1. **State the problem:** We are given the function $f(x) = \sqrt{2x^2 - 9x - 5}$ and asked to analyze it. 2. **Understand the domain:** Since $f(x)$ involves a square root, the expression inside must be non-negative: $$2x^2 - 9x - 5 \geq 0$$ 3. **Find the roots of the quadratic inside the root:** Solve $$2x^2 - 9x - 5 = 0$$ Using the quadratic formula: $$x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} = \frac{9 \pm \sqrt{81 + 40}}{4} = \frac{9 \pm \sqrt{121}}{4} = \frac{9 \pm 11}{4}$$ 4. **Calculate the roots:** - $x = \frac{9 + 11}{4} = \frac{20}{4} = 5$ - $x = \frac{9 - 11}{4} = \frac{-2}{4} = -\frac{1}{2}$ 5. **Determine intervals where the quadratic is non-negative:** Since the leading coefficient $2$ is positive, the parabola opens upward. The quadratic is non-negative outside the roots: $$x \leq -\frac{1}{2} \quad \text{or} \quad x \geq 5$$ 6. **Domain of $f(x)$:** $$\boxed{(-\infty, -\frac{1}{2}] \cup [5, \infty)}$$ 7. **Behavior of the function:** - At $x = -\frac{1}{2}$ and $x=5$, $f(x) = 0$ (square root of zero). - For $x$ in the domain, $f(x)$ is real and non-negative. 8. **Graph description:** The function starts at a positive value near $x = -3$ (which is in the domain $x \leq -\frac{1}{2}$), decreases to zero at $x = -\frac{1}{2}$, then is undefined between $-\frac{1}{2}$ and $5$, and then starts again at zero at $x=5$ and increases for $x > 5$. **Final answer:** The domain of $f(x)$ is $$(-\infty, -\frac{1}{2}] \cup [5, \infty)$$ and the function is defined and real-valued only on these intervals.