A comprehensive function question, good addition

1) The function image opening is downward, there are m2-2 0

and the axis of symmetry x=2m (m 2-2)=2, the vertices of the quadratic function of m=-1y=-x+4x+n=-(x-2) 2+n+4 are on a straight line, and x=2 is substituted into a straight line, and y=2n+4=2, n=-2 is obtained

The analytic formula of this function is y=-x 2+4x-2

2) Let a(x1,y1), b(x2,y2), and the analytical formula of the translated function is y=-(x-s) 2+t=-x 2+2sx-s 2+t

x1-x2|^2=(x1+x2)^2-4x1x2=4s^2+4(t-s^2)=(2ym)^2=4t^2

The solution yields t=0 or 1

When t=0, there is no triangle ABM

When t=1, substitute the straight line to get s=0

y=-x^2+1

It is obtained by the function y=-x 2+4x-2=-(x-2) 2+2 by 2 units to the left and one unit down.

1)1..The coefficient of the quadratic term obtained from the opening downward is less than 0, that is, m 2-2 0, and the solution m is greater than - root number 2, which is less than root number 2

2..From the axis of symmetry, -b 2a=2 is obtained, and the values of a,b (quadratic and primary coefficients) are substituted to obtain m as 2 or -1, and m=-1 is obtained by combining the above

So now the formula becomes y=-x 2+4x+n

3..Knowing that the fixed-point abscissa is 2 (the axis of symmetry is 2), we get y=2 by substituting x=2 into the analytic formula of the line, so (4ac-b 2) 4a=2, and bring a, b, and c into it, and we get n=-2

—y=-x^2+4x-2

2) Translate the quadratic function image, and the coefficient of the quadratic term remains unchanged, so let the analytical formula of the translated function be y=-x 2+bx+c

am=bm is determined, so m is a right angle, at this time the vertex coordinates m(b 2,(4c+b 2) 4), according to some properties of the isosceles triangle, the coordinates of a can be obtained as (b 2-(4c+b 2) 4,0), b(b 2+(4c+b 2) 4,0), using Vedr's theorem, x1 x2=c a, b 2 4=-c, and m's coordinates are substituted into the analytic formula of the straight line, which is reduced to b + 4 = 4c + b 2 , joint , solution obtained, b = 0, c = 1

The analytic formula is found out, how to move you can do it yourself! What an annoying question.

1. Because the abscissa of point p is 1, according to the proportional function, the ordinate can be known to be root number three, then, k is 1 * root number three = root number three.

2. Let the coordinates of point a be (x, root number three x), then point b is (x, root number three x), point c is (1 x, root number three x), ab = root number three x-1 x = root number three (x-1 x), ac = x-1 x, because the value of ac ab is unchanged, angle a is a right angle, so the degree of angle b is certain. And because ab ac = root number three, so according to the trigonometric function, we can see that b = 30°

3. When BC bisects ABP, ABP=2 ABC=60°, because A=30°, BP AP, so YBC=- root number 3 3X+4 root number 3 3, so B: (3, root number 3 3), then A: (3, 3 root number 3).