How do you find vertical, horizontal and oblique asymptotes for #f(x)=( 21 x^2 ) / ( 3 x + 7)#?

1 Answer
Dec 6, 2016

Answer:

The vertical asymptote is #x=-7/3#
The oblique asymptote is #y=7x-49/3#
No horizontal asymptote.

Explanation:

The domain of #f(x)# is #D_f(x)=RR-{-7/3} #

The denominator must be #!=0#, #=>#, #x!=-7/3#

So, the vertical asymptote is #x=-7/3#

As the degree of the numerator is #># then the degree of the denominator, we expect an oblique asymptote.

Let's do a long division

#color(white)(aaaa)##21x^2##color(white)(aaaaaaaaaaa)##∣##3x+7#

#color(white)(aaaa)##21x^2+49x##color(white)(aaaaaa)##∣##7x-49/3#

#color(white)(aaaaaaa)##0-49x#

#color(white)(aaaaaaaaa)##-49x-343/3#

#color(white)(aaaaaaaaaaaaa)##0+343/3#

So,

#(21x^2)/(3x+7)=7x-49/3+(343/3)/(3x+7)#

The oblique asymptote is #y=7x-49/3#

To calculate the limits of #x->+-oo#, you take the terms of highest degree in the numerator and the denominator

#lim_(x->+-oo)f(x)=lim_(x->+-oo)(21x^2)/(3x)=lim_(x->+-oo)7x=+-oo#

So, there is no horizontal asymptote.

graph{(y-(21x^2)/(3x+7))(y-7x+49/3)=0 [-139.1, 127.9, -106, 27.5]}