求几个一元二次函数的顶点坐标y=5x²+8xy=-4x²-8xy=-1/5x²-5x+1y=6x²-x+1

Question

求几个一元二次函数的顶点坐标
y=5x²+8x y=-4x²-8x y=-1/5x²-5x+1 y=6x²-x+1

Answer 1

配方即得到顶点
1）y=5x²+8x=5(x²+8x/5)=5[(x+4/5)²-16/25]=5(x+4/5)²-16/5
顶点为(-4/5,-16/5)
2)y=-4x²-8x=-4(x²+2x)=-4[(x+1)²-1]=-4(x+1)²+4
顶点为(-1,4)
3)y=-1/5x²-5x+1=-1/5(x²+25x)+1=-1/5[(x+25/2)²-625/4]+1=-1/5(x+25/2)²+129/4
顶点为(-25/2,129/4)
4)y=6x²-x+1=6(x²-x/6)+1=6[(x-1/12)²-1/144]+1=6(x-1/12)²+23/24
顶点为(1/12,23/24)