How do you find the asymptotes for #f(x)=x^2/(x+5)#?

1 Answer
Aug 2, 2017

Answer:

The vertical asymptote is #x=-5#
The slant asymptote is #y=x-5#
There is no horizontal asymptote

Explanation:

The vertical asymptotes are calculated by performing the limits

#lim_(x->-5^(-))f(x)=lim_(x->-5^(-))x^2/(x+5)= 25/(0^-) = -oo#

#lim_(x->-5^(+))f(x)=lim_(x->-5^(+))x^2/(x+5)= 25/(0^+) = +oo#

The vertical asymptote is #x=-5#

We perform a long division to calculate the slant asymptote

#color(white)(aaaa)##x+5##color(white)(aaaa)|##x^2+0x+0##|##color(white)(aaaa)##x-5#

#color(white)(aaaaaaaaaaaaa)##x^2+5x#

#color(white)(aaaaaaaaaaaaaa)##0-5x#

#color(white)(aaaaaaaaaaaaaaaa)##-5x-25#

#color(white)(aaaaaaaaaaaaaaaaaaa)##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 slant asymptote is #y=x-5#

To determine the horizontal asymptote, we calculate

#lim_(x->-oo)f(x)=lim_(x->-oo)x^2/(x+5)=-oo#

#lim_(x->+oo)f(x)=lim_(x->+oo)x^2/(x+5)=+oo#

There is no horizontal asymptote

graph{(y-x^2/(x+5))(y-x+5)=0 [-58.5, 58.53, -29.27, 29.28]}