How do you graph and find the discontinuities of $(x^2-25)/(x^2+5x)$ ?

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Answer 1 · 2015-08-24T12:32:39

We first factorise.

$=((x+5)(x-5))/(x(x+5))$

The disconituities happen when the denominator $=0$
So at $x=0andx=-5$
The discontinuity at $x=5$ is removable as both limits go to the same value:

$lim_(x->1^-) f(x)=lim_(x->1^+) f(x)=2$

In other cases we can cancel out the $(x+5)$ 's:

$=(x-5)/x=x/x-5/x=1-5/x$

This means that if $x$ gets larger $+or-$ the function goes to $1$ (from up or down)

graph{1-5/x [-16.02, 16, -8, 8.02]}

How do you graph and find the discontinuities of (x^2-25)/(x^2+5x)(x^2-25)/(x^2+5x)?