2006 AMC 12B Problems/Problem 12

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 $\boxed{b=4}$ .

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

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2006 AMC 12B Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See also