如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点. (1)求如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点.(1)求a的值(2
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![如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点. (1)求如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点.(1)求a的值(2](/uploads/image/z/12562085-29-5.jpg?t=%E5%A6%82%E5%9B%BE%2C%E6%8A%9B%E7%89%A9%E7%BA%BFy1%3D-ax%26sup2%3B-ax%3D1%E7%BB%8F%E8%BF%87%E7%82%B9P%EF%BC%88-1%2F2%2C9%2F8%EF%BC%89%2C%E4%B8%94%E4%B8%8E%E6%8A%9B%E7%89%A9%E7%BA%BFy2%3Dax%26sup2%3B-ax-1%E7%9B%B8%E4%BA%A4%E4%BA%8EA%2CB%E4%B8%A4%E7%82%B9.+%EF%BC%881%EF%BC%89%E6%B1%82%E5%A6%82%E5%9B%BE%2C%E6%8A%9B%E7%89%A9%E7%BA%BFy1%3D-ax%26sup2%3B-ax%3D1%E7%BB%8F%E8%BF%87%E7%82%B9P%EF%BC%88-1%2F2%2C9%2F8%EF%BC%89%2C%E4%B8%94%E4%B8%8E%E6%8A%9B%E7%89%A9%E7%BA%BFy2%3Dax%26sup2%3B-ax-1%E7%9B%B8%E4%BA%A4%E4%BA%8EA%2CB%E4%B8%A4%E7%82%B9.%EF%BC%881%EF%BC%89%E6%B1%82a%E7%9A%84%E5%80%BC%EF%BC%882)
如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点. (1)求如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点.(1)求a的值(2
如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点. (1)求
如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点.
(1)求a的值
(2)设y1=-ax²-ax+1与x轴分别交于M,N两点(点M在点N的左边),y2=a²-ax-1与x轴分别交于E,F两点(点E在点F的左边),观察M,N,E,F四点的坐标,写出一条正确的结论,并通过计算说明
(3)设A,B两点的横坐标分别记为xA,xB,若在x轴上右一动点Q(x,0),且xA≤x≤xB,过点Q作一条垂直于x轴的直线,与两条抛物线分别交于C,D两点,试问,当x为何值时,线段CD有最大值?其最大值为多少?
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如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点. (1)求如图,抛物线y1=-ax²-ax=1经过点P(-1/2,9/8),且与抛物线y2=ax²-ax-1相交于A,B两点.(1)求a的值(2
(1)抛物线y1=-ax²-ax+1经过点P(-1/2,9/8)
将x=-1/2 y=9/8代入得
9/8=-a/4+a/2+1
1/8=a/4
a=32
(2)
y1=-32x²-32x+1
根据韦达定理得
M+N=-1
y2=32x²-32x-1
E+F=1
所以得出M+N+E+F=0 |ME|=|NF|
(3)
求出xA,xB
y=-32x²-32x+1
y=32x²-32x-1
-32x²-32x+1=32x²-32x-1
64x²=2
x=±√2/8
所以x的取值范围是
-√2/8≤x≤√2/8
x²≤1/32
当Q点横坐标为x时,分别代入两方程
y1=-32x²-32x+1
y2=32x²-32x-1
|CD|=|y1-y2|
=|-32x²-32x+1-(32x²-32x-1)|
=|-64x²+2|
因为x²≤1/32 ,所以-64x²+2≥0
|CD|=-64x²+2
所以当x=0时,线段CD有最大值,最大值是2
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