1. The problem is to draw the graph of the quadratic function $$y = x^2 + x - 2$$ for values of $$x$$ from $$-3$$ to $$3$$. 2. The formula for a quadratic function is $$y = ax^2 + bx + c$$ where $$a$$, $$b$$, and $$c$$ are constants. Here, $$a=1$$, $$b=1$$, and $$c=-2$$. 3. To understand the graph, find the vertex and roots. 4. The vertex $$x$$-coordinate is given by $$x = -\frac{b}{2a} = -\frac{1}{2 \times 1} = -\frac{1}{2}$$. 5. Substitute $$x = -\frac{1}{2}$$ into the function to find the vertex $$y$$-coordinate: $$y = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 2 = \frac{1}{4} - \frac{1}{2} - 2 = \frac{1}{4} - \frac{1}{2} - 2$$ 6. Simplify the vertex $$y$$-coordinate: $$\frac{1}{4} - \frac{1}{2} - 2 = \frac{1}{4} - \frac{2}{4} - \frac{8}{4} = \frac{1 - 2 - 8}{4} = \frac{-9}{4} = -2.25$$ 7. The vertex is at $$\left(-\frac{1}{2}, -2.25\right)$$. 8. Find the roots by solving $$x^2 + x - 2 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-2)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm \sqrt{9}}{2}$$ 9. Simplify the roots: $$x = \frac{-1 \pm 3}{2}$$ 10. Calculate each root: $$x_1 = \frac{-1 + 3}{2} = \frac{2}{2} = 1$$ $$x_2 = \frac{-1 - 3}{2} = \frac{-4}{2} = -2$$ 11. The roots are $$x = 1$$ and $$x = -2$$. 12. Plotting points for $$x$$ from $$-3$$ to $$3$$: - For $$x = -3$$: $$y = (-3)^2 + (-3) - 2 = 9 - 3 - 2 = 4$$ - For $$x = -2$$: $$y = (-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0$$ - For $$x = -1$$: $$y = (-1)^2 + (-1) - 2 = 1 - 1 - 2 = -2$$ - For $$x = 0$$: $$y = 0^2 + 0 - 2 = -2$$ - For $$x = 1$$: $$y = 1^2 + 1 - 2 = 1 + 1 - 2 = 0$$ - For $$x = 2$$: $$y = 2^2 + 2 - 2 = 4 + 2 - 2 = 4$$ - For $$x = 3$$: $$y = 3^2 + 3 - 2 = 9 + 3 - 2 = 10$$ 13. The graph is a parabola opening upwards with vertex at $$\left(-\frac{1}{2}, -2.25\right)$$ and roots at $$x = -2$$ and $$x = 1$$. Final answer: The graph of $$y = x^2 + x - 2$$ is a parabola opening upwards with vertex $$\left(-\frac{1}{2}, -2.25\right)$$ and roots at $$x = -2$$ and $$x = 1$$, plotted for $$x$$ in $$[-3,3]$$.