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

Jul 3, 2017

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

#### Explanation:

$\frac{3 {x}^{2} + 2 x - 1}{x + 1}$, Vertical asymptote is $x + 1 = 0 \mathmr{and} x = - 1$ .But

$\frac{3 {x}^{2} + 2 x - 1}{x + 1} = \frac{\left(3 x - 1\right) \cancel{\left(x + 1\right)}}{\cancel{\left(x + 1\right)}}$, 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 $y = 3 x - 1$

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

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

Jul 3, 2017

$y = 3 x - 1$

#### Explanation:

$\text{factorising the numerator}$

$\frac{\left(3 x - 1\right) \cancel{\left(x + 1\right)}}{\cancel{\left(x + 1\right)}} = 3 x - 1$

$\text{the removal of the factor } \left(x + 1\right)$

$\text{from the numerator/denominator indicates a hole at x = - 1}$

$\text{the graph of " (3x^2+2x-1)/(x+1)" simplifies to the line}$

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