Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (alternative)
(→‎Solution)
Line 5: Line 5:
 
== 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).
  
Line 24: Line 25:
  
 
<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 12:33, 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_8&oldid=38789"