# How do you find the slant asymptote of y = (3x^2 + 2x - 3 )/( x - 1)?

Nov 14, 2016

The slant asymptote is $y = 3 x + 5$

#### Explanation:

The degree of the numerator $>$ the degree of the denominator, we expect a slant asymptote.

Just do a long division

$\textcolor{w h i t e}{a a a a}$$3 {x}^{2} + 2 x - 3$$\textcolor{w h i t e}{a a a a}$∣$x - 1$

$\textcolor{w h i t e}{a a a a}$$3 {x}^{2} - 3 x$$\textcolor{w h i t e}{a a a a a a a}$∣$3 x + 5$

$\textcolor{w h i t e}{a a a a a a}$$0 + 5 x - 3$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 5 x - 5$

$\textcolor{w h i t e}{a a a a a a a a a a a}$$0 + 2$

$y = \frac{3 {x}^{2} + 2 x - 3}{x - 1} = 3 x + 5 + \frac{2}{x - 1}$

Therefore, the slant asymptote is $y = 3 x + 5$

graph{(y-(3x^2+2x-3)/(x-1))(y-3x-5)=0 [-32.04, 32.92, -9.17, 23.3]}