Question

已知关于x的一元二次方程ax平方+bx+c=3的一个根为x,=2，且二次函数y=ax平方+bx+c的对称轴是直线x=2,

则抛物线的顶点坐标为？ 要详细过程！！

Answer 1

即x=2
y=ax²+bx+c=3
所以过(2,3)
顶点就是对称轴x=2上
所以顶点是(2,3)

Answer 2

1、3x^2+19xy+6y^2
=3x^2+18xy+xy+6y^2
=3x(x+6y)+y(x+6y)
=(x+6y)(3x+y)

2、ax^n+1-ax^n-20ax^n-1
=ax^(n-1)(x^2-x-20)
=ax^(n-1)(x-5)(x+4)

3、(1-x^2)(1-y^2)+4xy
=x^2y^2-x^2-y^2+1+2xy+2xy
= (x^2y^2+2xy+1)-(x^2-2xy+y^2)
=(xy+1)^2-(x-y)^2
=(xy+x-y+1)(xy-x+y+1)

Answer 3

3x^2+19xy+6y^2
=3x^2+18xy+xy+6y^2
=3x(x+6y)+y(x+6y)
=(x+6y)(3x+y)