Question

What are the vertical and horizontal asymptotes for the function f (x) =2Bx-6%20%7D%7Bx%5E%7B3%7D-1%7D" id="TexFormula1" title="\frac{x^{2}+x-6 }{x^{3}-1}" alt="\frac{x^{2}+x-6 }{x^{3}-1}" align="absmiddle" class="latex-formula">
A. vertical asymptote: x = 1

horizontal asymptote: none

B. vertical asymptote: x = 1

horizontal asymptote: y = 0

C. vertical asymptote: x = –2, x = 3

horizontal asymptote: y = 0

D. vertical asymptote: x = –2, x = –3

horizontal asymptote: none

Answer 1

Answer:

B. vertical asymptote: x = 1

horizontal asymptote: y = 0

Step-by-step explanation:

1) Vertical asymptotes of a function are determined by what input of x makes the denominator equal 0. So, let's set the denominator, $x^3-1$, equal to 0 and solve for x:

$x^3-1= 0\\x^3 = 1\\\sqrt[3]{x^3} = \sqrt[3]{1} \\x = 1$

Thus, the vertical asymptote is x = 1.

2) If the degree of the polynomial in the denominator is greater than the one on the top, the horizontal asymptote is automatically y = 0. Thus, the horizontal asymptote is y = 0.

Answer 2

Answer:

B. x=1, y=0

Step-by-step explanation:

Vertical asymptote: denominator approximate 0

(x²+x-6)/(x³-1)   ---> ±∞           as x ---> 1

x³-1 = 0

x³ = 1

x = ∛1 = 1

Horizontal asymptote: If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is always the x axis, i.e. the line y = 0

degree of the numerator: 2

degree of the denominator: 3

2<3

Horizontal asymptote: y = 0