Let's find the vertical, horizontal, and slant (oblique) asymptotes for the function [tex]\( f(x) = \frac{3x^3 + 20x^2 + 14x - 65}{x^2 + 8x + 15} \)[/tex].
### Step 1: Find the Vertical Asymptotes
Vertical asymptotes occur where the denominator is equal to zero (given the numerator is not zero at these points). So, we need to solve for the roots of the denominator:
[tex]\[ x^2 + 8x + 15 = 0 \][/tex]
Factoring the quadratic equation, we get:
[tex]\[ (x + 5)(x + 3) = 0 \][/tex]
Thus, the roots are:
[tex]\[ x = -5 \quad \text{and} \quad x = -3 \][/tex]
These are the values where the function has vertical asymptotes. Hence, the vertical asymptotes are:
[tex]\[ x = -5 \quad \text{and} \quad x = -3 \][/tex]
### Step 2: Find the Horizontal Asymptotes
To find the horizontal asymptotes, we compare the degrees of the numerator and the denominator.
- The degree of the numerator [tex]\( 3x^3 + 20x^2 + 14x - 65 \)[/tex] is 3.
- The degree of the denominator [tex]\( x^2 + 8x + 15 \)[/tex] is 2.
Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote.
### Step 3: Find the Slant (Oblique) Asymptotes
A slant asymptote occurs when the degree of the numerator is exactly one more than the degree of the denominator. In this case, the degree of the numerator is 3 and the degree of the denominator is 2, so we do have a slant asymptote.
To find the equation of the slant asymptote, we perform polynomial long division of [tex]\( 3x^3 + 20x^2 + 14x - 65 \)[/tex] by [tex]\( x^2 + 8x + 15 \)[/tex].
The result simplifies to:
[tex]\[ y = 3x + 4 \][/tex]
Thus, the slant asymptote is:
[tex]\[ y = 3x + 4 \][/tex]
### Summary of Asymptotes
- Vertical asymptotes: [tex]\( x = -5 \)[/tex] and [tex]\( x = -3 \)[/tex]
- Horizontal asymptote: None
- Slant asymptote: [tex]\( y = 3x + 4 \)[/tex]
