How do you find the vertical, horizontal and slant asymptotes of: # f(x)=(2x - 2 )/( (x-1)(x^2 + x - 1))#?

1 Answer
Oct 31, 2016

Answer:

The vertical asymptotes are #x=(-1-sqrt5)/2#
and #x=(-1+sqrt5)/2#
The horizontal asymptote is #y=0#
There is no slant symptote

Explanation:

You can start by simplifying #f(x)#
#f(x)=(2x-2)/((x-1)(x^2+x-1))=(2(cancel(x-1)))/(cancel(x-1)(x^2+x-1))#
So #f(x)=2/(x^2+x-1)#
As we cannot divide by #0#, so #x^2+x-1!=0#
The roots of this equation is #(-b+-sqrt(b^2-4ac))/(2a)#
So the roots are #(-1+-sqrt(1+5))/2=(-1+-sqrt5)/2#
So the vertical asymptotes are #x=(-1-sqrt5)/2#
and #x=(-1+sqrt5)/2#

And the limit of #f(x)# as #x->oo# is #0#
so the horizontal asymptote is #y=0#
graph{2/(x^2+x-1) [-7.474, 8.33, -4.27, 3.63]}