x - 2 < 0
x < 2
For the quadratic to be less then zero then 4/3 < x < 2
b(i)
dy/dx = 20x - 6x^2 - 16
For the tangent to be parallel to the x-axis the gradient must be zero. at P
Substituting in P (2)
20(2) - 6(2)^2 - 16 = 40 - 24 - 16 = 0
So the tangent is parallel to the x-axis at P
(ii)
Then tangent at P has the equation y = 3
The slope at Q = 20(3) - 6(3)^2 - 16 = 60 - 54 - 16 = -10
Therefore the normal slope is 1/10
Eq'n of the normal at Q is
y --1 = 1( x -3) / 10
y + 1 = x/10 - 3/10
y = x/10 - 13/10
Since P & Q intersect at R then
y = 3
y = x/10 = 13/10
3 = x /10 -13/10
30 = x - 13
x = 43
When x = 43
y = 43/10 - 13/10
y = 30/10
y = 3
So R (x,y) = ( 43, 3)
I hope that helps 'me'!!!!
Ha!! Ha!! Ha!!
by 'you' (lenpol7)