1. **Problem Statement:** We are given the quadratic function $$y = -2x^2 + 2x + 24$$ defined for $$-4 \leq x \leq 4$$. We need to: - Draw the graph. - Solve the equations: i) $$-2x^2 + 2x + 24 = 0$$ ii) $$-2x^2 - 2x + 20 = 0$$ iii) $$-2x^2 + 2x + 20 = 0$$ - Determine the nature of the turning point. 2. **Formula and Important Rules:** The general quadratic equation is $$ax^2 + bx + c = 0$$. The solutions (roots) are given by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ The discriminant $$\Delta = b^2 - 4ac$$ determines the nature of roots: - If $$\Delta > 0$$, two distinct real roots. - If $$\Delta = 0$$, one real root (repeated). - If $$\Delta < 0$$, no real roots (complex roots). The turning point (vertex) of $$y = ax^2 + bx + c$$ is at: $$x = -\frac{b}{2a}$$ and the corresponding $$y$$ value is: $$y = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c$$ 3. **Graph Description:** Since $$a = -2 < 0$$, the parabola opens downward. 4. **Find the turning point:** $$x = -\frac{b}{2a} = -\frac{2}{2 \times (-2)} = -\frac{2}{-4} = \frac{1}{2}$$ Calculate $$y$$ at $$x=\frac{1}{2}$$: $$y = -2\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) + 24 = -2\times \frac{1}{4} + 1 + 24 = -\frac{1}{2} + 1 + 24 = 24.5$$ So the turning point is at $$\left(\frac{1}{2}, 24.5\right)$$, which is a maximum point because the parabola opens downward. 5. **Solve i) $$-2x^2 + 2x + 24 = 0$$:** Here, $$a = -2$$, $$b = 2$$, $$c = 24$$. Calculate discriminant: $$\Delta = 2^2 - 4(-2)(24) = 4 + 192 = 196$$ Since $$\Delta > 0$$, two real roots. Calculate roots: $$x = \frac{-2 \pm \sqrt{196}}{2 \times -2} = \frac{-2 \pm 14}{-4}$$ - For $$+14$$: $$x = \frac{-2 + 14}{-4} = \frac{12}{-4} = -3$$ - For $$-14$$: $$x = \frac{-2 - 14}{-4} = \frac{-16}{-4} = 4$$ So roots are $$x = -3$$ and $$x = 4$$. 6. **Solve ii) $$-2x^2 - 2x + 20 = 0$$:** Here, $$a = -2$$, $$b = -2$$, $$c = 20$$. Calculate discriminant: $$\Delta = (-2)^2 - 4(-2)(20) = 4 + 160 = 164$$ Since $$\Delta > 0$$, two real roots. Calculate roots: $$x = \frac{-(-2) \pm \sqrt{164}}{2 \times -2} = \frac{2 \pm \sqrt{164}}{-4}$$ Approximate $$\sqrt{164} \approx 12.81$$: - For $$+12.81$$: $$x = \frac{2 + 12.81}{-4} = \frac{14.81}{-4} = -3.70$$ - For $$-12.81$$: $$x = \frac{2 - 12.81}{-4} = \frac{-10.81}{-4} = 2.70$$ So roots are approximately $$x = -3.70$$ and $$x = 2.70$$. 7. **Solve iii) $$-2x^2 + 2x + 20 = 0$$:** Here, $$a = -2$$, $$b = 2$$, $$c = 20$$. Calculate discriminant: $$\Delta = 2^2 - 4(-2)(20) = 4 + 160 = 164$$ Since $$\Delta > 0$$, two real roots. Calculate roots: $$x = \frac{-2 \pm \sqrt{164}}{2 \times -2} = \frac{-2 \pm 12.81}{-4}$$ - For $$+12.81$$: $$x = \frac{-2 + 12.81}{-4} = \frac{10.81}{-4} = -2.70$$ - For $$-12.81$$: $$x = \frac{-2 - 12.81}{-4} = \frac{-14.81}{-4} = 3.70$$ So roots are approximately $$x = -2.70$$ and $$x = 3.70$$. 8. **Nature of the turning point:** Since $$a = -2 < 0$$, the parabola opens downward, so the turning point at $$\left(\frac{1}{2}, 24.5\right)$$ is a maximum point. **Final answers:** - Turning point: maximum at $$\left(\frac{1}{2}, 24.5\right)$$. - i) Roots: $$x = -3$$ and $$x = 4$$. - ii) Roots: approximately $$x = -3.70$$ and $$x = 2.70$$. - iii) Roots: approximately $$x = -2.70$$ and $$x = 3.70$$.