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

##### 2 Answers

**Straight line ( slant asymptote)**,

#### Explanation:

disconinuity is removable.

The degree of numerator is

By long division we have slant asymptote as

The graph will show a straight line (slant asymptote)

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

#### 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]}