# 2006 AMC 10A Problems/Problem 8

## Problem

A parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$. What is $c$? $\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$

## Solution 1

Substitute the points $(2,3)$ and $(4,3)$ into the given equation for $(x,y)$.

Then we get a system of two equations: $3=4+2b+c$ $3=16+4b+c$

Subtracting the first equation from the second we have: $0=12+2b$ $b=-6$

Then using $b=-6$ in the first equation: $0=1+-12+c$ $c=\boxed{\textbf{(E) }11}$.

### Solution 1.1

Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely $(3,2)$. Thus, the form of the equation of the parabola is $y - 2 = (x - 3)^2$. Expanding this out, we find that $c=\boxed{\textbf{(E) }11}$.

## Solution 2

The points given have the same $y$-value, so the vertex lies on the line $x=\frac{2+4}{2}=3$.

The $x$-coordinate of the vertex is also equal to $\frac{-b}{2a}$, so set this equal to $3$ and solve for $b$, given that $a=1$: $x=\frac{-b}{2a}$ $3=\frac{-b}{2}$ $6=-b$ $b=-6$

Now the equation is of the form $y=x^2-6x+c$. Now plug in the point $(2,3)$ and solve for $c$: $y=x^2-6x+c$ $3=2^2-6(2)+c$ $3=4-12+c$ $3=-8+c$ $c=\boxed{\textbf{(E) }11}$.

## Solution 3

Substituting y into the two equations, we get: $3=x^2+bx+c$

Which can be written as: $x^2+bx+c-3=0$ $4$ and $2$ are the solutions to the quadratic. Thus: $c-3=4\times2$ $c-3=8$ $c=\boxed{\textbf{(E) }11}$.

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