How do you find vertical, horizontal and oblique asymptotes for #f(x) =( 10x^2)/(x-4)#?

1 Answer
Jun 8, 2018

Answer:

Vertical asymptote at #x = 4#, no horizontal asymptote and
slant asymptote is
#y= 10x +40#

Explanation:

#f(x)= (10 x^2)/(x-4)#

The vertical asymptotes will occur at those values of #x# for which

the denominator is equal to zero. #x-4=0 or x=4#

Thus, the graph will have a vertical asymptote at #x = 4#

The degree of the numerator is #2# and the degree of the

denominator is #1#. Since the larger degree occurs in the

numerator, the graph will have no horizontal asymptote.

Since the degree of the numerator is greater by margin of #1#

we have a slant asymptote , found by quotient on long division.

#f(x)= (10 x^2)/(x-4)= (10 x +40) + 160/(x-4) #

Slant asymptote is #y= 10x +40#. [Ans]