Art of Problem Solving

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

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


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


Then we get a system of two equations:
Then we get a system of two equations:
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<math>0=1+-12+c</math>
<math>0=1+-12+c</math>


<math>c=11</math>. E is the answer.
<math>c=11</math>. So <math>\mathrm{(E) \ }</math> is the answer.


== See Also ==
== See Also ==
*[[2006 AMC 10A Problems]]
*[[2006 AMC 10A Problems]]

Revision as of 10:26, 31 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 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:

$0=1+-12+c$

$c=11$. So $\mathrm{(E) \ }$ is the answer.

See Also