1. **Problem Statement:** We need to sketch the graph of the quadratic function $$f(x) = -2x^2 + 5x + 2$$ and find its important features: vertex, axis of symmetry, and intercepts. 2. **Formula and Rules:** - The vertex of a parabola given by $$f(x) = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. - The axis of symmetry is the vertical line through the vertex, $$x = -\frac{b}{2a}$$. - The y-intercept is found by evaluating $$f(0)$$. - The x-intercepts (roots) are found by solving $$f(x) = 0$$ using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Calculate the vertex:** - Here, $$a = -2$$, $$b = 5$$, $$c = 2$$. - Calculate $$x$$ coordinate of vertex: $$x = -\frac{5}{2 \times -2} = -\frac{5}{-4} = \frac{5}{4} = 1.25$$ - Calculate $$y$$ coordinate of vertex: $$f(1.25) = -2(1.25)^2 + 5(1.25) + 2 = -2(1.5625) + 6.25 + 2 = -3.125 + 6.25 + 2 = 5.125$$ 4. **Axis of symmetry:** - The axis of symmetry is the vertical line $$x = 1.25$$. 5. **Find y-intercept:** - Evaluate $$f(0)$$: $$f(0) = -2(0)^2 + 5(0) + 2 = 2$$ - So the y-intercept is at $$(0, 2)$$. 6. **Find x-intercepts:** - Solve $$-2x^2 + 5x + 2 = 0$$ using quadratic formula: $$x = \frac{-5 \pm \sqrt{5^2 - 4(-2)(2)}}{2(-2)} = \frac{-5 \pm \sqrt{25 + 16}}{-4} = \frac{-5 \pm \sqrt{41}}{-4}$$ - Calculate approximate roots: $$\sqrt{41} \approx 6.403$$ $$x_1 = \frac{-5 + 6.403}{-4} = \frac{1.403}{-4} = -0.35075$$ $$x_2 = \frac{-5 - 6.403}{-4} = \frac{-11.403}{-4} = 2.85075$$ - So the x-intercepts are approximately at $$(-0.35, 0)$$ and $$(2.85, 0)$$. 7. **Summary:** - Vertex: $$(1.25, 5.125)$$ - Axis of symmetry: $$x = 1.25$$ - Y-intercept: $$(0, 2)$$ - X-intercepts: $$(-0.35, 0)$$ and $$(2.85, 0)$$ The parabola opens downward because $$a = -2 < 0$$, so the vertex is a maximum point. This completes the analysis and sketching of the quadratic function.