Answer :
To determine the vertical and horizontal asymptotes of the function [tex]\( f(x) = \frac{x^2 + x - 6}{x^3 - 1} \)[/tex], we need to follow a methodical approach:
### Step 1: Finding the Vertical Asymptotes
Vertical asymptotes occur where the denominator equals zero, causing the function to be undefined or to approach infinity. To find them, we need to solve the equation [tex]\( x^3 - 1 = 0 \)[/tex].
1. Solve [tex]\( x^3 - 1 = 0 \)[/tex]:
[tex]\[ x^3 - 1 = 0 \implies x^3 = 1 \][/tex]
We solve for [tex]\( x \)[/tex] by finding the cube roots of 1. These roots are:
[tex]\[ x = 1, \quad x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i, \quad x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
So, the vertical asymptotes are at [tex]\( x = 1 \)[/tex], and the complex solutions [tex]\( x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \)[/tex] and [tex]\( x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \)[/tex].
Therefore, the only real vertical asymptote is:
[tex]\[ \boxed{x = 1} \][/tex]
### Step 2: Finding the Horizontal Asymptote
Horizontal asymptotes describe the behavior of the function as [tex]\( x \)[/tex] approaches infinity or negative infinity. These are determined by comparing the degrees of the numerator and the denominator.
1. Identify the degrees of the numerator and the denominator:
- The degree of the numerator [tex]\( x^2 + x - 6 \)[/tex] is 2.
- The degree of the denominator [tex]\( x^3 - 1 \)[/tex] is 3.
2. Compare the degrees:
- If the degree of the numerator is less than the degree of the denominator, as [tex]\( x \)[/tex] approaches [tex]\(\pm \infty\)[/tex], the function value approaches 0.
Since the degree of the numerator (2) is less than the degree of the denominator (3), the horizontal asymptote is:
[tex]\[ \boxed{y = 0} \][/tex]
### Conclusion
The function [tex]\( f(x) = \frac{x^2 + x - 6}{x^3 - 1} \)[/tex] has a vertical asymptote at [tex]\( x = 1 \)[/tex] and a horizontal asymptote at [tex]\( y = 0 \)[/tex]. Hence, the correct answer is:
Vertical asymptote: [tex]\( x = 1 \)[/tex] \\
Horizontal asymptote: [tex]\( y = 0 \)[/tex]
[tex]\[ \boxed{\text{vertical asymptote: } x=1 \text{ horizontal asymptote: } y=0} \][/tex]
### Step 1: Finding the Vertical Asymptotes
Vertical asymptotes occur where the denominator equals zero, causing the function to be undefined or to approach infinity. To find them, we need to solve the equation [tex]\( x^3 - 1 = 0 \)[/tex].
1. Solve [tex]\( x^3 - 1 = 0 \)[/tex]:
[tex]\[ x^3 - 1 = 0 \implies x^3 = 1 \][/tex]
We solve for [tex]\( x \)[/tex] by finding the cube roots of 1. These roots are:
[tex]\[ x = 1, \quad x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i, \quad x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \][/tex]
So, the vertical asymptotes are at [tex]\( x = 1 \)[/tex], and the complex solutions [tex]\( x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \)[/tex] and [tex]\( x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \)[/tex].
Therefore, the only real vertical asymptote is:
[tex]\[ \boxed{x = 1} \][/tex]
### Step 2: Finding the Horizontal Asymptote
Horizontal asymptotes describe the behavior of the function as [tex]\( x \)[/tex] approaches infinity or negative infinity. These are determined by comparing the degrees of the numerator and the denominator.
1. Identify the degrees of the numerator and the denominator:
- The degree of the numerator [tex]\( x^2 + x - 6 \)[/tex] is 2.
- The degree of the denominator [tex]\( x^3 - 1 \)[/tex] is 3.
2. Compare the degrees:
- If the degree of the numerator is less than the degree of the denominator, as [tex]\( x \)[/tex] approaches [tex]\(\pm \infty\)[/tex], the function value approaches 0.
Since the degree of the numerator (2) is less than the degree of the denominator (3), the horizontal asymptote is:
[tex]\[ \boxed{y = 0} \][/tex]
### Conclusion
The function [tex]\( f(x) = \frac{x^2 + x - 6}{x^3 - 1} \)[/tex] has a vertical asymptote at [tex]\( x = 1 \)[/tex] and a horizontal asymptote at [tex]\( y = 0 \)[/tex]. Hence, the correct answer is:
Vertical asymptote: [tex]\( x = 1 \)[/tex] \\
Horizontal asymptote: [tex]\( y = 0 \)[/tex]
[tex]\[ \boxed{\text{vertical asymptote: } x=1 \text{ horizontal asymptote: } y=0} \][/tex]
