1. The problem is to understand why $f(-3) = \frac{21}{2}$ for the function $f(x) = \frac{1}{2}x^2 - 2x$. 2. The formula used is $f(x) = \frac{1}{2}x^2 - 2x$. This means for any input $x$, you square $x$, multiply by $\frac{1}{2}$, then subtract $2$ times $x$. 3. Substitute $x = -3$ into the function: $$f(-3) = \frac{1}{2}(-3)^2 - 2(-3)$$ 4. Calculate $(-3)^2$: $$(-3)^2 = 9$$ 5. Multiply by $\frac{1}{2}$: $$\frac{1}{2} \times 9 = \frac{9}{2}$$ 6. Calculate $-2(-3)$: $$-2 \times (-3) = 6$$ 7. Add the two results: $$\frac{9}{2} + 6$$ 8. Convert $6$ to a fraction with denominator $2$ to add easily: $$6 = \frac{12}{2}$$ 9. Add the fractions: $$\frac{9}{2} + \frac{12}{2} = \frac{21}{2}$$ 10. So, $f(-3) = \frac{21}{2}$ because the function's formula and arithmetic steps lead to this value. This explains why the answer is $\frac{21}{2}$.