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A quartic Bézier curve is determined by five control points, $\mathbf{p}_{0}, \mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3},$ and $\mathbf{p}_{4} :$$$\begin{aligned} \mathbf{x}(t)=(1-t)^{4} \mathbf{p}_{0} &+4 t(1-t)^{3} \mathbf{p}_{1}+6 t^{2}(1-t)^{2} \mathbf{p}_{2} \\ &+4 t^{3}(1-t) \mathbf{p}_{3}+t^{4} \mathbf{p}_{4} \quad \text { for } 0 \leq t \leq 1 \end{aligned}$$Construct the quartic basis matrix $M_{B}$ for $\mathbf{x}(t)$
$M_{B}=\left[\begin{array}{ccccc}{1} & {-4} & {6} & {-4} & {1} \\ {0} & {4} & {-12} & {12} & {-4} \\ {0} & {0} & {6} & {-12} & {6} \\ {0} & {0} & {0} & {4} & {-4} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right]$
Calculus 3
Chapter 8
The Geometry of Vector Spaces
Section 6
Curves and Surfaces
Vectors
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all right, You can actually see the answer to the question over here on my screen. But I've already put the points into my virtual graphing calculator here. If I press stat and one for edit, there are the points and you put them in by doing left and right, left button and the right button and down and up. And you can just enter them it like this. Three enter for enter. If I wanted to get that negative seven, I just go over and up. Negative seven. Enter. Okay, they're all in there. So now I go back to stat, I go to calculate by going over, and then I scrolled down until I get two cubic regression six. Enter. And there it is. I think I just demonstrated for you that it is possible to make mistakes. So, um, that is not as far as I can tell. The correct answer. Um oh, I'm supposed to use quart IQ regression. So now I understand my mistake. Oops. I just want press the stop button over to calculate go down a quart IQ, regression it. Enter, hit, enter. And now I have the correct cancer. So that's a demonstration of If you make a mistake, just ah, think again. Go back to the beginning and consider all that you've done. Um, because there is a mistake somewhere, and I believe in you. You confined.
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