Difference between revisions of "2006 AMC 12B Problems/Problem 12"
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== Problem == | == Problem == | ||
+ | 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>b</math>? | ||
− | == Solution == | + | <math>\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p</math> |
+ | |||
+ | == Solution 1== | ||
+ | Substituting <math>(0,-p)</math>, we find that <math>y = -p = a(0)^2 + b(0) + c = c</math>, so our parabola is <math>y = ax^2 + bx - p</math>. | ||
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+ | The x-coordinate of the vertex of a parabola is given by <math>x = p = \frac{-b}{2a} \Longleftrightarrow a = \frac{-b}{2p}</math>. Additionally, substituting <math>(p,p)</math>, we find that <math>y = p = a(p)^2 + b(p) - p \Longleftrightarrow ap^2 + (b-2)p = \left(\frac{-b}{2p}\right)p^2 + (b-2)p = p\left(\frac b2-2\right) = 0</math>. Since it is given that <math>p \neq 0</math>, then <math>\frac{b}{2} = 2 \Longrightarrow b = 4\ \mathrm{(D)}</math>. | ||
+ | |||
+ | ==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<cmath> | ||
+ | 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 == | ||
− | + | {{AMC12 box|year=2006|ab=B|num-b=11|num-a=13}} | |
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
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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