Difference between revisions of "2006 AMC 10A Problems/Problem 8"
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<math>x^2+bx+c-3=0</math> | <math>x^2+bx+c-3=0</math> | ||
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<math>c-3=4\times2</math> | <math>c-3=4\times2</math> | ||
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<math>c=11</math> | <math>c=11</math> | ||
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== See also == | == See also == |
Revision as of 12:41, 4 July 2020
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 conic 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, arelutions to the quadratic. Thus:
YO FACE
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.