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Difference between revisions of "2006 AMC 10A Problems/Problem 8"
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== Problem ==
 
== Problem ==
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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>?
  
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<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>
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== Solution ==
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== See Also ==
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== Solution 1 ==
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*[[2006 AMC 10A Problems]]
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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:
 +
 
 +
<math>3=4+2b+c</math>
 +
 
 +
<math>3=16+4b+c</math>
 +
 
 +
Subtracting the first equation from the second we have:
 +
 
 +
<math>0=12+2b</math>
 +
 
 +
<math>b=-6</math>
 +
 
 +
Then using <math>b=-6</math> in the first equation:
 +
 
 +
<math>0=1+-12+c</math>
 +
 
 +
<math>c=\boxed{\textbf{(E) }11}</math>.
 +
 
 +
=== Solution 1.1 ===
 +
 
 +
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>.
 +
 
 +
== Solution 2 ==
 +
 
 +
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>
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 +
<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>
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 +
<math>3=4-12+c</math>
 +
 
 +
<math>3=-8+c</math>
 +
 
 +
<math>c=\boxed{\textbf{(E) }11}</math>.
 +
 
 +
== Solution 3 ==
 +
Substituting y into the two equations, we get:
 +
 
 +
<math>3=x^2+bx+c</math>
 +
 
 +
Which can be written as:
 +
 
 +
<math>x^2+bx+c-3=0</math>
 +
 
 +
<math>4</math> and <math>2</math> are the solutions to the quadratic. Thus:
 +
 
 +
<math>c-3=4\times2</math>
 +
 
 +
<math>c-3=8</math>
 +
 
 +
<math>c=\boxed{\textbf{(E) }11}</math>.
 +
 
 +
== See also ==
 +
{{AMC10 box|year=2006|ab=A|num-b=7|num-a=9}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:10, 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 (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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