# How do you find vertical, horizontal and oblique asymptotes for (-2x^2-6x+3)/(3x+9)?

Apr 27, 2018

The vertical asymptote is x = -3.
The horizontal asymptote is y = $y = \left(- \frac{2}{3}\right) x$

#### Explanation:

Vertical asymptote(s)
To find any vertical asymptotes, factor the numerator and the denominator. Any factor in the denominator containing an x that that does not cancel out when reducing, will give you a vertical asymptote. To find the vertical asymptote(s), set the remaining factor(s) in the denominator after reducing equal to zero and solve for x.

(Note: Any factor that does cancel out gives you a hole in the graph.) In this problem, the numerator will not factor, except to factor a negative out, so the expression will not reduce. We can factor the denominator to $3 \left(x + 3\right)$. Set x + 3 equal to zero and solve for x. So we have x + 3 = 0, and solving for x we get x = -3. So the vertical asymptote is x = -3.

Horizontal or oblique asymptote:
To determine if there is a horizontal asymptote or an oblique asymptote, compare the degree of the numerator to the degree of the denominator.

a)If the degree of the denominator is greater than the degree of the numerator, then you have a horizontal asymptote of y = 0.

b)If the degree of the numerator is equal to the degree of the denominator, then you have a horizontal asymptote. You find the equation of the asymptote by writing
In this problem, the degree of the numerator is larger so we have an oblique asymptote. We need to divide $- 2 {x}^{2} - 6 x + 3$ by
$3 x + 9$
When we divide, we get $\left(- \frac{2}{3}\right) x$ with a remainder of 3. Ignore the remainder and set the quotient equal to y. so the oblique asymptote is .