How do you find vertical, horizontal and oblique asymptotes for #(12x^3-x) /( 6x^2+4)#?

Question

Serenity Johnson · Accepted Answer

No Horizontal, or vertical [Asymptotes]
oblique asymptote of y= 2x

No HA, because numerator degree is higher than denominator degree, No vertical asymptotes, because the denominator does not factor,

Elias Ackerman · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for $ x $. Any values of $ x $ that make the denominator zero will result in vertical asymptotes.

To find horizontal asymptotes, 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 $ y = 0 $. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients of both polynomials to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find oblique asymptotes, perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient obtained will be the equation of the oblique asymptote.

So, for $ f(x) = \frac{12x^3 - x}{6x^2 + 4} $:

1. Vertical Asymptotes: Set the denominator $ 6x^2 + 4 $ equal to zero and solve for $ x $.
$$ 6x^2 + 4 = 0 $$
$$ x^2 = -\frac{4}{6} $$
$$ x^2 = -\frac{2}{3} $$
Since $ x^2 $ cannot be negative, there are no real solutions. Therefore, there are no vertical asymptotes.

2. Horizontal Asymptotes: Compare the degrees of the numerator and denominator.
The degree of the numerator is 3, and the degree of the denominator is 2.
Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

3. Oblique Asymptotes: Perform polynomial long division or synthetic division to divide $ 12x^3 - x $ by $ 6x^2 + 4 $.
$$ \frac{12x^3 - x}{6x^2 + 4} = 2x - \frac{8x - 1}{6x^2 + 4} $$
The equation $ y = 2x $ represents the oblique asymptote.