问题 填空题

若函数y=a(x-h)2+k的图象经过原点,最小值为8,且形状与抛物线y=-2x2-2x+3相同,

则此函数关系式______.

答案

y=-2x2+8xy=-2x2-8x

函数图象经过原点,可得等式ah2+k=0;已知最小值8,可得k=8;根据抛物线形状相同可知a=-2,从而可求h.

解:∵函数y=a(x-h)2+k的图象经过原点,把(0,0)代入解析式,得:ah2+k=0,

∵最大值为8,即函数的开口向下,a<0,顶点的纵坐标k=8,

又∵形状与抛物线y=-2x2-2x+3相同,

∴二次项系数a=-2,

把a=-2,k=8代入ah2+k=0中,得h=±2,

∴函数解析式是:y=-2(x-2)2+8或y=-2(x+2)2+8,

即:y=-2x2+8x或y=-2x2-8x.

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