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 general form of 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 roots (x-intercepts) by solving $$x^2 + x - 2 = 0$$. 4. Factor the quadratic: $$x^2 + x - 2 = (x + 2)(x - 1) = 0$$ 5. Set each factor to zero: $$x + 2 = 0 \Rightarrow x = -2$$ $$x - 1 = 0 \Rightarrow x = 1$$ 6. The roots are $$x = -2$$ and $$x = 1$$, so the graph crosses the x-axis at these points. 7. Find the y-intercept by evaluating $$y$$ at $$x=0$$: $$y = 0^2 + 0 - 2 = -2$$ 8. The vertex of the parabola is at $$x = -\frac{b}{2a} = -\frac{1}{2 \times 1} = -\frac{1}{2}$$. 9. Calculate the y-coordinate of the vertex: $$y = (-\frac{1}{2})^2 + (-\frac{1}{2}) - 2 = \frac{1}{4} - \frac{1}{2} - 2 = -\frac{9}{4} = -2.25$$ 10. The vertex is at $$\left(-\frac{1}{2}, -2.25\right)$$, which is the minimum point since $$a=1 > 0$$. 11. Plot points for $$x$$ values from $$-3$$ to $$3$$: - $$x=-3$$: $$y = 9 - 3 - 2 = 4$$ - $$x=-2$$: $$y = 4 - 2 - 2 = 0$$ - $$x=-1$$: $$y = 1 - 1 - 2 = -2$$ - $$x=0$$: $$y = -2$$ - $$x=1$$: $$y = 1 + 1 - 2 = 0$$ - $$x=2$$: $$y = 4 + 2 - 2 = 4$$ - $$x=3$$: $$y = 9 + 3 - 2 = 10$$ 12. Connect these points smoothly to form a parabola opening upwards. Final answer: The graph of $$y = x^2 + x - 2$$ is a parabola with roots at $$x=-2$$ and $$x=1$$, vertex at $$\left(-\frac{1}{2}, -2.25\right)$$, and y-intercept at $$-2$$.