To solve this problem, let’s analyze the rational function given:

$f(x) = \frac{2x^2 - 9x - 5}{2x^2 + 13x + 15}$

1. Finding the y-intercept:

   • To find the y-intercept, set $x = 0$ in $f(x)$: $f(0) = \frac{2(0)^2 - 9(0) - 5}{2(0)^2 + 13(0) + 15} = \frac{-5}{15} = -\frac{1}{3}$
   • Y-intercept is at $\left(0, -\frac{1}{3}\right)$.

2. Finding the x-intercepts:

   • X-intercepts occur when the numerator is zero, so we solve $2x^2 - 9x - 5 = 0$.
   • Factor or use the quadratic formula: $x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} = \frac{9 \pm \sqrt{81 + 40}}{4} = \frac{9 \pm \sqrt{121}}{4} = \frac{9 \pm 11}{4}$
   • This gives $x = \frac{20}{4} = 5$ and $x = \frac{-2}{4} = -\frac{1}{2}$.
   • X-intercepts are at $(5, 0)$ and $\left(-\frac{1}{2}, 0\right)$.

3. Finding the vertical asymptotes:

   • Vertical asymptotes occur where the denominator is zero, so solve $2x^2 + 13x + 15 = 0$.
   • Factor the quadratic: $2x^2 + 13x + 15 = (2x + 3)(x + 5) = 0$
   • This gives $x = -\frac{3}{2}$ and $x = -5$.
   • Vertical asymptotes are at $x = -\frac{3}{2}$ and $x = -5$.

Summary of Results

1. Y-intercept at $\left(0, -\frac{1}{3}\right)$
2. X-intercepts at $(5, 0)$ and $\left(-\frac{1}{2}, 0\right)$
3. Vertical asymptotes at $x = -\frac{3}{2}$ and $x = -5$

Would you like more details or have any questions?

Here are some related questions to explore further:

1. How do we determine horizontal asymptotes for rational functions?
2. Can this function have oblique asymptotes?
3. How do changes in the numerator affect the x-intercepts?
4. What happens to the graph near the vertical asymptotes?
5. How can we determine if there are any holes in this function?

Tip: For rational functions, vertical asymptotes occur where the denominator equals zero and the numerator is non-zero.