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

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== Problem ==
 
== Problem ==
A parabola with equation <math>\displaystyle y=x^2+bx+c</math> passes through the points (2,3) and (4,3).  What is <math>\displaystyle c</math>?
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A [[parabola]] with equation <math>y=x^2+bx+c</math> passes through the points <math> (2,3) </math> and <math> (4,3) </math>.  What is <math>c</math>?
  
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11 </math>
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<math> \textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11 </math>
== Solution ==
 
  
Substitute the points (2,3) and (4,3) into the first equation for (x,y).
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== Solution 1 ==
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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:
 
Then we get a system of two equations:
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<math>0=1+-12+c</math>
 
<math>0=1+-12+c</math>
  
<math>c=11</math>. E is the answer.
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<math>c=\boxed{\textbf{(E) }11}</math>.
  
--[[User:Someperson01|Someperson01]] 17:35, 15 July 2006 (EDT)
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=== Solution 1.1 ===
  
== See Also ==
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Alternatively, notice that since the equation is that of a conic 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=\boxed{\textbf{(E) }11}</math>.
*[[2006 AMC 10A Problems]]
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== Solution 2 ==
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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>.
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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>:
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<math>x=\frac{-b}{2a}</math>
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<math>3=\frac{-b}{2}</math>
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<math>6=-b</math>
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<math>b=-6</math>
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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>:
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<math>y=x^2-6x+c</math>
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<math>3=2^2-6(2)+c</math>
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<math>3=4-12+c</math>
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<math>3=-8+c</math>
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<math>c=\boxed{\textbf{(E) }11}</math>.
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== Solution 3 ==
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Substituting y into the two equations, we get:
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<math>3=x^2+bx+c</math>
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Which can be written as:
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<math>x^2+bx+c-3=0</math>
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<math>4</math> and <math>2</math> are the solutions to the quadratic. Thus:
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<math>c-3=4\times2</math>
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<math>c-3=8</math>
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<math>c=\boxed{\textbf{(E) }11}</math>.
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== See also ==
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{{AMC10 box|year=2006|ab=A|num-b=7|num-a=9}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:10, 16 December 2021

Problem

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

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

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:

$0=1+-12+c$

$c=\boxed{\textbf{(E) }11}$.

Solution 1.1

Alternatively, notice that since the equation is that of a conic 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 $c=\boxed{\textbf{(E) }11}$.

Solution 2

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 $c$:

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

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

$3=4-12+c$

$3=-8+c$

$c=\boxed{\textbf{(E) }11}$.

Solution 3

Substituting y into the two equations, we get:

$3=x^2+bx+c$

Which can be written as:

$x^2+bx+c-3=0$

$4$ and $2$ are the solutions to the quadratic. Thus:

$c-3=4\times2$

$c-3=8$

$c=\boxed{\textbf{(E) }11}$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
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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