Ind the slope m of the tangent to the curve y = 9 + 4x2 − 2x3 at the point where x = a. m = 8a−6a2 (b) find equations of the tangent lines at the points (1, 11) and (2, 9). y(x) = 2x+11 (at the point (1, 11)) y(x) = (at the point (2, 9))

Find the derivative:-
y' = 8x - 6x^2
This is the slope in terms of x.
When x = a , the slope is
8a - 6a^2

(b)  equation of tangent line at x1,y1 
is  y- y1  =  (8x - 6x^2)( x - x1) , so at (1,11) it is
y - 11 = (8(1) - 6((1)^2) ) (x - 1)
y =  2(x - 1) + 11
y = 2x + 9 (answer)

At  (2,9) 
y - 9 = (16-24)(x - 2)
y - 9 = -8x + 16
y = -8x + 25  answer