1. **State the problem:** Find all x-intercepts of the function $$f(x) = \frac{5x^2 + 3x}{x^2 - 2x - 63}$$. The x-intercepts occur where the function equals zero, i.e., where the numerator is zero and the denominator is not zero. 2. **Formula and rules:** To find x-intercepts, solve $$f(x) = 0$$ which means solving $$\frac{5x^2 + 3x}{x^2 - 2x - 63} = 0$$. This happens when the numerator $$5x^2 + 3x = 0$$ and the denominator $$x^2 - 2x - 63 \neq 0$$. 3. **Solve numerator:** $$5x^2 + 3x = 0$$ Factor out $$x$$: $$x(5x + 3) = 0$$ Set each factor to zero: $$x = 0$$ or $$5x + 3 = 0$$ Solve second: $$5x = -3$$ $$x = -\frac{3}{5}$$ 4. **Check denominator for these x-values:** $$x^2 - 2x - 63 = 0$$ Factor: $$ (x - 9)(x + 7) = 0$$ So denominator is zero at $$x = 9$$ and $$x = -7$$. 5. **Verify x-intercepts:** Neither $$x=0$$ nor $$x=-\frac{3}{5}$$ is equal to 9 or -7, so both are valid x-intercepts. 6. **Final answer:** The x-intercepts are at points: $$\boxed{(0, 0)}$$ and $$\boxed{\left(-\frac{3}{5}, 0\right)}$$. There are two x-intercepts, not one.