How do you find vertical, horizontal and oblique asymptotes for #(2x-2) /( 2x+2 )#?

1 Answer
May 8, 2016

Answer:

vertical asymptote x = -1
horizontal asymptote y = 1

Explanation:

Begin by factorising numerator/denominator

#rArr( cancel(2) (x-1))/(cancel(2) (x+1))=(x-1)/(x+1)#

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve: x + 1 = 0 → x = -1 is the asymptote

Horizontal asymptotes occur as # lim_(x to +- oo) , f(x) to 0 #

divide terms on numerator/denominator by x

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

as #x to +- oo , f(x) to (1-0)/(1+0)#

#rArry=1" is the asymptote "#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) hence there are no oblique asymptotes.
graph{(x-1)/(x+1) [-10, 10, -5, 5]}