已知抛物线y=x2-(m2+5)x+2m2+6. (1)求证:无论m为何值,抛物线与x轴必有两第二小题用交点式算得∵A(2,0),B(m2+3,0)∴d=AB=m2+1正确答案为d=AB=m2+1但是用对称轴为-b/2a=(m2+5)/2 则AB应为(m2+5)/2
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![已知抛物线y=x2-(m2+5)x+2m2+6. (1)求证:无论m为何值,抛物线与x轴必有两第二小题用交点式算得∵A(2,0),B(m2+3,0)∴d=AB=m2+1正确答案为d=AB=m2+1但是用对称轴为-b/2a=(m2+5)/2 则AB应为(m2+5)/2](/uploads/image/z/10795714-34-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy%3Dx2-%EF%BC%88m2%2B5%EF%BC%89x%2B2m2%2B6%EF%BC%8E+%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E6%97%A0%E8%AE%BAm%E4%B8%BA%E4%BD%95%E5%80%BC%2C%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%B8%8Ex%E8%BD%B4%E5%BF%85%E6%9C%89%E4%B8%A4%E7%AC%AC%E4%BA%8C%E5%B0%8F%E9%A2%98%E7%94%A8%E4%BA%A4%E7%82%B9%E5%BC%8F%E7%AE%97%E5%BE%97%E2%88%B5A%EF%BC%882%2C0%EF%BC%89%2CB%EF%BC%88m2%2B3%2C0%EF%BC%89%E2%88%B4d%3DAB%3Dm2%2B1%E6%AD%A3%E7%A1%AE%E7%AD%94%E6%A1%88%E4%B8%BAd%3DAB%3Dm2%2B1%E4%BD%86%E6%98%AF%E7%94%A8%E5%AF%B9%E7%A7%B0%E8%BD%B4%E4%B8%BA-b%2F2a%3D%28m2%2B5%29%2F2++++%E5%88%99AB%E5%BA%94%E4%B8%BA%28m2%2B5%29%2F2)
已知抛物线y=x2-(m2+5)x+2m2+6. (1)求证:无论m为何值,抛物线与x轴必有两第二小题用交点式算得∵A(2,0),B(m2+3,0)∴d=AB=m2+1正确答案为d=AB=m2+1但是用对称轴为-b/2a=(m2+5)/2 则AB应为(m2+5)/2
已知抛物线y=x2-(m2+5)x+2m2+6. (1)求证:无论m为何值,抛物线与x轴必有两
第二小题用交点式算得∵A(2,0),B(m2+3,0)
∴d=AB=m2+1
正确答案为d=AB=m2+1
但是用对称轴为-b/2a=(m2+5)/2 则AB应为(m2+5)/2乘2=m2+5)
为什么两种方法做出来会不一样?
已知抛物线y=x2-(m2+5)x+2m2+6. (1)求证:无论m为何值,抛物线与x轴必有两第二小题用交点式算得∵A(2,0),B(m2+3,0)∴d=AB=m2+1正确答案为d=AB=m2+1但是用对称轴为-b/2a=(m2+5)/2 则AB应为(m2+5)/2
y=x2-(m2+5)x+2m2+6
(1)△恒大雨零,且A(2,0)带入y=x2-(m2+5)x+2m2+6恒成立;
(2)-b/2a=(m2+5)/2=d/2-2
d=AB=m2+1
(3)d=10 m=3
假定存在p点坐标(x,x2-14x+24)
PA^2=(X-2)^2+(x2-14x+24)^2
PB^2=(X-12)^2+(x2-14x+24)^2
PA^2+PB^2=100
(X-2)^2+(x2-14x+24)^2+(X-12)^2+(x2-14x+24)^2=100
(x2-14x+24)^2+2(x2-14x+24)+100=100
x2-14x+26=0
X=7±√92
y=
存在