left(0 , 0\right)$

$f ' \left(x\right) = \frac{2 x}{x + 5} - {x}^{2} / {\left(x + 5\right)}^{2}$

$= \frac{\left(2 x\right) \left(x + 5\right) - {x}^{2}}{x + 5} ^ 2$

$= \frac{{x}^{2} + 10 x}{x + 5} ^ 2$

$= \frac{x \left(x + 10\right)}{x + 5} ^ 2$

So $f ' \left(x\right) = 0$ when $x = 0$ and $x = - 10$

So there is a local minimum at $\left(0 , 0\right)$ and a local maximum at $\left(- 10 , f \left(- 10\right)\right) = \left(- 10 , - 20\right)$

graph{x^2/(x+5) [-86.4, 73.6, -45.1, 34.9]}