Difference between revisions of "2006 AMC 10A Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | A [[parabola]] with equation <math> | + | 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> \ | + | <math> \textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11 </math> |
| − | |||
| − | + | == Solution 1 == | |
| − | Substitute the points (2,3) and (4,3) into the given equation for (x,y). | + | 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: | ||
| Line 24: | Line 23: | ||
<math>0=1+-12+c</math> | <math>0=1+-12+c</math> | ||
| − | <math>c= | + | <math>c=\boxed{\textbf{(E) }11}</math>. |
| − | === Solution | + | === Solution 1.1 === |
| − | Alternatively, notice that since the equation is that of a | + | 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 points given have the same <math>y</math>-value, so the vertex lies on the line <math>x=\frac{2+4}{2}=3</math>. | ||
| Line 54: | Line 53: | ||
<math>3=-8+c</math> | <math>3=-8+c</math> | ||
| − | <math>\boxed{c= | + | <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 == | == See also == | ||
| Line 60: | Line 76: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:10, 16 December 2021
Problem
A parabola with equation 
passes through the points 
and 
. What is 
?
Solution 1
Substitute the points and into the given equation for 
.
Then we get a system of two equations:
Subtracting the first equation from the second we have:
Then using in the first equation:
.
Solution 1.1
Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely 
. Thus, the form of the equation of the parabola is 
. Expanding this out, we find that .
Solution 2
The points given have the same 
-value, so the vertex lies on the line 
.
The 
-coordinate of the vertex is also equal to 
, so set this equal to 
and solve for 
, given that 
:
Now the equation is of the form 
. Now plug in the point and solve for :
.
Solution 3
Substituting y into the two equations, we get:
Which can be written as:
and 
are the solutions to the quadratic. Thus:
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
