Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = (2x^2 + x + 2) /( x + 1)`?

Answer 1

Vertical asymptote at `x=-1`
No horizontal asymptote
Slant asymptote: `f(x)=2x-1`

Explanation:

Given: `f(x)=(2x^2+2x+2)/(x+1)`

`f(x)` is undefined when `(x+1)=0`
giving us the vertical asymptote of `x=-1`

`lim_(xrarroo) f(x) rarr oo`
and
`lim_(xrarr-oo) f(x) rarr -oo`
so there is no horizontal asymptote.

Since the degree of the numerator is greater than the degree of the denominator,
we can divide the denominator into the numerator to get a slant asysmptote:
`f(x)=(2x^2+x+2)div(x+1)=(2x-1) +3/(x+1)`
So the slant asymptote is `f(x)=2x-1`

Answer 2

Slant Asymptote: `y=2x-1`

Vertical Asymptote: `x=-1`

Horizontal Asymptote: None

Explanation:

The given function is `y=(2x^2+x+2)/(x+1)`

To find the slant asymptote, divide numerator by the denominator of the given rational function.
`" " " " " "underline(" "2x-1" " " " " ")`
`x+1""|~" "2x^2+x+2`
`" " " " " "underline(2x^2+2x" " " " " ")`
`" " " " " " " " " "-x+2`
`" " " " " " " " " "underline(-x-1)`
`" " " " " " " " " " " " " " "+3`

The result of the division is

`y=2x-1+3/(x+1)`

The whole number part of the quotient which is `2x-1` becomes the right side part of the linear equation

`color(red)(y=2x-1)`

which is the `color(red)("Slant Asymptote")`

To solve for the Vertical Asymptote, use the divisor and equate to zero

`x+1=0`

and the Vertical Asymptote is

`color(red)(x=-1 " is the Vertical Asymptote")`

There is `color(red)("No Horizontal Asymptote")`

Kindly see the graph of the function `y=(2x^2+x+2)/(x+1)` colored blue, the slant asymptote `y=2x-1` colored red, vertical asymptote `x=-1` colored green below.

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