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

Question

Christian Antunez · Accepted Answer

The vertical asymptote is x = -3.
The horizontal asymptote is y = # y = (-2/3)x# 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(x+3)#. 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 y = (lead coefficient of numerator) / (lead coefficient of denominator) c)If the degree of the numerator is greater than the degree of the denominator, then you have an oblique asymptote. You find the equation of the asymptote by dividing the denominator into the numerator either by long division. In this problem, the degree of the numerator is larger so we have an oblique asymptote. We need to divide #-2x^2-6x+3# by #3x+9# When we divide, we get #(-2/3)x # with a remainder of 3. Ignore the remainder and set the quotient equal to y. so the oblique asymptote is .

Serenity Jones · Answer

undefined To find the vertical, horizontal, and oblique asymptotes of the function $ \frac{-2x^2-6x+3}{3x+9} $:

1. **Vertical Asymptotes:** Vertical asymptotes occur where the denominator of the fraction equals zero and the numerator does not. To find them, set the denominator $ 3x + 9 $ equal to zero and solve for $ x $.

$$ 3x + 9 = 0 $$
$$ x = -3 $$

So, there is a vertical asymptote at $ x = -3 $.

2. **Horizontal Asymptote:** To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, the degree of the numerator is 2 and the degree of the denominator is 1. Therefore, there is no horizontal asymptote.

3. **Oblique Asymptote (Slant Asymptote):** If the degree of the numerator is exactly one more than the degree of the denominator, then there is an oblique asymptote. To find it, perform polynomial long division or use synthetic division to divide the numerator by the denominator. The quotient will be the equation of the oblique asymptote.

Performing polynomial long division or synthetic division:

$$ \frac{-2x^2-6x+3}{3x+9} = -\frac{2}{3}x - 1 $$

So, the equation of the oblique asymptote is $ y = -\frac{2}{3}x - 1 $.