Difference between revisions of "2006 AMC 10A Problems/Problem 8"
Math Kirby (talk | contribs) m (→Problem) |
Math Kirby (talk | contribs) m (→Solution 1) |
||
Line 7: | Line 7: | ||
=== Solution 1 === | === 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: |
Revision as of 22:58, 5 September 2011
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
:
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 |