Difference between revisions of "2006 AMC 10A Problems/Problem 8"
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<math>\boxed{c=11 \text{(E)}}</math> | <math>\boxed{c=11 \text{(E)}}</math> | ||
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| + | === Solution 4 === | ||
| + | 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> | ||
| + | |||
| + | 4, 2, are the solutions to the quadratic. Thus: | ||
| + | |||
| + | <math>c-3=4\times2</math> | ||
| + | |||
| + | <math>c-3=8</math> | ||
| + | |||
| + | <math>c=11</math> | ||
== See also == | == See also == | ||
Revision as of 15:17, 17 January 2017
Contents
Problem
A parabola with equation 
passes through the points 
and 
. What is 
?
Solution
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:
is the answer.
Solution 2
Alternatively, notice that since the equation is that of a monic parabola, the vertex is likely 
. Thus, the form of the equation of the parabola is 
. Expanding this out, we find that 
.
Solution 3
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 4
Substituting y into the two equations, we get:
Which can be written as:
4, 2, 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. 
