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

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