How do you graph $y = ln (5 - x^{2})$ $y= ln(5 -x^2)$ ?

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Answer 1 · 2016-06-30T16:44:43

See the explanation and the graph.

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

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

So, $x = \pm \sqrt{5}$ $x=+-sqrt 5$ 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;

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

$(\pm 1, 1.39) (0, 1.61)$ $(+-1, 1.39) (0,1.61)$ .

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]}.

How do you graph y= ln(5 -x^2)y=ln(5−x2)y= ln(5 -x^2)?