1. **State the problem:** Sketch the graph of the function $f(x) = -x^2 - 3x - 2$. 2. **Domain:** The domain of any quadratic function is all real numbers, so $\text{Domain} = (-\infty, \infty)$. 3. **Limits at the endpoints and asymptotes:** Since the function is a polynomial, there are no vertical or horizontal asymptotes. As $x \to \pm \infty$, $f(x) \to -\infty$ because the leading coefficient is negative. 4. **First derivative:** Calculate $f'(x)$ to find critical points and increasing/decreasing intervals. $$f'(x) = \frac{d}{dx}(-x^2 - 3x - 2) = -2x - 3$$ Set $f'(x) = 0$ to find critical points: $$-2x - 3 = 0 \implies -2x = 3 \implies x = -\frac{3}{2}$$ 5. **Second derivative:** Calculate $f''(x)$ to determine concavity. $$f''(x) = \frac{d}{dx}(-2x - 3) = -2$$ Since $f''(x) = -2 < 0$, the graph is concave down everywhere. 6. **Variation table:** - For $x < -\frac{3}{2}$, $f'(x) = -2x - 3 > 0$ (since $-2x$ is positive and dominates), so $f$ is increasing. - For $x > -\frac{3}{2}$, $f'(x) < 0$, so $f$ is decreasing. Thus, $x = -\frac{3}{2}$ is a maximum point. 7. **Intercepts:** - **y-intercept:** Set $x=0$: $$f(0) = -0 - 0 - 2 = -2$$ - **x-intercepts:** Solve $f(x) = 0$: $$-x^2 - 3x - 2 = 0 \implies x^2 + 3x + 2 = 0$$ Factor: $$ (x + 1)(x + 2) = 0 \implies x = -1, -2$$ 8. **Supplementary points:** Calculate $f(-1.5)$ (vertex): $$f\left(-\frac{3}{2}\right) = -\left(-\frac{3}{2}\right)^2 - 3\left(-\frac{3}{2}\right) - 2 = -\frac{9}{4} + \frac{9}{2} - 2 = -2.25 + 4.5 - 2 = 0.25$$ 9. **Sketch the curve:** - The parabola opens downward (concave down). - Vertex at $\left(-\frac{3}{2}, 0.25\right)$ is the maximum point. - Crosses the x-axis at $x = -1$ and $x = -2$. - Crosses the y-axis at $y = -2$. Final answer: The graph is a downward-opening parabola with vertex at $\left(-\frac{3}{2}, 0.25\right)$, x-intercepts at $-1$ and $-2$, and y-intercept at $-2$.