PLEASE HELP!

Of course, I'd be happy to help you graph the function \(f(x) = \frac{x^2 - 3x}{2x^2 - 3x - 9}\) step by step. Let's go through each of your requests:

**1. Graphing the Function and Identifying Intercepts:**

To graph the function, follow these steps:

a. **Find the Intercepts:**
   
   Start by finding the x-intercepts and y-intercept of the function.

   **X-Intercepts:** These occur when \(f(x) = 0\). Set the numerator \(x^2 - 3x\) equal to 0 and solve for x:
   
   \(x^2 - 3x = 0\)
   \(x(x - 3) = 0\)
   
   This gives us two possible x-intercepts: \(x = 0\) and \(x = 3\).

   **Y-Intercept:** This occurs when \(x = 0\). Substitute \(x = 0\) into the function:
   
   \(f(0) = \frac{0^2 - 3 \cdot 0}{2 \cdot 0^2 - 3 \cdot 0 - 9} = \frac{0}{-9} = 0\)
   
   So, the y-intercept is at the point (0, 0).

b. **Plotting the Intercepts:**
   
   Plot the x-intercepts at (0, 0) and (3, 0), and plot the y-intercept at (0, 0).

**2. Identifying Holes:**

Holes occur when factors in the numerator and denominator cancel each other out. Factor both the numerator and the denominator:

Numerator: \(x(x - 3)\)
Denominator: \(2x^2 - 3x - 9 = (2x - 3)(x +3)\)

You can see that the numerator and the denominator have no common factors. This means there are no holes in the graph,

**3. Asymptotes:**

Vertical Asymptote: Vertical asymptotes occur where the denominator is equal to zero but the numerator is not. In this case, the vertical asymptote occurs at \(x = \frac{3}{2}\) and \(x=-3\) because that's where \(2x^2 - 3x - 9 = 0\) and the numerator doesn't cancel it out.

Horizontal Asymptote: To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since both have the same degree (degree 2), the horizontal asymptote can be found by dividing the leading coefficients of both:

Horizontal asymptote: \(y = \frac{2}{2} = 1\)

Now you can plot the graph using the information gathered:

- Plot the x-intercepts: (0, 0) and (3, 0)
- Plot the y-intercept: (0, 0)
- Draw vertical asymptotes at \(x = \frac{3}{2}\) and \(x=-3\)
- Draw horizontal asymptote at \(y = 1\)

Remember that as you approach the vertical asymptote, the function values will tend to positive or negative infinity, depending on which side you're approaching from. Similarly, the function will get closer and closer to the horizontal asymptote as \(x\) goes to positive or negative infinity.

With these steps, you should be able to sketch the graph of the function \(f(x)\) and include the requested points, holes, and asymptotes.