How do you find vertical, horizontal and oblique asymptotes for # f(x)= (5x-15)/(2x+4)#?

1 Answer
Apr 23, 2016

Answer:

vertical asymptote x = -2
horizontal asymptote #y = 5/2 #

Explanation:

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

solve : 2x + 4 = 0 → x = -2 is the asymptote.

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

divide terms on numerator/denominator by x

#((5x)/x - 15/x)/((2x)/x + 4/x) = (5-15/x)/(2+4/x) #

as # xto+-oo , y to (5-0)/(2+0)#

# rArr y = 5/2 " is the asymptote " #

Oblique asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here hence there are no oblique asymptotes.
graph{(5x-15)/(2x+4) [-14.24, 14.24, -7.11, 7.13]}