The vertex of the parabola y = AX2 + BX + C is on the Y-axis and passes through two points - (- 1,3), (- 2,6)

The vertex of the parabola y = AX2 + BX + C is on the Y-axis and passes through two points - (- 1,3), (- 2,6)

The parabola of the vertex on the Y axis is symmetric about the Y axis,
It can be set as y = ax ^ 2 + B,
And then (- 1,3) and (- 2,6),
The equations are as follows
3=a+b
6=4a+b,
The solution is a = 1, B = 2,
The expression: y = x ^ 2 + 2

If the vertex of the parabola y = ax ^ 2 + BX + C is on the Y axis and passes through (- 1,3), (- 2,6), then the expression is

∵ the vertex of the parabola is on the y-axis
∴b=0
Bring two in. It's time
a=1 c=2
The expression is y = x ^ 2 + 2

Given that the parabola y = ax & # 178; + BX + C passes through one, two and three quadrants, then a 0, b 0, C 0, B & # 178; - 4ac 0
Why can't it be greater than 0, and 1.2.3 for = 0
b＞0，c＞0......＞0

Here, because a function passes through one, two and three quadrants, it means that it is definitely not a quadratic equation of one variable. It also excludes constant value function and positive proportion function, so only a function of one degree can pass through three quadrants. Therefore, the original formula is equivalent to y = KX + B (K ≠ 0). In this way, B in the parabola is 0. Because passing through one or three quadrants is an upward trend, a > 0, and passing through the second term line means intercept C > 0

If > 0, b > 0, C > 0, b-4ac > 0, then the parabola y = ax + BX + C passes through the fourth quadrant

If a > 0, b > 0, C > 0, B & # 178; - 4ac > 0,
Then the parabola y = ax & # 178; + BX + C passes through the first, second and third quadrants