1. Problem: Given a linear function $f(x)$ with $f(1) = -3$ and $f(-2) = 4$, find $f(10)$. 2. Since $f(x)$ is linear, it can be written as $f(x) = mx + b$, where $m$ is the slope and $b$ is the y-intercept. 3. Use the two points $(1, -3)$ and $(-2, 4)$ to find the slope $m$: $$m = \frac{f(1) - f(-2)}{1 - (-2)} = \frac{-3 - 4}{1 + 2} = \frac{-7}{3} = -\frac{7}{3}$$ 4. Substitute $m$ and one point into $f(x) = mx + b$ to find $b$: $$-3 = -\frac{7}{3} \times 1 + b \implies b = -3 + \frac{7}{3} = -\frac{9}{3} + \frac{7}{3} = -\frac{2}{3}$$ 5. The function is: $$f(x) = -\frac{7}{3}x - \frac{2}{3}$$ 6. Find $f(10)$: $$f(10) = -\frac{7}{3} \times 10 - \frac{2}{3} = -\frac{70}{3} - \frac{2}{3} = -\frac{72}{3} = -24$$ Answer: $f(10) = -24$, which corresponds to option C.