1. **State the problem:** We need to fill in the table for the quadratic functions $y = x^2$, $y = 2x^2$, $y = \frac{1}{2}x^2$, and $y = -2x^2$ at given values of $x$. 2. **Recall the formula:** The quadratic function is $y = ax^2 + bx + c$. Here, $b=0$ and $c=0$, so the functions simplify to $y = ax^2$ with different values of $a$. 3. **Calculate each value:** - For $y = x^2$, simply square $x$. - For $y = 2x^2$, multiply $x^2$ by 2. - For $y = \frac{1}{2}x^2$, multiply $x^2$ by $\frac{1}{2}$. - For $y = -2x^2$, multiply $x^2$ by $-2$. 4. **Fill the table:** | $x$ | $x^2$ | $2x^2$ | $\frac{1}{2}x^2$ | $-2x^2$ | |---|---|---|---|---| | -3 | 9 | $2 \times 9 = 18$ | $\frac{1}{2} \times 9 = 4.5$ | $-2 \times 9 = -18$ | | -2 | 4 | $2 \times 4 = 8$ | $\frac{1}{2} \times 4 = 2$ | $-2 \times 4 = -8$ | | -1 | 1 | $2 \times 1 = 2$ | $\frac{1}{2} \times 1 = 0.5$ | $-2 \times 1 = -2$ | | 0 | 0 | $2 \times 0 = 0$ | $\frac{1}{2} \times 0 = 0$ | $-2 \times 0 = 0$ | | 1 | 1 | $2 \times 1 = 2$ | $\frac{1}{2} \times 1 = 0.5$ | $-2 \times 1 = -2$ | | 2 | 4 | $2 \times 4 = 8$ | $\frac{1}{2} \times 4 = 2$ | $-2 \times 4 = -8$ | | 3 | 9 | $2 \times 9 = 18$ | $\frac{1}{2} \times 9 = 4.5$ | $-2 \times 9 = -18$ | 5. **Summary:** Changing the coefficient $a$ in $y = ax^2$ stretches or compresses the parabola vertically and can flip it upside down if $a$ is negative. **Final table:** | $x$ | $x^2$ | $2x^2$ | $\frac{1}{2}x^2$ | $-2x^2$ | |---|---|---|---|---| | -3 | 9 | 18 | 4.5 | -18 | | -2 | 4 | 8 | 2 | -8 | | -1 | 1 | 2 | 0.5 | -2 | | 0 | 0 | 0 | 0 | 0 | | 1 | 1 | 2 | 0.5 | -2 | | 2 | 4 | 8 | 2 | -8 | | 3 | 9 | 18 | 4.5 | -18 |