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]}