How do you find all the asymptotes for #(x^2 + x + 3) /( x-1)#?
1 Answer
Oct 6, 2015
Simplify:
#(x^2+x+3)/(x-1) = x + 2 + 5/(x-1)#
to find the oblique asymptote
Explanation:
Let
#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
So
graph{(y - (x^2+x+3)/(x-1))(y - x - 2)(0.999x - 1) = 0 [-21.19, 24.42, -9.2, 13.63]}