To determine the vertical and horizontal asymptotes for the function $f(x) = \frac{-9x + 5}{3x - 3}$, let's go through each step.

Vertical Asymptote

Vertical asymptotes occur where the denominator equals zero (assuming the numerator does not also become zero at that point).

1. Set the denominator equal to zero: $3x - 3 = 0$
2. Solve for $x$: $x = 1$

So, there is a vertical asymptote at $x = 1$.

Horizontal Asymptote

For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator:

• If the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.

1. Here, the degrees of both the numerator and the denominator are 1.
2. The leading coefficient of the numerator is $-9$, and for the denominator, it is $3$.

Thus, the horizontal asymptote is: $y = \frac{-9}{3} = -3$

Summary:

• Vertical Asymptote: $x = 1$
• Horizontal Asymptote: $y = -3$

You should graph these asymptotes by drawing a dashed vertical line at $x = 1$ and a dashed horizontal line at $y = -3$ on the graph.

Do you need further explanation on any part of this solution, or have any questions?

1. How would the asymptotes change if the degrees of the numerator and denominator were different?
2. What are the implications of having no horizontal asymptote?
3. How would we find slant asymptotes for functions with higher-degree numerators?
4. Can rational functions have both vertical and oblique asymptotes?
5. How does one verify these asymptotes graphically?

Tip: Asymptotes represent limits of behavior for the function, showing where it heads as $x$ approaches specific values or infinity.