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

Revision as of 21:58, 5 September 2011

Problem

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

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

Solution

Solution 1

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.

Solution 2

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 $c = 11$ .

Solution 3

The points given have the same $y$ -value, so the vertex lies on the line $x=\frac{2+4}{2}=3$ .

The $x$ -coordinate of the vertex is also equal to $\frac{-b}{2a}$ , so set this equal to $3$ and solve for $b$ , given that $a=1$ :

$x=\frac{-b}{2a}$

$3=\frac{-b}{2}$

$6=-b$

$b=-6$

Now the equation is of the form $y=x^2-6x+c$ . Now plug in the point $(2,3)$ and solve for $c$ :

$y=x^2-6x+c$

$3=2^2-6(2)+c$

$3=4-12+c$

$3=-8+c$

$\boxed{c=11 \text{(E)}}$

@@ Line 7: / Line 7: @@
 === Solution 1 ===
-Substitute the points (2,3) and (4,3) into the given equation for (x,y).
+Substitute the points <math> (2,3) </math> and <math> (4,3) </math> into the given equation for <math> (x,y) </math>.
 Then we get a system of two equations:

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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