How do you find the asymptotes for #(2x-4)/(x^2-4)#?

Question

Lincoln Anderson · Accepted Answer

vertical asymptote x = -2
horizontal asymptote y = 0

The first step is to factorise and simplify the function #rArr(2 cancel((x-2)))/(cancel((x-2)) (x+2))=2/(x+2 # 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 + 2 = 0 → x = -2 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo) , f(x) to 0# divide terms on numerator/denominator by x #rArr (2/x)/(x/x+2/x)=(2/x)/(1+2/x)# as #xto+-oo, y to 0/(1+0)# #rArry=0" is the asymptote"# graph{2/(x+2) [-10, 10, -5, 5]}

Greyson Brown · Answer

undefined To find the asymptotes for the function $ \frac{2x - 4}{x^2 - 4} $, you first factor the denominator. $ x^2 - 4 $ factors to $ (x + 2)(x - 2) $. The asymptotes will occur where the denominator equals zero, but the numerator doesn't. Therefore, you have vertical asymptotes at $ x = -2 $ and $ x = 2 $. There are no horizontal asymptotes in this case because the degree of the numerator is less than the degree of the denominator. However, there may be an oblique (slant) asymptote if the degree of the numerator is exactly one less than the degree of the denominator, but in this case, the numerator's degree is lower.