find vertical and horizontal Asymptotes

Question

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

Answer 1

First you would factor the numerator and denominator

(x+2)(x-1)

x(x-2)(x-1)

The vertical asymptotes come from setting the bottom pieces each equal to zero

x=0 VERTICAL ASYMPTOTE

x-2=0 adding 2 to both sides

you'd get x=2 VERTICAL ASYMPTOTE

x-1=0 add 1 to both sides

x=1 VERTICAL ASYMPTOTE

To get the HORIZONTAL ASYMPTOTE we look at the end behavior of the numerator & denominator

in this case that would be

x²/x³

As the function approaches infinity, the bottom gets much larger than the top, so the function approaches

a HORIZONTAL ASYMPTOTE of y=0