1. The problem is to sketch the function $y = g(x) = (x + 1)(x - 2) = x^2 - x - 2$. 2. This is a quadratic function in standard form $y = ax^2 + bx + c$ where $a = 1$, $b = -1$, and $c = -2$. 3. Important properties of quadratic functions: - The graph is a parabola. - Since $a > 0$, the parabola opens upwards. - The vertex can be found using the formula $x = -\frac{b}{2a}$. - The roots (x-intercepts) are the values of $x$ where $y=0$. 4. Find the roots by factoring: $$g(x) = (x + 1)(x - 2) = 0$$ So, $x = -1$ or $x = 2$. 5. Find the vertex: $$x = -\frac{b}{2a} = -\frac{-1}{2 \times 1} = \frac{1}{2}$$ Substitute back to find $y$: $$g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 2 = \frac{1}{4} - \frac{1}{2} - 2 = -\frac{9}{4}$$ 6. The vertex is at $\left(\frac{1}{2}, -\frac{9}{4}\right)$. 7. The y-intercept is found by evaluating $g(0)$: $$g(0) = (0 + 1)(0 - 2) = 1 \times (-2) = -2$$ 8. Summary: - Roots at $x = -1$ and $x = 2$ - Vertex at $\left(\frac{1}{2}, -\frac{9}{4}\right)$ - Y-intercept at $(0, -2)$ 9. Sketch the parabola using these points and the fact it opens upward. Final answer: The parabola $y = x^2 - x - 2$ has roots at $x = -1$ and $x = 2$, vertex at $\left(\frac{1}{2}, -\frac{9}{4}\right)$, and opens upward.