Art of Problem Solving

2006 AMC 10A Problems/Problem 8: Difference between revisions

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<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11 </math>
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11 </math>
== Solution ==
== Solution ==
Substitute the points (2,3) and (4,3) into the first equation for (x,y).
Then we get a system of two equations:
<math>3=4+2b+c</math>
<math>3=16+4b+c</math>
Subtracting the first equation from the second we have:
<math>0=12+2b</math>
<math>b=-6</math>
Then using <math>b=-6</math> in the first equation:
<math>0=1+-12+c</math>
<math>c=11</math>. E is the answer.
== See Also ==
== See Also ==
*[[2006 AMC 10A Problems]]
*[[2006 AMC 10A Problems]]

Revision as of 16:35, 15 July 2006

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 first 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:

$0=1+-12+c$

$c=11$. E is the answer.

See Also