00:01
For this problem, we want to find the xy points where the tangent is horizontal or vertical, given parametric equations, x equals t to the third minus 3t and y equals t squared minus 8.
00:15
First, we recall that the tangent is horizontal whenever the derivative, dy over d x, equals zero.
00:23
And the tangent is vertical if the derivative is undefined or whenever, the denominator is zero.
00:33
So our first step would be to find dy over dx and because we're dealing with parametric equations we recall that dy over dx this is equal to dy over dt all over dx over dt so if you take the derivative of x with respect to t we have dx over dt which is equal to 3 t squared minus 3, and dy over dt, this is equal to 2t.
01:12
So then our dy over dx, this is equal to 2t all over 3t squared minus 3.
01:24
So if we set this to 0, we have 2t all over 3t squared minus 0, that leaves us.
01:33
To 2 t equals 0 or that's t equals 0 which tells us that the tangent is horizontal whenever t is 0 so that'll be that will be when x equals 0 to the third minus 3 times 0 or 0 and y will be 0 squared minus 8 or that negative 8.
01:59
So the point there where the tangent is horizontal is 0 negative 8...