AMC12 2006 B

AMC12 2006 B · Q12

AMC12 2006 B · Q12. It mainly tests Quadratic equations, Graphs (coordinate plane).

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

抛物线 $y=ax^2+bx+c$ 的顶点为 $(p,p)$，$y$-截距为 $(0,-p)$，其中 $p\ne 0$。$b$ 等于多少？

(A) $-p$ $-p$

(B) 0 0

(D) 4 4

(E) $p$ $p$

Answer

Correct choice: (D)

正确答案：（D）

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

代入 $(0,-p)$，得 $y=-p=a(0)^2+b(0)+c=c$，所以抛物线为 $y=ax^2+bx-p$。抛物线顶点的 $x$ 坐标为 $x=p=\frac{-b}{2a} \Longleftrightarrow a=\frac{-b}{2p}$。另外代入 $(p,p)$，得 $y=p=a(p)^2+b(p)-p \Longleftrightarrow ap^2+(b-2)p=\left(\frac{-b}{2p}\right)p^2+(b-2)p=p\left(\frac b2-2\right)=0$。由于 $p\neq 0$，所以 $\frac{b}{2}=2 \Longrightarrow b=4\ \mathrm{(D)}$.

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