By a man
k=6
4a+2b-2=3 ( 1)
tanADE= 1/2
yA/(xA-xD)= 1/2
3/(2-xD)= 1/2
xD=-4
The coordinates of point D are: (-4,0)
16a-4b-2=0 (2)
(1) and (2) simultaneously solve:
a = 3/8; b= 1
The quadratic function is y = 3/8x 2+x-2.
The inverse proportional function is: y=6/x
2) Coordinate of point B: (0, -2)
The area of DMBE is equal to the sum of the areas of triangle BDE and AMB. Because BDE is a fixed point, only M is a moving point, and the area of triangle BDE is unchanged. When the area of triangle AMB is the largest, the area of quadrilateral is the largest, and the problem is transformed into the problem of finding the maximum distance from moving point M to fixed straight line AB.
Obviously, the straight AB equation is
x/4+y/2=- 1
x+2y+4=0
d=|xm+2ym+4|/√5
Where XM and ym are the coordinates of m points, and m satisfies the parabolic equation, and it is obtained by substitution:
d=|x+3/4x^2+2x-4+4|/√5=|3x^2+ 12x|*4/√5
The quadratic function 3x 2+ 12x takes the extreme value when x=- 12/6=-2 (minimum value, its absolute value is maximum).
......
3) When the abscissa is the same as point M, it is only the ordinate of straight line AB, and the AB equation can be found, so it is easy to find.
I'm sure you can handle the rest yourself. If you have any questions, please ask again. Please adopt it if it helps.