Sketch the graph of the rational function by hand. As sketching aids, check for intercepts, vertical asymptotes, horizontal asymptotes, and holes. Show complete work on your worksheet! \( f(x)=\frac{x^{2}-9}{x^{2}-x-12} \) Reduced function: \( f_{\text {red }}(x)=\frac{x-3}{x-4} \) Domain: x= VA: There is/are vertical asymptote(s) at Hole(s): There are no vertical asymptotes. O There is a hole at OThere is no hole in this function. O-intercept(s): OThere is/are \( x \)-intercept(s) at There are no \( x \)-intercepts.

To sketch the graph of the rational function \( f(x) = \frac{x^2 - 9}{x^2 - x - 12} \), let's analyze the components step by step. 1. **Factor the Function:** - The numerator \( x^2 - 9 \) factors as \( (x - 3)(x + 3) \). - The denominator \( x^2 - x - 12 \) factors as \( (x - 4)(x + 3) \). Thus, we rewrite the function as: \[ f(x) = \frac{(x - 3)(x + 3)}{(x - 4)(x + 3)} \] 2. **Reduce the Function:** - We can simplify the function since \( (x + 3) \) is in both the numerator and the denominator: \[ f_{\text{red}}(x) = \frac{x - 3}{x - 4} \] 3. **Domain:** - The domain excludes values that make the denominator zero. The denominator \( (x - 4)(x + 3) \) is zero at \( x = 4 \) and \( x = -3 \). Therefore, the domain is: \[ x \in (-\infty, -3) \cup (-3, 4) \cup (4, \infty) \] 4. **Vertical Asymptotes (VA):** - Since the original function has a factor \( (x + 3) \) in both the numerator and the denominator that cancels, there are no vertical asymptotes at that point. The vertical asymptote occurs where the remaining denominator equals zero: \[ \text{There is a vertical asymptote at } x = 4. \] 5. **Holes:** - A hole exists where we canceled out a factor: \[ \text{There is a hole at } x = -3. \] 6. **O-Intercept(s):** - To find the x-intercepts, set the numerator of the reduced function equal to zero: \[ x - 3 = 0 \Rightarrow x = 3. \] \[ \text{There is an } x\text{-intercept at } x = 3. \] 7. **Horizontal Asymptote:** - The degrees of the numerator and denominator are the same (both are degree 1 after simplification). The horizontal asymptote is determined by the ratio of the leading coefficients: \[ y = \frac{1}{1} = 1. \] **Summary of Features:** - Domain: \( x \in (-\infty, -3) \cup (-3, 4) \cup (4, \infty) \) - Vertical Asymptote at \( x = 4 \) - Hole at \( x = -3 \) - x-intercept at \( x = 3 \) - Horizontal Asymptote at \( y = 1 \) ### Final Thoughts With all these features identified, you can plot the graph accordingly by marking the intercept, asymptotes, and holes. Enjoy sketching!