已知抛物线y=-0.5x+(5-m)x+m-3的对称轴是y轴，求抛物线的顶点坐标

对称轴y轴→b=0→m=5→y=-0.5x^2+2
→y(x=0)=2→顶点(0,2)

y = -1/2x^2 + (5-m)x + m-3

y = -1/2[x^2 - 2(5-m)x-2(m-3)] = -1/2(x - (5-m))^2 - 1/2((5-m)^2 - 2(m-3))

顶点为(5-m,-1/2((5-m)^2-2(m-3)))

因为对称轴为y轴,所以顶点横坐标5-m = 0, m = 5

所以顶点为(0,2)