Difference between revisions of "2006 AMC 12B Problems/Problem 12"
I like pie (talk | contribs) (Added problem) |
(solution) |
||
Line 1: | Line 1: | ||
== 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>? | + | 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 == | ||
− | {{ | + | 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>. |
+ | |||
+ | 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}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Revision as of 22:11, 23 April 2008
Problem
The parabola has vertex and -intercept , where . What is ?
Solution
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 .
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 |