How do you find vertical, horizontal and oblique asymptotes for #( 4x^5)/(x^3-1)#?

Question

Benjamin Achtenberg · Accepted Answer

This function has a vertical asymptote #x=1# and no horizontal or oblique asymptotes.

Given: #f(x) = (4x^5)/(x^3-1)# Take note of this: #(4x^5)/(x^3-1)=(4x^5-4x^2+4x^2)/(x^3-1)# #color(white)((4x^5)/(x^3-1))=(4x^2(x^3-1)+4x^2)/(x^3-1)# #color(white)((4x^5)/(x^3-1))=4x^2+(4x^2)/(x^3-1)# #color(white)((4x^5)/(x^3-1))=4x^2+(4x^2)/((x-1)(x^2+x+1))# The only real zero of the denominator is #x=1#, which is not a zero of the numerator. Hence, #f(x)# has a vertical asymptote at #x=1# Additionally, take note of this: #lim_(x->+-oo) (4x^2)/(x^3-1) = lim_(x->+-oo) (4/x)/(1-1/x^3) = 0/(1-0) = 0# Hence #f(x)# is asymptotic to the parabola #4x^2# and has no horizontal or oblique asymptotes. graph{ [-5, 5, -35, 75]} / (x^3-1)

Sadie Ackerman · Answer

undefined To find the vertical asymptotes of a rational function, set the denominator equal to zero and solve for $x$. Vertical asymptotes occur where the denominator is zero, but the numerator is not zero.

For the given function $\frac{4x^5}{x^3 - 1}$, set the denominator equal to zero and solve for $x$:

$$x^3 - 1 = 0$$

This equation can be factored using the difference of cubes formula:

$$x^3 - 1 = (x - 1)(x^2 + x + 1)$$

Setting each factor equal to zero:

$$x - 1 = 0 \implies x = 1$$
$$x^2 + x + 1 = 0$$

Since the quadratic equation $x^2 + x + 1 = 0$ has no real roots, there is only one vertical asymptote at $x = 1$.

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

For the given function, the degree of the numerator is $5$ and the degree of the denominator is $3$. Since the degree of the numerator is greater, there is no horizontal asymptote.

To find oblique asymptotes (also known as slant asymptotes), divide the numerator by the denominator using polynomial long division or synthetic division. If the resulting quotient has a finite limit as $x$ approaches positive or negative infinity, then the oblique asymptote is the equation of the quotient polynomial.

Performing polynomial long division or synthetic division, the quotient is a degree $2$ polynomial, which means it doesn't have a finite limit as $x$ approaches positive or negative infinity. Therefore, there is no oblique asymptote for the given function.