2006 AMC 10A Problem 8

Each point gives an equation, and we have two unknowns, $b$ and $c$.

1. Put in $(2, 3)$: $3 = 2^2 + 2b + c$, that is $3 = 4 + 2b + c$.
2. Put in $(4, 3)$: $3 = 4^2 + 4b + c$, that is $3 = 16 + 4b + c$.
3. Both left sides are $3$, so subtracting the first equation from the second makes $c$ disappear:
$$0 = 12 + 2b, \qquad b = -6.$$
4. Put $b = -6$ into the first equation:
$$3 = 4 - 12 + c = -8 + c, \qquad c = 11.$$
5. Check with the second point: $16 - 24 + 11 = 3$. Correct.

The answer is $\boxed{\textbf{(E)}\ 11}$.

When a curve passes through given points, each point gives one equation. Here both points have the same height, so subtracting the equations wipes out $c$ and leaves only $b$. More generally, when two equations share a term, subtract them to get rid of it.

Use symmetry. A parabola is symmetric about a vertical line through its lowest point. The points $(2,3)$ and $(4,3)$ are at the same height, so that line is halfway between them, at $x = 3$. The parabola is then $y = (x - 3)^2 + k$ for some number $k$. At $x = 2$: $1 + k = 3$, so $k = 2$. Expanding, $y = x^2 - 6x + 9 + 2 = x^2 - 6x + 11$, so $c = 11$.

The two given points share the height $y = 3$. Setting $y = 3$ in the parabola's equation,
$$x^2 + bx + (c - 3) = 0,$$
and this quadratic is satisfied by $x = 2$ and $x = 4$. A monic quadratic with roots $2$ and $4$ has constant term equal to the product of the roots, so
$$c - 3 = 2 \cdot 4 = 8 \quad\Longrightarrow\quad c = 11.$$
(For completeness, the root sum gives $b = -(2 + 4) = -6$, and indeed $4 - 12 + 11 = 3$.)

The answer is $\boxed{\textbf{(E)}\ 11}$.