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Explain why a cubic Bézier curve is completely determined by $\mathbf{x}(0), \mathbf{x}^{\prime}(0), \mathbf{x}(1),$ and $\mathbf{x}^{\prime}(1)$

$\mathbf{p}_{0}=\mathbf{x}(0), \mathbf{p}_{3}=\mathbf{x}(1), \mathbf{p}_{1}=\frac{1}{3} \mathbf{x}^{\prime}(0)+\mathbf{p}_{0}, \mathbf{p}_{2}=\mathbf{p}_{3}-\frac{1}{3} \mathbf{x}^{\prime}(1)$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 6

Curves and Surfaces

Vectors

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Okay, so let's say we have a cubic death of X equals Hey X cubed plus B Thanks. Squared plus see X close d Okay. And we once and just pretend like a B and C and G or just numbers. And we'LL try to find the inflection points and just see what happens. And of course we are assuming that a is not zero. So it's a proper cubic a zero. This would actually be quadratic and we studied quadratic completely isn't quadratic is completely in a previous problem. So this is a proper cubic a non equal to zero. And so if we want to find inflection points, we need to take second derivative six a x plus two B All right, but then when is if the crime equal to zero? This is a line away. The second derivative is a line, so we know it's going to pass through do and A is not zero. So it's it's a line with slope not equal to zero. So absolutely the second group. But it is going to have exactly ones here, and that's going to be a X equals. Um so we just said this equal to zero. We want X to be negative B over three. You're always gonna have an inflection point for every proper cubic and X equals negative B over three.

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