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>?
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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>?
  
 
<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>
  
 
== Solution ==
 
== Solution ==
{{solution}}
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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>.
  
 
== 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]]

Revision as of 22:11, 23 April 2008

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

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 $\frac{b}{2} = 2 \Longrightarrow b = 4\ \mathrm{(D)}$.

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
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