1. **Problem Statement:** Sketch the graph of the quadratic function $f(x) = -2x^2 + 5x + 2$ and find its vertex, axis of symmetry, and intercepts. 2. **Formula and Rules:** The quadratic function is in the form $f(x) = ax^2 + bx + c$ where $a = -2$, $b = 5$, and $c = 2$. - The vertex formula is given by $x = -\frac{b}{2a}$. - The axis of symmetry is the vertical line $x = -\frac{b}{2a}$. - The y-intercept is $f(0) = c$. - The x-intercepts are found by solving $ax^2 + bx + c = 0$. 3. **Find the vertex:** Calculate $x$ coordinate of vertex: $$x = -\frac{5}{2 \times (-2)} = -\frac{5}{-4} = \frac{5}{4} = 1.25$$ Calculate $y$ coordinate of vertex by substituting $x=1.25$ into $f(x)$: $$f(1.25) = -2(1.25)^2 + 5(1.25) + 2 = -2(1.5625) + 6.25 + 2 = -3.125 + 6.25 + 2 = 5.125$$ So, vertex is at $\left(1.25, 5.125\right)$. 4. **Axis of symmetry:** The axis of symmetry is the vertical line: $$x = 1.25$$ 5. **Find the y-intercept:** Substitute $x=0$: $$f(0) = 2$$ So, y-intercept is at $(0, 2)$. 6. **Find the x-intercepts:** Solve $-2x^2 + 5x + 2 = 0$. Multiply both sides by $-1$ to simplify: $$\cancel{-1} \times (-2x^2 + 5x + 2) = \cancel{-1} \times 0 \Rightarrow 2x^2 - 5x - 2 = 0$$ Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-2)}}{2(2)} = \frac{5 \pm \sqrt{25 + 16}}{4} = \frac{5 \pm \sqrt{41}}{4}$$ Calculate approximate values: $$x_1 = \frac{5 + 6.4031}{4} = \frac{11.4031}{4} = 2.8508$$ $$x_2 = \frac{5 - 6.4031}{4} = \frac{-1.4031}{4} = -0.3508$$ So, x-intercepts are approximately at $(2.85, 0)$ and $(-0.35, 0)$. 7. **Summary:** - Vertex: $\left(1.25, 5.125\right)$ - Axis of symmetry: $x = 1.25$ - Y-intercept: $(0, 2)$ - X-intercepts: approximately $(2.85, 0)$ and $(-0.35, 0)$ This completes the sketching and identification of important features of the quadratic function $f(x) = -2x^2 + 5x + 2$.