Difference between revisions of "2006 AMC 10A Problems/Problem 8"

(Solution)
Line 4: Line 4:
 
<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