《2025年悦然好学生必开卷九年级数学全一册华师大版长春专版》

22. （9分）如图，两条抛物线$y_1 = -x^2 + 4$，$y_2 = -\frac{1}{5}x^2 + bx + c$相交于$A$，$B$两点，点$A$在$x$轴负半轴上，且为抛物线$y_2$的最高点.
（1）求抛物线$y_2$的表达式和点$B$的坐标；
（2）点$C$是抛物线$y_1$上$A$，$B$之间的一点，过点$C$作$x$轴的垂线交$y_2$于点$D$，当线段$CD$取最大值时，求$S_{\triangle BCD}$.

答案： 22.
(1)当$y_1 = 0$时，$-x^2 + 4 = 0$，
解得$x = 2$或$x = -2$，
又点A在$x$轴负半轴上，
$\therefore A(-2,0)$.
$\because$点A$(-2,0)$是抛物线$y_2$的最高点，
$\therefore \frac{b}{2 × (-\frac{1}{5})} = -2,\therefore b = -\frac{4}{5}$.
把A$(-2,0)$代入$y_2 = -\frac{1}{5}x^2 - \frac{4}{5}x + c$，
解得$c = -\frac{4}{5}$，
$\therefore$抛物线$y_2$的表达式为$y_2 = -\frac{1}{5}x^2 - \frac{4}{5}x - \frac{4}{5}$.
由$\begin{cases} y = -x^2 + 4, \\ y = -\frac{1}{5}x^2 - \frac{4}{5}x - \frac{4}{5}, \end{cases}$
解得$\begin{cases} x_1 = -2, \\ y_1 = 0, \end{cases} \begin{cases} x_2 = 3, \\ y_2 = -5. \end{cases}$
$\because A(-2,0),\therefore B(3,-5)$.
(2)由题意得$CD = y_1 - y_2 = -x^2 + 4 - (-\frac{1}{5}x^2 - \frac{4}{5}x - \frac{4}{5})$，
即$CD = -\frac{4}{5}x^2 + \frac{4}{5}x + \frac{24}{5}$.
当$x = \frac{1}{2}$时，$CD$最大$= -\frac{4}{5} × \frac{1}{4} + \frac{4}{5} × \frac{1}{2} + \frac{24}{5} = 5$，
$\therefore S_{\triangle BCD} = \frac{1}{2} × 5 × (3 - \frac{1}{2}) = \frac{25}{4}$.