唔識做功課38

It may be a little bit messy, sorry about that.
a) 0=-x^2+8x-7
0=x^2-8x+7
x=(-(-8)+sqrt((-8)^2-4*1*7))/(2*1) or (-(-8)-sqrt((-8)^2-4*1*7))/(2*1)
x=(8+sqrt(64-28))/2 or (8-sqrt(64-28))/2
x=(8+sqrt(36))/2 or (8-sqrt(36))/2
x=(8+6)/2 or (8-6)/2 (these steps are just using the quadratic equation)
x=7 or 1
the coordinates of A and B are (7,0) or (1,0)

b) the area of triangle AMB
=AB*height*0.5
=AB*q*0.5
(Because q>0, M lies on Quadrant I or II, So that the height is the value of q (q-0) )
=(7-1)*q*0.5
=3q

c)To find the max area of the triangle, we must the max height as the base can't be changed
y=-x^2+8x-7
y=-(x^2-8x+7)
y=-(x^2-8x+16-16+7)
y=-((x-4)^2-9)
y=-(x-4)^2+9
so that the highest point is (4,9) (I suppose you know what's going on)
the max area
=3q
=3*9
=27 sq.unit

If you still have questions, please feel free to ask.