1. **State the problem:** We are given the differential equation $$\frac{dy}{dx} = x - 1$$ and asked to analyze it through several parts including slope field, tangent line, and particular solution. 2. **Slope field:** The slope at any point $(x,y)$ is given by $$m = x - 1$$. This means the slope depends only on $x$, not on $y$. For 25 points, we can consider a grid of $x$ and $y$ values (e.g., $x = -2, -1, 0, 1, 2$ and $y = -2, -1, 0, 1, 2$) and draw small line segments with slope $x-1$ at each point. 3. **Sketch solution through $(-1,1)$:** At $x = -1$, slope is $-1 - 1 = -2$. The solution curve passing through $(-1,1)$ will have slope $-2$ there and follow the differential equation's slope field. 4. **Equation of tangent line at $(-1,1)$:** The slope at $x = -1$ is $$m = -1 - 1 = -2$$. Using point-slope form: $$y - 1 = -2(x + 1)$$ Simplify: $$y - 1 = -2x - 2$$ $$y = -2x - 1$$ 5. **Estimate $f(0.3)$ using tangent line:** Substitute $x = 0.3$: $$f(0.3) \approx -2(0.3) - 1 = -0.6 - 1 = -1.6$$ 6. **Find particular solution with initial condition $f(-1) = 1$:** The differential equation is: $$\frac{dy}{dx} = x - 1$$ Integrate both sides: $$y = \int (x - 1) dx = \frac{x^2}{2} - x + C$$ Use initial condition $y(-1) = 1$: $$1 = \frac{(-1)^2}{2} - (-1) + C = \frac{1}{2} + 1 + C = \frac{3}{2} + C$$ Solve for $C$: $$C = 1 - \frac{3}{2} = -\frac{1}{2}$$ So the particular solution is: $$y = \frac{x^2}{2} - x - \frac{1}{2}$$