How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (x-4)/(x^2-8x+16)#?

1 Answer
Nov 2, 2016

Answer:

#x=4# is a Vertical Asymptote

#y=0# is a Horizontal Asymptote

There is no Slant Asymptote.

Explanation:

This function is defined for all # x in RR# except #x=4#

#f(x)=(x-4)/(x^2-2(4)x-4^2)#

#f(x)=(x-4)/(x-4)^2#

Domain of #f# is:#(-oo,4)U(4,+oo)#

The vertical Asymptote is computed by setting the denominator to zero

#x-4=0rArrcolor(blue)(x=4)# is a #color(blue)( V.A )#

Let us start computing the asymptote:

The degree of numerator is greater than that of the denominator

Therefore ,#color(blue)(y=0)# is a #color(blue)(H.A)#

There is no Slant asymptote.