Question

What are the vertical and horizontal asymptotes for the function f(x) = (x²(x - 6))/(x³ - 1)?

Answer 1

horizontal y=1 , vertical x=1

Step-by-step explanation:

`f(x)=(x^2(x-6))/(x^3-1)=(x^2(x-6))/((x-1)(x^2+x+1))\\\\horizontal:\quad\lim_(x\rightarrow\infty)(x^2(x-6))/((x-1)(x^2+x+1))=\lim_(x\rightarrow\infty)(x^3-6x)/(x^3-1)=\lim_(x\rightarrow\infty)(1-6/x^2)/(1-1/x^3)=1\\vertical:\quad \lim_(x\rightarrow 1)(x^2(x-6))/((x-1)(x^2+x+1))\rightarrow\infty`

Answer 2

The function f(x) = (x²(x - 6))/(x³ - 1) has no vertical asymptotes, as the factor causing the denominator to be zero is cancelled out by the numerator. The function has a horizontal asymptote at y = 1.

Step-by-step explanation:

The function given is f(x) = (x²(x - 6))/(x³ - 1). To find the vertical asymptotes of this function, you need to determine the values of x for which the denominator is zero since these are the points where the function is undefined. Factor the denominator as x³ - 1 = (x - 1)(x² + x + 1). Notice that x = 1 makes the denominator zero, but since it also cancels out the (x - 1) factor in the numerator, there is no vertical asymptote at x = 1. The other factors do not produce real zeros, thus there are no vertical asymptotes for this function.

For the horizontal asymptote, the degrees of the polynomials in the numerator and denominator indicate the behavior of the function at infinity. Since both the numerator and denominator are cubic polynomials, the horizontal asymptote can be found by dividing the leading coefficients, which are both 1. Therefore, the horizontal asymptote is at y = 1.