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2006 AMC 10A Problem 8

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2006 AMC 10A Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Quadratics · difficulty 2 of 5 · about 2 min · position P1-10

A parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$ ?

A)
$2$
B)
$5$
C)
$7$
D)
$10$
E)
$11$

Answer

Key insight

Both points have y = 3, so 2 and 4 are roots of x^2 + bx + (c − 3) = 0; Vieta gives c − 3 = 8.

Solution

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 $\boxed{\textbf{(E)}\ 11}$ .

Why this works

Two points at the same height on a parabola are the two solutions of "parabola $=$ that height," so Vieta's formulas apply directly and the constant term pops out with no system to solve. More generally, whenever a problem hands you the roots of a quadratic, reach for sum and product before substituting.

Alternative approach

By symmetry the vertex sits midway, at $x = 3$ , so $-\tfrac{b}{2} = 3$ and $b = -6$ . Then $(2, 3)$ gives $3 = 4 - 12 + c$ , so $c = 11$ . Or subtract the two substituted equations $3 = 4 + 2b + c$ and $3 = 16 + 4b + c$ to get $b = -6$ first.

The trap

A sign slip when back-substituting b = −6, such as 3 = 4 + 12 + c, which gives c = −13 or c = 5 (choice (B)).

Common mistakes

A sign slip when back-substituting $b = -6$ , such as $3 = 4 + 12 + c$ , which gives $c = -13$ or $c = 5$ (choice (B)).
Forgetting to move the $3$ to the left side, so Vieta is applied to $x^2 + bx + c$ and $c = 8$ is reported.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Exploit symmetry to reduce work or pair up objects