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

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== Solution ==
== Solution ==


=== Solution 1 ===
Substitute the points (2,3) and (4,3) into the given equation for (x,y).
Substitute the points (2,3) and (4,3) into the given equation for (x,y).


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<math>c=11 \Longrightarrow \mathrm{(E)}</math> is the answer.
<math>c=11 \Longrightarrow \mathrm{(E)}</math> is the answer.
=== Solution 2 ===


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>.
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>.
=== Solution 3 ===
The points given have the same <math>y</math>-value, so the vertex lies on the line <math>x=\frac{2+4}{2}=3</math>.
The <math>x</math>-coordinate of the vertex is also equal to <math>\frac{-b}{2a}</math>, so set this equal to <math>3</math> and solve for <math>b</math>, given that <math>a=1</math>:
<math>x=\frac{-b}{2a}</math>
<math>3=\frac{-b}{2}</math>
<math>6=-b</math>
<math>b=-6</math>
Now the equation is of the form <math>y=x^2-6x+c</math>.  Now plug in the point <math>(2,3)</math> and solve for <math>c</math>:
<math>y=x^2-6x+c</math>
<math>3=2^2-6(2)+c</math>
<math>3=4-12+c</math>
<math>3=-8+c</math>
<math>\boxed{c=11 \text{(E)}}</math>


== See also ==
== See also ==

Revision as of 11:33, 30 May 2011

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

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 $c$:

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

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

$3=4-12+c$

$3=-8+c$

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

See also

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
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