How do you graph #f(x)=3/(x^2(x+5))# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Lucy
Jul 15, 2018

Below

Explanation:

For vertical asymptotes, look at the denominator. #x^2(x+5) !=0# because the graph will be undefined

#x^2(x+5) !=0# means that the vertical asymptotes are #x=0# and #x=-5# when solving for #x#

For the horizontal asymptote, look at the degree of the numerator and denominator

If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is #y=0#.

Or you can think of it as if you put some numbers into your #f(x)#, you will notice that #x^2(x+5)# will be a lot bigger than #3#. Hence, when you divide a small number by a big number, #3/(x^2(x+5)) ->0#

For your x and y intercepts,
sub #y=0# for x intercepts,
#0=3/(x^2(x+5))#
#0=3# which isn't true so no x-intercepts

sub #x=0# for y intercepts which cannot occur since #x=0# is an asymptote

Therefore, there are no intercepts

Below is the graph. You can see that the endpoints of the graphs approaches the asymptotes #y=0#, #x=0# and #x=-5#
graph{3/(x^2(x+5) [-10, 10, -5, 5]}

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