How do you find the asymptotes for # y=(x+3)/(2x-8)#?

1 Answer
Dec 23, 2015

Answer:

Explanation is given below

Explanation:

#y=(x+3)/(2x-8)#

For this problem, vertical asymptotes are found by equating the denominator to zero and solving for x.

# 2x-8 =0#
#2x=8#
#x=8/2#
#x=4# Equation of vertical asymptote.

For horizontal asymptote, check out the degree of numerator and denominator. If the degree of the numerator is same as the degree of denominator.

The Horizontal asymptote is got by dividing the lead coefficients of numerator and denominator.

For example if #y=(ax+b)/(cx+d)#

Vertical asymptote is got by solving for #x# from #cx+d=0#
The Horizontal asymptote is got by #y=a/c# as both numerator and denominator are of degree 1 and their lead coefficient are #a# and #c# respectively.

For our problem #y=(x+3)/(2x-8)#

The horizontal asymptote would be #y=1/2#

Note:
If the degree of the numerator is greater than the degree of denominator then there is no Horizontal Asymptote.

If the degree of the denominator is greater than the degree of numerator then #y=0# is the Horizontal Asymptote.