It is known that the vertex m of the parabola y = AX2 + BX + C with downward opening is in the second quadrant And through the points a (1,0), B (0,1) (1) Please judge the value range of real number a and explain the reason (2) Let the other intersection of the parabola and the x-axis be c. when the area of △ AMC is 25 / 16 times of the area of △ ABC, a is obtained

It is known that the vertex m of the parabola y = AX2 + BX + C with downward opening is in the second quadrant And through the points a (1,0), B (0,1) (1) Please judge the value range of real number a and explain the reason (2) Let the other intersection of the parabola and the x-axis be c. when the area of △ AMC is 25 / 16 times of the area of △ ABC, a is obtained

(1) A is less than 0
(2) From (0,1), C is 1
From (1,0), B = - 1-A
Because the area of △ AMC is 25 / 16 times that of △ ABC
So, (4ac-b ^ 2) / (4a) = 25 / 16
Substituting B and C into a = - 4 or a = - 1 / 4
And because the vertex is in the second quadrant
So a = - 4 is not suitable
The final result is a = - 1 / 4

When m is the value, the vertex of the parabola y = x & # 178; + MX + M-1 is on the coordinate axis

y=x²+mx+m-1=(x+0.5m)²-0.25m²+m-1
-0.25m²+m-1=0
m=2

The vertex of parabola y = x2-2ax + 9 is on the coordinate axis. Try to find out the value of A

A = plus or minus 3
X = - B / 2a, that is, this point is the vertex
A is 1 in this problem
B is - 2A
In other words, when x takes a, the vertex is on the coordinate axis and Y is 0
Then 0 = a2-2axa + 9
A = plus or minus 3

When a < 0, find the quadrant of the vertex of the parabola y = x2 + 2aX + 1 + 2A2

∵ y = x2 + 2aX + 1 + 2A2 = (x + a) 2 + A2 + 1, the vertex coordinates of the parabola are (- A, A2 + 1), ∵ a ＜ 0, ∵ a ＞ 0, and ∵ A2 + 1 ＞ 0, the vertex of the parabola is in the first quadrant