1. **State the problem:** Find all x-intercepts of the function $$f(x) = \frac{2x + 7}{2x^2 + 25x + 63}$$. 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 rule:** To find x-intercepts of a rational function $$\frac{N(x)}{D(x)}$$, solve $$N(x) = 0$$ and ensure $$D(x) \neq 0$$ at those points. 3. **Solve numerator:** Set numerator equal to zero: $$2x + 7 = 0$$ $$2x = -7$$ $$x = \frac{-7}{2}$$ 4. **Check denominator at $$x = -\frac{7}{2}$$:** $$D\left(-\frac{7}{2}\right) = 2\left(-\frac{7}{2}\right)^2 + 25\left(-\frac{7}{2}\right) + 63$$ Calculate step-by-step: $$= 2 \times \frac{49}{4} - \frac{175}{2} + 63$$ $$= \frac{98}{4} - \frac{175}{2} + 63$$ $$= 24.5 - 87.5 + 63 = 0$$ Since denominator equals zero at $$x = -\frac{7}{2}$$, the function is undefined there, so no x-intercept at this point. 5. **Conclusion:** There are no x-intercepts because the numerator zero makes the denominator zero, causing a vertical asymptote, not an x-intercept. **Final answer:** No x-intercepts exist for this function.