已知二次函数Y=(2m-1)x²-(5m+3)x+3m+5(1)证明:m为任何实数时它的图象必与x轴相较于两点
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![已知二次函数Y=(2m-1)x²-(5m+3)x+3m+5(1)证明:m为任何实数时它的图象必与x轴相较于两点](/uploads/image/z/1811400-24-0.jpg?t=%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0Y%3D%EF%BC%882m-1%29x%26%23178%3B-%285m%2B3%29x%2B3m%2B5%281%29%E8%AF%81%E6%98%8E%EF%BC%9Am%E4%B8%BA%E4%BB%BB%E4%BD%95%E5%AE%9E%E6%95%B0%E6%97%B6%E5%AE%83%E7%9A%84%E5%9B%BE%E8%B1%A1%E5%BF%85%E4%B8%8Ex%E8%BD%B4%E7%9B%B8%E8%BE%83%E4%BA%8E%E4%B8%A4%E7%82%B9)
已知二次函数Y=(2m-1)x²-(5m+3)x+3m+5(1)证明:m为任何实数时它的图象必与x轴相较于两点
已知二次函数Y=(2m-1)x²-(5m+3)x+3m+5(1)证明:m为任何实数时它的图象必与x轴相较于两点
已知二次函数Y=(2m-1)x²-(5m+3)x+3m+5(1)证明:m为任何实数时它的图象必与x轴相较于两点
y=(2m-1)x²-(5m+3)x+3m+5
与x轴相交:y=0
(2m-1)x²-(5m+3)x+3m+5=0
Δ=[-(5m+3)]²-4(2m-1)(3m+5)
=25m²+30m+9-24m²-28m+20
=m²+2m+1+28
=(m+1)²+28
∵无论m取何值,Δ>=28>0
∴无论m取何值,二次函数y=(2m-1)x²-(5m+3)x+3m+5必与x轴有2个交点.