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When a
first quadrant
When a is less than 0, which quadrant is the parabola y = x ^ 2 + 2aX + 2A ^ 2
Why
y=x²+2ax+2a²
=(x+a)²+a²
When x = - A, A0
y=a²>0
The axis of symmetry is x = - a A0
So the image is in quadrants one, two, four
If the vertex of parabola y = x ^ 2-2ax + A ^ 2 + A + 1 is in the second quadrant, the value range of constant a is
A-11 C -1
∵ axis of symmetry x = B / - 2A
∴x=a
∵ the function is in the second quadrant
∴a＜0
So exclude BCD
So choose a
The image of the function y = 2x + 1 passes through the vertex of the parabola y = x2-2ax + 1 to find a
The vertex of the parabola y = x2-2ax + 1 is (a, 1-A & sup2;) (according to the vertex formula - B / 2a, 4ac-b & sup2. / 2a), which is brought into y = 2x + 1, a = 0 or a = - 2 can be obtained
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