How do you find vertical, horizontal and oblique asymptotes for #(x^3+5x^2)/(x^2-25)#?

Question

2036 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-07T07:47:04

The vertical asymptote is #x=5#
The oblique asymptote is #y=x+5#
No horizontal asymptote

The denominator is

#x^2-25=(x+5)(x-5)#

and the numerator is

#x^3+5x^2=x^2(x+5)#

Therefore,

#(x^3+5x^2)/(x^2-25)=(x^2cancel(x+5))/(cancel(x+5)(x-5))#

#=x^2/(x-5)#

Let #f(x)=x^2/(x-5)#

As we cannot divide by #0#, #x!=5#

The vertical asymptote is #x=5#

The degree of the numerator is #># than the degree of the denominator, there is an oblique asymptote.

Let 's do a long division

#color(white)(aaaa)##x^2##color(white)(aaaaaaaaaaaa)##|##x-5#

#color(white)(aaaa)##x^2-5x##color(white)(aaaaaaaa)##|##x+5#

#color(white)(aaaaa)##0+5x#

#color(white)(aaaaaaa)##+0+25#

Therefore,

#f(x)=x+5+25/(x-5)#

#lim_(x->+oo)(f(x)-(x+5))=lim_(x->+oo)25/(x-5)=0^+#

#lim_(x->-oo)(f(x)-(x+5))=lim_(x->-oo)25/(x-5)=0^-#

The oblique asymptote is #y=x+5#

graph{(y-(x^2)/(x-5))(y-x-5)(y+50x-250)=0 [-65.86, 65.94, -32.9, 32.9]}