# How do you find vertical, horizontal and oblique asymptotes for [(3x^2) + 14x + 4] / [x+2]?

Dec 27, 2016

Vertical: $\uparrow x = - 2 \downarrow$
Oblique: $y = 3 x + 8$, in both directions. See the graphs.

#### Explanation:

$\left(y \left(x + 2\right) - 3 {x}^{2}\right) - 14 x - 4 = 0$

Reorganizing to the form (ax+bx+c((a'x+b/y+c')=k,

$\left(x + 2\right) \left(y - 3 x - 8\right) = - 12$

This represents a hyperbola with asymptotes

$\left(x + 2\right) \left(y - 3 x - 8\right)$=0

The first graph is asymptotes inclusive and the second is for the

hyperbola, sans asynptotes.

Note: The second degree equation

$a {x}^{2} + 2 h x y + b {y}^{2} + \ldots = 0$ represents a hyperbola, if $a b - {h}^{2} < 0$. It

is so for the given equation.

graph{(y-3x-8)(x+2)((y-3x-8)(x+2)+14)=02 [-80, 80, -40, 40]}

graph{(y-3x-8)(x+2)+14=0 [-80, 80, -40, 40]}