00:01
In this question, we are asked to find the points where the tangent line to the given parameter curve is horizontal and vertical.
00:07
So first, let's find the slope of the tangent line.
00:13
It equals to dy or dt, divide by dx over dt.
00:20
D .y or dt equals to 2t and dx over dt equals to 3t squared minus 3.
00:28
Or we can write this as 2t divided by 3 multiplied by t minus 1 times t plus 1.
00:35
So this is a slope of the tangent line.
00:40
Now recall that the tangent line is horizontal if dy or dx is 0.
00:55
The tangent line is vertical if dy or dx is either infinity or negative infinity.
01:09
Now let's start with the horizontal tangent line.
01:16
We want dy or dx to be equal to 0 and dx equals to 2t divided by 3 t minus 1 times t plus 1.
01:33
And we want this to be equal to 0...