How do you graph #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)#.

As #x to +-sqrt 5, y to -oo#.

So, #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): (+-2.2,.-1.83) (+-2.1, -.53) (+-2, 0)#

# (+-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]}.