Difference between revisions of "2006 AMC 10A Problems/Problem 8"
Dairyqueenxd (talk | contribs) (→Problem) |
Dairyqueenxd (talk | contribs) (→Solution 3) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 4: | Line 4: | ||
<math> \textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11 </math> | <math> \textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11 </math> | ||
| − | + | == Solution 1 == | |
| − | |||
| − | |||
Substitute the points <math> (2,3) </math> and <math> (4,3) </math> into the given equation for <math> (x,y) </math>. | Substitute the points <math> (2,3) </math> and <math> (4,3) </math> into the given equation for <math> (x,y) </math>. | ||
| Line 25: | Line 23: | ||
<math>0=1+-12+c</math> | <math>0=1+-12+c</math> | ||
| − | <math>c= | + | <math>c=\boxed{\textbf{(E) }11}</math>. |
| − | === Solution | + | === Solution 1.1 === |
| − | Alternatively, notice that since the equation is that of a conic 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>. | + | Alternatively, notice that since the equation is that of a conic 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=\boxed{\textbf{(E) }11}</math>. |
| − | + | == Solution 2 == | |
The points given have the same <math>y</math>-value, so the vertex lies on the line <math>x=\frac{2+4}{2}=3</math>. | The points given have the same <math>y</math>-value, so the vertex lies on the line <math>x=\frac{2+4}{2}=3</math>. | ||
| Line 55: | Line 53: | ||
<math>3=-8+c</math> | <math>3=-8+c</math> | ||
| − | <math>\boxed{ \ | + | <math>c=\boxed{\textbf{(E) }11}</math>. |
| − | + | == Solution 3 == | |
Substituting y into the two equations, we get: | Substituting y into the two equations, we get: | ||
| Line 66: | Line 64: | ||
<math>x^2+bx+c-3=0</math> | <math>x^2+bx+c-3=0</math> | ||
| − | 4 | + | <math>4</math> and <math>2</math> are the solutions to the quadratic. Thus: |
<math>c-3=4\times2</math> | <math>c-3=4\times2</math> | ||
| Line 72: | 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. 
