1. **Problem Statement:** Complete the table for the function $f(x) = -2x^2 + 5x + 3$ for the domain $-2 \leq x \leq 5$ and determine the two roots of the quadratic equation. 2. **Formula and Rules:** The quadratic function is given by: $$f(x) = ax^2 + bx + c$$ where $a = -2$, $b = 5$, and $c = 3$. The roots of the quadratic equation $ax^2 + bx + c = 0$ can be found using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Table Completion Verification:** Given the table values: - For $x = -2$, $f(-2) = -2(-2)^2 + 5(-2) + 3 = -8 - 10 + 3 = -15$ - For $x = -1$, $f(-1) = -2(-1)^2 + 5(-1) + 3 = -2 - 5 + 3 = -4$ - For $x = 0$, $f(0) = 0 + 0 + 3 = 3$ - For $x = 1$, $f(1) = -2(1)^2 + 5(1) + 3 = -2 + 5 + 3 = 6$ - For $x = 2$, $f(2) = -8 + 10 + 3 = 5$ - For $x = 3$, $f(3) = -18 + 15 + 3 = 0$ - For $x = 4$, $f(4) = -32 + 20 + 3 = -9$ - For $x = 5$, $f(5) = -50 + 25 + 3 = -22$ All values match the table. 4. **Axis of Symmetry:** The axis of symmetry for a quadratic $ax^2 + bx + c$ is: $$x = -\frac{b}{2a} = -\frac{5}{2(-2)} = \frac{5}{4} = 1.25$$ 5. **Finding the Roots:** Calculate the discriminant: $$\Delta = b^2 - 4ac = 5^2 - 4(-2)(3) = 25 + 24 = 49$$ Since $\Delta > 0$, there are two real roots: $$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-5 \pm 7}{2(-2)}$$ Calculate each root: - Root 1: $$x_1 = \frac{-5 + 7}{-4} = \frac{2}{-4} = -0.5$$ - Root 2: $$x_2 = \frac{-5 - 7}{-4} = \frac{-12}{-4} = 3$$ 6. **Summary:** - The function values for $x$ from $-2$ to $5$ are as given. - The axis of symmetry is $x = 1.25$. - The roots of the quadratic equation are $x = -0.5$ and $x = 3$. These roots correspond to the points where the graph crosses the x-axis.