Find the slope of the tangent to the curve y=3+4x^2-x^3 at the point where x=a. If anyone knows how to solve this problem and can give step-by-step instructions, that would be great. Thanks.
y = 3 + 4x^2 - x^3
The derivative of this is the slope at any point, a
So
y' = 0 + 8x - 3x^2 simplify
y' = -3x^2 + 8x
so...the slpoe at x = a is
-3(a)^2 + 8a
+1904 
