1. **Problem statement:** Given the value table: $$\begin{array}{c|ccccccccc} x & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 4 & 1 & 0 & 1 & 4 & 9 & 16 & 25 & 36 \end{array}$$ Find the quadratic function $f(x)$ that fits these points. 2. **Recognize the pattern:** The values $0,1,4,9,16,25,36$ are perfect squares: $0^2,1^2,2^2,3^2,4^2,5^2,6^2$. 3. **Hypothesize the function form:** Since $f(-2) = 0$, and the values increase symmetrically, try the form: $$f(x) = (x + 2)^2$$ 4. **Check the function against the table:** - For $x = -4$: $f(-4) = (-4 + 2)^2 = (-2)^2 = 4$ ✓ - For $x = -3$: $f(-3) = (-3 + 2)^2 = (-1)^2 = 1$ ✓ - For $x = -2$: $f(-2) = 0^2 = 0$ ✓ - For $x = -1$: $f(-1) = 1^2 = 1$ ✓ - For $x = 0$: $f(0) = 2^2 = 4$ ✓ - For $x = 1$: $f(1) = 3^2 = 9$ ✓ - For $x = 2$: $f(2) = 4^2 = 16$ ✓ - For $x = 3$: $f(3) = 5^2 = 25$ ✓ - For $x = 4$: $f(4) = 6^2 = 36$ ✓ All values match perfectly. 5. **Conclusion:** The function is $$f(x) = (x + 2)^2$$ This is a parabola with vertex at $(-2,0)$.