Question

How do you find vertical, horizontal and oblique asymptotes for `(3x^2 + 2x - 1 )/ (x + 1)`?

Answer 1

Straight line ( slant asymptote), `y=3x-1`

Explanation:

`(3x^2+2x-1)/(x+1)`, Vertical asymptote is `x+1= 0 or x= -1` .But

`(3x^2+2x-1)/(x+1) =( (3x-1)cancel( (x+1)))/cancel((x+1))`, this

disconinuity is removable.

The degree of numerator is `1` more than dgree of denominator , so horizontal asymptote is absent.

By long division we have slant asymptote as `y=3x-1`

The graph will show a straight line (slant asymptote) `y=3x-1` only

graph{(3x^2+2x-1)/(x+1) [-10, 10, -5, 5]} [Ans]

Answer 2

`y=3x-1`

Explanation:

`"factorising the numerator"`

`((3x-1)cancel((x+1)))/cancel((x+1))=3x-1`

`"the removal of the factor " (x+1)`

`"from the numerator/denominator indicates a hole at x = - 1"`

`"the graph of " (3x^2+2x-1)/(x+1)" simplifies to the line"`

`y=3x-1 " which has no hole"`
graph{(3x^2+2x-1)/(x+1) [-10, 10, -5, 5]}