设函数f(x)=ax^2+bx+c((a≠0),满足f(x+1)=f(-x-3),且f(-2)>f(2),解不等式f(-2x^2+2x-3)>f(x^2+4x+3)因为f(x+1)=f(-x-3) 所以f(1)=f(-3)所以f(x)对称轴为x=-1又因为f(-2)>f(2) 因为-2比2距离对称轴更近 显然 a=-1-2x^2+2x-3=-(x-1
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![设函数f(x)=ax^2+bx+c((a≠0),满足f(x+1)=f(-x-3),且f(-2)>f(2),解不等式f(-2x^2+2x-3)>f(x^2+4x+3)因为f(x+1)=f(-x-3) 所以f(1)=f(-3)所以f(x)对称轴为x=-1又因为f(-2)>f(2) 因为-2比2距离对称轴更近 显然 a=-1-2x^2+2x-3=-(x-1](/uploads/image/z/1635325-61-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dax%5E2%2Bbx%2Bc%28%28a%E2%89%A00%29%2C%E6%BB%A1%E8%B6%B3f%28x%2B1%29%3Df%28-x-3%29%2C%E4%B8%94f%28-2%29%3Ef%282%29%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28-2x%5E2%2B2x-3%29%3Ef%28x%5E2%2B4x%2B3%29%E5%9B%A0%E4%B8%BAf%28x%2B1%29%3Df%28-x-3%29+%E6%89%80%E4%BB%A5f%281%29%3Df%28-3%29%E6%89%80%E4%BB%A5f%28x%29%E5%AF%B9%E7%A7%B0%E8%BD%B4%E4%B8%BAx%3D-1%E5%8F%88%E5%9B%A0%E4%B8%BAf%28-2%29%3Ef%282%EF%BC%89+%E5%9B%A0%E4%B8%BA-2%E6%AF%942%E8%B7%9D%E7%A6%BB%E5%AF%B9%E7%A7%B0%E8%BD%B4%E6%9B%B4%E8%BF%91+%E6%98%BE%E7%84%B6+a%3D-1-2x%5E2%2B2x-3%3D-%28x-1)
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设函数f(x)=ax^2+bx+c((a≠0),满足f(x+1)=f(-x-3),且f(-2)>f(2),解不等式f(-2x^2+2x-3)>f(x^2+4x+3)因为f(x+1)=f(-x-3) 所以f(1)=f(-3)所以f(x)对称轴为x=-1又因为f(-2)>f(2) 因为-2比2距离对称轴更近 显然 a=-1-2x^2+2x-3=-(x-1
设函数f(x)=ax^2+bx+c((a≠0),满足f(x+1)=f(-x-3),且f(-2)>f(2),解不等式f(-2x^2+2x-3)>f(x^2+4x+3)
因为f(x+1)=f(-x-3) 所以f(1)=f(-3)
所以f(x)对称轴为x=-1
又因为f(-2)>f(2) 因为-2比2距离对称轴更近 显然 a=-1
-2x^2+2x-3=-(x-1/2)^2-13/4f(x^2+4x+3)
★显然:-2x^2+2x-3比x^2+4x+3距离对称轴更近一些
-1-(-2x^2+2x-3)
设函数f(x)=ax^2+bx+c((a≠0),满足f(x+1)=f(-x-3),且f(-2)>f(2),解不等式f(-2x^2+2x-3)>f(x^2+4x+3)因为f(x+1)=f(-x-3) 所以f(1)=f(-3)所以f(x)对称轴为x=-1又因为f(-2)>f(2) 因为-2比2距离对称轴更近 显然 a=-1-2x^2+2x-3=-(x-1
首先可以确定对称轴是X=-1,由于f(-2)>f(2),可以确定a-1时根据单调性解-2x^2+2x-3>x^2+4x+3.当x
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