Home > Precalculus > Asymptotes

How do you find the asymptotes for #{-9 x^3 - 9 x^2 + 28 x + 9 }/{3 x^2 + 6 x + 2}#?

Answer 1

Sadie Alexander

Oblique asymptotes y= -3x +3

Vertical asymptotes #x=(-3+-sqrt 3)/3#

On division the given expression appears like this

(-3x +3)+ # (16x+3)/(3x^2 +6x +2)#

Oblique asymptote is given by y=-3x+3

Vertical asymptotes are given by #3x^2 +6x+2=0#, that is #x= (-6+- sqrt(36-12))/6#

#x=(-3+-sqrt 3)/3#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Chloe Alexander

To find the asymptotes of the rational function (\frac{-9x^3 - 9x^2 + 28x + 9}{3x^2 + 6x + 2}), you need to perform polynomial division to determine if there are any horizontal or slant asymptotes. After polynomial division, if the degree of the numerator is less than the degree of the denominator, there will be a horizontal asymptote. If the degrees are the same, there will be a slant asymptote.

Performing polynomial division, you get:

[\frac{-9x^3 - 9x^2 + 28x + 9}{3x^2 + 6x + 2} = -3x - 3 + \frac{5x + 3}{3x^2 + 6x + 2}]

Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at (y = -3x - 3).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.