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2006 AMC 10A Problems/Problem 8

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Problem

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

\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11

Solution

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=11 \Longrightarrow \mathrm{(E)} is the answer.

Alternatively, notice that since the equation is that of a monic 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 = 11.

See also

2006 AMC 10A (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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