find the gradient of the curve 2t^4+3t^3-t+4 at the points (0,4) and (1,8)

Question

all method must be shown so that i will understand it very well please

Answer 1

First, differentiate the function: 8t^3+9t^2-1 which gives us the gradient at any point. 

When t=0, the function is 4 and when t=1, the function is 2+3-1+4=8 which confirms that the given points are on the curve.

We only need the t value to find the gradient at the given points, so at t=0 the gradient is -1 and at t=1 it's 8+9-1=16.