The problem shown in your image asks to find the equation of the slant asymptote of the rational function:

$f(x) = \frac{8x^2 - 16x}{4x - 16}$

Steps to find the slant asymptote:

1. Perform polynomial long division since the degree of the numerator (quadratic) is higher than the degree of the denominator (linear). This will help us extract the quotient, which represents the slant asymptote.

Polynomial long division:

• Divide the leading term of the numerator $8x^2$ by the leading term of the denominator $4x$:

  $\frac{8x^2}{4x} = 2x$

• Multiply $2x$ by the denominator $4x - 16$, and subtract the result from the original numerator:

  $(8x^2 - 16x) - (2x(4x - 16)) = (8x^2 - 16x) - (8x^2 - 32x) = 16x$

• Now divide $16x$ by $4x$:

  $\frac{16x}{4x} = 4$

• Multiply $4$ by $4x - 16$ and subtract from $16x$:

  $16x - (4(4x - 16)) = 16x - (16x - 64) = 64$

• The remainder is $64$, which is irrelevant for finding the slant asymptote.

Conclusion:

The quotient from the division is $2x + 4$. This is the equation of the slant asymptote.

Thus, the slant asymptote is: $y = 2x + 4$

This matches the result already entered in the image.

Would you like more details on any of the steps? Here are a few related questions to explore:

1. How do you know if a rational function has a slant asymptote?
2. What happens if the degree of the numerator is less than the denominator?
3. How do you interpret the remainder in polynomial division when finding asymptotes?
4. What are the differences between vertical, horizontal, and slant asymptotes?
5. How do slant asymptotes affect the graph of a rational function?

Tip: A rational function will have a slant asymptote only if the degree of the numerator is exactly one higher than the degree of the denominator.