# How do you graph f(x)=(3x^2+2)/(x+1) using holes, vertical and horizontal asymptotes, x and y intercepts?

Feb 18, 2018

V.A $x = - 1$
H.A none
S.A $y = 3 x - 3$
no $x$-intercept
$y$-intercept is $\left(0 , 2\right)$
no holes

#### Explanation:

• V.A are the zeros (undefined points) of the denominator
$x + 1 = 0$
$x = - 1$
vertical asymptotes are $x = - 1$

• H.A if the degrees of the numerator is equal to the degree of denominator,
if the numerators degree > 1 + degree of denominator, there is a slant asymptote
degree of numerator is 2, degree of denominator is 1
$y = m x + b$
$\frac{3 {x}^{2} + 2}{x + 1}$
$= 3 x + \frac{- 3 x + 2}{x + 1}$
$= 3 x - 3 + \frac{5}{x + 1}$
$y = 3 x - 3$
slant asymptote is $y = 3 x - 3$

• $x$-intercept is a point on the graph where $y = 0$
$\frac{3 {x}^{2} + 2}{x + 1} = 0$ no solution for $x \in \mathbb{R}$
no $x$-intercept

• $y$-intercept is the point on the graph where $x = 0$
$y = \frac{3 \cdot {0}^{2} + 2}{0 + 1}$
$y = 2$
$y$-intercept is $\left(0 , 2\right)$

• no holes because the denominator doesn't cancel out