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2006 AMC 12B Problems/Problem 12

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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 $y = p = a(p)^2 + b(p) - p \Longleftrightarrow ap^2 + (b-2)p = \left(\frac{-b}{2p}\right)p^2 + (b-2)p = p\left(\frac b2-2\righ...$ . Since it is given that $p \neq 0$ , then $\frac{b}{2} = 2 \Longrightarrow b = 4\ \mathrm{(D)}$ .

See also

2006 AMC 12B (Problems • 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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12B_Problems/Problem_12"

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