How do you find the asymptotes for #(x^2 + x + 3 )/( x-1)#?

1 Answer
Jun 21, 2016

Answer:

Vertical asymptote #x = 1#

Oblique asymptote #y = x+2#

Explanation:

#f(x) = (x^2+x+3)/(x-1)#

#= (x^2-x+2x-2+5)/(x-1)#

#= ((x+2)(x-1)+5)/(x-1)#

#= x+2+5/(x-1)#

As #x->+-oo# the term #5/(x-1)->0#

So there is an oblique (slant) asymptote #y = x+2#

When #x=1# the denominator of #f(x)# becomes #0# while the numerator is non-zero. So #f(x)# is undefined at #x=1# and has a vertical asymptote there.

graph{(y - (x^2+x+3)/(x-1))(y - (x+2)) = 0 [-39, 41, -17.04, 22.96]}