2006 AMC 10A Problems/Problem 8: Difference between revisions
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<math>0=1+-12+c</math> | <math>0=1+-12+c</math> | ||
<math>c=11 | <math>c=11 \Longrightarrow \mathrm{(E)}</math> is the answer. | ||
== | Alternatively, notice that since the equation is that of a monic 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 = 11</math>. | ||
== See also == | |||
{{AMC10 box|year=2006|ab=A|num-b=7|num-a=9}} | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 16:19, 6 March 2007
Problem
A parabola with equation 
passes through the points (2,3) and (4,3). What is 
?
Solution
Substitute the points (2,3) and (4,3) into the given equation for (x,y).
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.
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 
.
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 | ||
