Question

How do you identify the oblique asymptote of `f(x) = (2x^2+3x+8)/(x+3)`?

Answer 1

Oblique Asymptote is `y=2x-3`
Vertical Asymptote is `x=-3`

Explanation:

from the given:

`f(x)=(2x^2+3x+8)/(x+3)`

perform long division so that the result is

`(2x^2+3x+8)/(x+3)=2x-3+17/(x+3)`

Notice the part of the quotient

`2x-3`

equate this to `y` like as follows

`y=2x-3` this is the line which is the Oblique Asymptote

And the divisor `x+3` be equated to zero and that is the Vertical asymptote

`x+3=0` or `x=-3`

You can see the lines `x=-3` and `y=2x-3` and the graph of
`f(x)=(2x^2+3x+8)/(x+3)`
graph{(y-(2x^2+3x+8)/(x+3))(y-2x+3)=0[-60,60,-30,30]}

God bless...I hope the explanation is useful..