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

Question

Eva Abramson · Accepted Answer

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

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]}

Kennedy Adams · Answer

undefined To find the asymptotes for the function f(x) = x^2/(x+5), we need to consider both vertical and horizontal asymptotes.

Vertical asymptotes occur when the denominator of the function becomes zero. In this case, the denominator is (x+5), so the vertical asymptote is x = -5.

To determine the horizontal asymptote, we compare the degrees of the numerator and denominator. Since the degree of the numerator (2) is less than the degree of the denominator (1), there is no horizontal asymptote.

Therefore, the only asymptote for the function f(x) = x^2/(x+5) is the vertical asymptote x = -5.