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

Revision as of 17:19, 6 March 2007

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 $c = 11$ .

2006 AMC 10A (Problems • Answer Key • Resources)
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
All AMC 10 Problems and Solutions

@@ Line 23: / Line 23: @@
 <math>0=1+-12+c</math>
-<math>c=11</math>.  So <math>\mathrm{(E) \ }</math> is the answer.
+<math>c=11 \Longrightarrow \mathrm{(E)}</math> is the answer.
-== See Also ==
+Alternatively, notice that since the equation is that of a monic parabola, the vertex is likely <math>(3,2)</math>. Thus, the form of the equation of the parabola is <math>y - 2 = (x - 3)^2</math>. Expanding this out, we find that <math>c = 11</math>.
-*[[2006 AMC 10A Problems]]
-*[[2006 AMC 10A Problems/Problem 7|Previous Problem]]
+== See also ==
+{{AMC10 box|year=2006|ab=A|num-b=7|num-a=9}}
-*[[2006 AMC 10A Problems/Problem 9|Next Problem]]
 [[Category:Introductory Algebra Problems]]

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

Revision as of 17:19, 6 March 2007

Problem

Solution

See also