To find the slant (oblique) asymptote of the given rational function:

$f(x) = \frac{8x^3 + 3x}{4x^2 + 15}$

we follow these steps:

Step 1: Determine the degrees of the numerator and denominator.

• The degree of the numerator $8x^3 + 3x$ is 3.
• The degree of the denominator $4x^2 + 15$ is 2.

Since the degree of the numerator (3) is exactly one greater than the degree of the denominator (2), there will be a slant asymptote.

Step 2: Perform polynomial long division.

We divide $8x^3 + 3x$ by $4x^2 + 15$.

Division process:

1. Divide the leading term of the numerator by the leading term of the denominator: $\frac{8x^3}{4x^2} = 2x$
2. Multiply $2x$ by the entire denominator $4x^2 + 15$: $2x \cdot (4x^2 + 15) = 8x^3 + 30x$
3. Subtract this from the numerator: $(8x^3 + 3x) - (8x^3 + 30x) = -27x$
4. Now divide the new term $-27x$ by $4x^2$. Since the degree of the remainder is less than the degree of the denominator, we stop here.

Thus, the quotient is $2x$, which represents the equation of the slant asymptote.

Step 3: Write the slant asymptote equation.

The equation of the slant asymptote is: $y = 2x$

This is the equation of the slant asymptote for the given rational function.

Would you like more details or have any further questions?

Here are some related questions for further exploration:

1. How do you perform polynomial long division in general?
2. What happens if the degree of the numerator is less than the degree of the denominator?
3. How do you find vertical asymptotes of a rational function?
4. Can a rational function have more than one slant asymptote?
5. How do horizontal asymptotes differ from slant asymptotes?

Tip: The slant asymptote appears when the degree of the numerator is exactly one more than the degree of the denominator.