Difference between revisions of "2006 AMC 12B Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
− | A parabola with the given equation and with vertex <math>(p,p)</math> must have equation <math>y=a(x-p)^2+p</math>. Because the <math>y</math>-intercept is <math>(0,-p)</math> and <math>p\ne 0</math>, it follows that <math>a=-2/p</math>. Thus | + | A parabola with the given equation and with vertex <math>(p,p)</math> must have equation <math>y=a(x-p)^2+p</math>. Because the <math>y</math>-intercept is <math>(0,-p)</math> and <math>p\ne 0</math>, it follows that <math>a=-2/p</math>. Thus<cmath> |
y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p, | y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p, | ||
− | + | </cmath> so <math>\boxed{b=4}</math>. | |
== See also == | == See also == |
Latest revision as of 22:20, 17 October 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 |
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