Question

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

Answer 1

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]}