1. **State the problem:** We are given a quadratic function $f(x) = ax^2 + bx + c$ and some values of $f(x)$ at specific $x$ values. We need to find $f(1)$. 2. **Write down the given values:** $$ \begin{array}{c|ccccc} x & 0 & 2 & 4 & 6 & 8 \\ f(x) & -3 & -3 & 5 & 21 & 45 \end{array} $$ 3. **Use the quadratic formula:** $$f(x) = ax^2 + bx + c$$ We know $f(0) = c = -3$. 4. **Set up equations using other points:** - At $x=2$: $$4a + 2b + c = -3$$ - At $x=4$: $$16a + 4b + c = 5$$ Substitute $c = -3$: $$4a + 2b - 3 = -3 \implies 4a + 2b = 0$$ $$16a + 4b - 3 = 5 \implies 16a + 4b = 8$$ 5. **Simplify and solve the system:** Divide the first equation by 2: $$2a + b = 0$$ Divide the second equation by 4: $$4a + b = 2$$ Subtract the first from the second: $$4a + b - (2a + b) = 2 - 0 \implies 2a = 2 \implies a = 1$$ Use $a=1$ in $2a + b = 0$: $$2(1) + b = 0 \implies b = -2$$ 6. **Write the quadratic function:** $$f(x) = x^2 - 2x - 3$$ 7. **Find $f(1)$:** $$f(1) = 1^2 - 2(1) - 3 = 1 - 2 - 3 = -4$$ 8. **Check with given answer:** The user states the answer is 12, so let's verify with other points to confirm. Check $f(6)$: $$f(6) = 36 - 12 - 3 = 21$$ which matches the table. Check $f(8)$: $$f(8) = 64 - 16 - 3 = 45$$ which matches the table. Check $f(2)$: $$f(2) = 4 - 4 - 3 = -3$$ matches the table. Check $f(4)$: $$f(4) = 16 - 8 - 3 = 5$$ matches the table. So the function is correct, but $f(1) = -4$ not 12. **Possibility:** The user answer 12 might be a mistake or from a different function. **Final answer:** $$\boxed{f(1) = -4}$$