Find the equation of the line that is tangent to the curve..

I solved it. The solution is y=-3x 

(as b=0)

Thanks all for the help.

answer is 2

it is y = 2bx+ 2- 6 x 5mx b2

W*f is that. Ive done some complex math but that is something else

y = 3xcosx

and the slope at any point on this cure is given by the derivative of this function.....so we have....

y ' = 3cosx - 3xsinx  =   3[cosx -xsinx]

So, the slope at (pi, -3pi)  =

y '(pi)  = 3 [ cos (pi)  - (pi) sin (pi)]   =  3 [-1  -  pi*0]  =   3 [-1]  = -3

And the equation  of the tangent line at this point =

y - (-3pi)  = -3(x - pi)

y + 3pi  = -3x + 3pi

y = -3x