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Difference between revisions of "2006 AMC 12B Problems/Problem 12"
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== Problem ==
 
== Problem ==
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{{problem}}
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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>b</math>?
  
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== Solution ==
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<math>\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p</math>
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{{solution}}
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== Solution 1==
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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>.
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==Solution 2==
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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>
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y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p,
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</cmath> so <math>\boxed{b=4}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=11|num-a=13}}
 
{{AMC12 box|year=2006|ab=B|num-b=11|num-a=13}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:20, 17 October 2020

Problem

The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$-intercept $(0,-p)$, where $p\ne 0$. What is $b$?

$\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p$

Solution 1

Substituting $(0,-p)$, we find that $y = -p = a(0)^2 + b(0) + c = c$, so our parabola is $y = ax^2 + bx - p$.

The x-coordinate of the vertex of a parabola is given by $x = p = \frac{-b}{2a} \Longleftrightarrow a = \frac{-b}{2p}$. Additionally, substituting $(p,p)$, we find that $y = p = a(p)^2 + b(p) - p \Longleftrightarrow ap^2 + (b-2)p = \left(\frac{-b}{2p}\right)p^2 + (b-2)p = p\left(\frac b2-2\right) = 0$. Since it is given that $p \neq 0$, then $\frac{b}{2} = 2 \Longrightarrow b = 4\ \mathrm{(D)}$.

Solution 2

A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$. Because the $y$-intercept is $(0,-p)$ and $p\ne 0$, it follows that $a=-2/p$. Thus\[y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p,\] so $\boxed{b=4}$.

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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