#### boylesg

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I get that the formula for a cubic equation is Y = aX3 + bX2 + cX + d

I get the concept of derivatives in calculus.

So the first derivative of a cubic equation has the form of Y = 3aX2 + 2bX + c and the second derivative has the form Y = 6aX + 4b.

So I remember from maths in the 80s that the derivative of a quadratic equation will give you the gradient of the curve at any point X.

**I have been trying to understand the process from written tutorials but have been finding it confusing. Perhaps a diagramatic description of the process would be easier to 'get'.**

*So I assume the second derivative of a cubic equation similarly gives you the gradient of the curve at an point X?*

Question 1: What does the first derivative of a cubic equation tell you about the curve if the second derivative gives you the gradient at any point X?

Question 2: How do you use the second derivative to calculate a series of lines that define a close match to the curve given an iteration interval between two control points?

Question 3: Given the control points, how do you work out the cubic equation that you can then work out the derivatives from?

Question 1: What does the first derivative of a cubic equation tell you about the curve if the second derivative gives you the gradient at any point X?

Question 2: How do you use the second derivative to calculate a series of lines that define a close match to the curve given an iteration interval between two control points?

Question 3: Given the control points, how do you work out the cubic equation that you can then work out the derivatives from?