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

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<math>0=1+-12+c</math>
 
<math>0=1+-12+c</math>
  
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<math>c=\boxed{\textbf{(E) }}</math>.
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<math>c=\boxed{\textbf{(E) }11}</math>.
  
 
=== Solution 1.1 ===
 
=== Solution 1.1 ===
  
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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) }}</math>.
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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>.
  
 
== Solution 2 ==
 
== Solution 2 ==
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<math>3=-8+c</math>
 
<math>3=-8+c</math>
  
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<math>c=\boxed{\textbf{(E) }}</math>.
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<math>c=\boxed{\textbf{(E) }11}</math>.
  
 
== Solution 3 ==
 
== Solution 3 ==
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<math>c-3=8</math>
 
<math>c-3=8</math>
  
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<math>c=11</math>
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<math>c=\boxed{\textbf{(E) }11}</math>.
  
 
== See also ==
 
== See also ==

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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