Question

二次函数的顶点坐标公式 b/2a,c-b^2/4a,是咋推算出来的哦,我咋推出来是 b/2a,c-b^2/4a^2

Answer 1

配方

y=ax^2+bx+c

=a(x^2+bx/a)+c

=a[x^2+2*(b/2a)*x+(b/2a)^2-(b/2a)^2]+c

=a(x+b/2a)^2-a*b^2/4a^2+c

=a(x+b/2a)^2-b^2/4a+4ac/4a

=a(x+b/2a)^2+(4ac-b^2)/(4a)

=a[x-(-b/2a)]^2+(4ac-b^2)/(4a)

所以顶点是[-b/2a,(4ac-b^2)/(4a)]

对称轴是x=-b/2a