00:01
For this problem, we want to determine the points on the curve where the tangent is horizontal or vertical.
00:07
Now here we are given parametric curves.
00:10
So to determine the points where the tangent is horizontal, we need to set dy over dt to zero and solver t.
00:21
Now, dy over dt that's equal to 6t squared plus 6t, t, and setting this to 0, we have, 0 equal to 6t squared plus 60.
00:34
We factor out 60, we have 0 equal to 6t times t plus 1.
00:41
We have t equals 0 or t equals negative 1.
00:47
So if t equals 0, the x value would be x equal to 2 times 0 cubed plus 3 times 0 squared minus 72 times 0 that's equal to 0 and for the y value that'll be 2 times 0 cubed plus 3 times 0 squared plus 5 that's equal to 5 so we have the point 0 5 and if t equals 1 we have x value equal to 2 times 1 cubed plus 3 times 1 squared minus 72 that's equal to negative 67 and for the y value we have 2 times 1 raised of the third power plus 3 times 1 squared plus 5 that's equal to 10.
01:37
So the point would be negative 67 10...