1. **State the problem:** We want to analyze and understand the function $$f(x) = \sqrt{2x^2 - 9x - 5}$$ and determine where it is defined and its graph shape. 2. **Domain of the function:** Since the function 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 square root:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-9$, $c=-5$. Calculate the discriminant: $$\Delta = (-9)^2 - 4 \times 2 \times (-5) = 81 + 40 = 121$$ Calculate the roots: $$x = \frac{9 \pm \sqrt{121}}{2 \times 2} = \frac{9 \pm 11}{4}$$ So, $$x_1 = \frac{9 - 11}{4} = \frac{-2}{4} = -\frac{1}{2}$$ $$x_2 = \frac{9 + 11}{4} = \frac{20}{4} = 5$$ 4. **Determine intervals where the quadratic is non-negative:** Since $a=2 > 0$, the parabola opens upwards. The quadratic is non-negative outside the roots: $$x \leq -\frac{1}{2} \quad \text{or} \quad x \geq 5$$ 5. **Domain of $f(x)$:** $$\boxed{(-\infty, -\frac{1}{2}] \cup [5, \infty)}$$ 6. **Graph shape description:** - The function is real and defined only on the intervals where the expression inside the root is non-negative. - The graph touches the x-axis at $x = -\frac{1}{2}$ and $x=5$ (where the inside expression is zero). - For $x \leq -\frac{1}{2}$ and $x \geq 5$, the function values are positive and increase as $|x|$ grows because the quadratic inside grows. 7. **Summary:** - The function is continuous on its domain. - It has two branches starting at $x = -\frac{1}{2}$ and $x=5$ on the x-axis and rising upwards.