1. **State the problem:** We have a linear function $f$ such that $f(0) = \frac{1}{2}$ and $f(-3) = -\frac{3}{2}$. We need to find the value of $f(3)$. 2. **Recall the form of a linear function:** A linear function can be written as $f(x) = mx + b$, where $m$ is the slope and $b$ is the y-intercept. 3. **Use the given information to find $b$:** Since $f(0) = \frac{1}{2}$, substituting $x=0$ gives $f(0) = m \cdot 0 + b = b = \frac{1}{2}$. 4. **Find the slope $m$ using the point $(-3, -\frac{3}{2})$:** Substitute $x = -3$ and $f(-3) = -\frac{3}{2}$ into $f(x) = mx + b$: $$ -\frac{3}{2} = m(-3) + \frac{1}{2} $$ 5. **Solve for $m$:** $$ -\frac{3}{2} - \frac{1}{2} = -3m $$ $$ -2 = -3m $$ $$ m = \frac{2}{3} $$ 6. **Write the function:** $$ f(x) = \frac{2}{3}x + \frac{1}{2} $$ 7. **Find $f(3)$:** $$ f(3) = \frac{2}{3} \times 3 + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2} = 2.5 $$ **Answer:** $f(3) = 2.5$, which corresponds to option C.