Let's solve these questions step by step.

Question 18: Finding the Slant Asymptote

To find the slant asymptote of the rational function: $f(x) = \frac{x^2 - 4x - 5}{x - 3}$ we need to perform polynomial long division since the degree of the numerator ($x^2$) is one higher than the degree of the denominator ($x$).

1. Divide $x^2$ by $x$: This gives $x$.
2. Multiply $x$ by $(x - 3)$: We get $x^2 - 3x$.
3. Subtract: $(x^2 - 4x - 5) - (x^2 - 3x) = -x - 5$.
4. Divide $-x$ by $x$: This gives $-1$.
5. Multiply $-1$ by $(x - 3)$: We get $-x + 3$.
6. Subtract: $(-x - 5) - (-x + 3) = -8$.

Thus, the quotient is: $f(x) = x - 1 + \frac{-8}{x - 3}$

As $x$ approaches infinity, the term $\frac{-8}{x - 3}$ approaches zero, so the slant asymptote is: $y = x - 1$

Answer to Question 18: D. $y = x - 1$

Question 19: Finding the Vertical Asymptote

The vertical asymptote occurs where the denominator of the rational function is zero. In this case, we set the denominator $x - 3 = 0$, which gives: $x = 3$

Answer to Question 19: E. $x = 3$

Would you like further clarification on any part of these solutions?

Related Questions:

1. How do you determine if a rational function has a slant asymptote?
2. Can you explain why the vertical asymptote only depends on the denominator?
3. How does polynomial long division work for finding slant asymptotes in general?
4. Are there cases where a rational function has both a slant and a horizontal asymptote?
5. What changes in the slant asymptote calculation if the numerator has a higher degree by more than one?

Tip:

Always check the degree of the numerator and denominator to decide whether the function has a horizontal, slant, or no asymptote.