Difference between revisions of "2006 AMC 12B Problems/Problem 12"
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− | The [[parabola]] <math>y=ax^2+bx+c</math> has [[vertex]] <math>(p,p)</math> and <math>y</math>-intercept <math>(0,-p)</math>, where <math>p\ne 0</math>. What is <math> | + | The [[parabola]] <math>y=ax^2+bx+c</math> has [[vertex]] <math>(p,p)</math> and <math>y</math>-intercept <math>(0,-p)</math>, where <math>p\ne 0</math>. What is <math>a</math>? |
<math>\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p</math> | <math>\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p</math> |
Revision as of 13:00, 4 September 2020
Contents
Problem
The parabola has vertex and -intercept , where . What is ?
Solution 1
Substituting , we find that , so our parabola is .
The x-coordinate of the vertex of a parabola is given by . Additionally, substituting , we find that . Since it is given that , then .
Solution 2
A parabola with the given equation and with vertex must have equation . Because the -intercept is and , it follows that . Thus so .
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.