What are the asymptotes of #y=x/(x^2-9)# and how do you graph the function?

1 Answer
Mar 20, 2017

The vertical asymptotes are #x=-3# and #x=3#
The horizontal asymptote is #y=0#
No oblique asymptote

Explanation:

We need

#a^2-b^2=(a+b)(a-b)#

We factorise the denominator

#x^2-9=(x+3)(x-3)#

#y=x/((x+3)(x-3))#

As we cannot divide by #0#, x!=3 and #x!=3#

The vertical asymptotes are #x=-3# and #x=3#

There are no oblique asymptotes as the degree of the numerator is #<# than the degree of the denominator

#lim_(x->-oo)y=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^-#

#lim_(x->+oo)y=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^+#

The horizontal asymptote is #y=0#

We can build a sign chart to have a general view of the graph

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaaaaaa)##0##color(white)(aaaaaaa)##+3##color(white)(aaaaaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaaa)##||##color(white)(aaa)##+#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaaa)##||##color(white)(aaa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaaa)##||##color(white)(aaa)##+#

#color(white)(aaaa)##y##color(white)(aaaaaaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaaa)##||##color(white)(aaa)##+#

The intercepts are #(0,0)#

#lim_(x->-3^-)y=-oo#

#lim_(x->-3^+)y=+oo#

#lim_(x->3^-)y=-oo#

#lim_(x->3^+)y=+oo#

Here is the graph

graph{(y-(x)/(x^2-9))(y)(y-1000(x+3))(y-1000(x-3))=0 [-18.05, 18.02, -9.01, 9.03]}