Difference between revisions of "2006 AMC 10A Problems/Problem 8"
Dairyqueenxd (talk | contribs) (→Solution 1) |
Dairyqueenxd (talk | contribs) (→Solution 3) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 53: | Line 53: | ||
<math>3=-8+c</math> | <math>3=-8+c</math> | ||
| − | <math>c=\boxed{\textbf{(E) }}</math>. | + | <math>c=\boxed{\textbf{(E) }11}</math>. |
== Solution 3 == | == Solution 3 == | ||
| Line 70: | Line 70: | ||
<math>c-3=8</math> | <math>c-3=8</math> | ||
| − | <math>c=11</math> | + | <math>c=\boxed{\textbf{(E) }11}</math>. |
== See also == | == See also == | ||
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. 
