# How do you graph y= ln(5 -x^2)?

##### 1 Answer
Jun 30, 2016

See the explanation and the graph.

#### Explanation:

For real y, $x \in \left(- \sqrt{5} , \sqrt{5}\right)$.

As $x \to \pm \sqrt{5} , y \to - \infty$.

So, $x = \pm \sqrt{5}$ represent vertical asymptotes

#y'=(-2x/(5-x^2) = 0, at x=0. There is no minimum.

Max y = ln 5, at x =0.

Some data for making the graph;

$\left(x , y\right) : \left(\pm 2.2 , . - 1.83\right) \left(\pm 2.1 , - .53\right) \left(\pm 2 , 0\right)$

$\left(\pm 1 , 1.39\right) \left(0 , 1.61\right)$ .

The graph, in its entirety, will look like a collar, with infinite arms

graph{y-ln (5-x^2)=0[-5 5 -10 1.61]}.