# Strange tangent for parametric

For the parametric equations x = t^3 - 3t and y = t^3 - 3t^2 I got that the graph has a vertical tangent when t is = to postive or negative one. And it is horizontal at t = 2. However, this implies that at the point (x,y) = (2, -4) the graph has both a vertical and horizontal tangent. How is this possible? Thanks

jedishrfu
Mentor
How are you computing the tangents?

Mark44
Mentor
Just show us your work, not an image of it, especially one that doesn't render.

Mark44
Mentor
For the parametric equations x = t^3 - 3t and y = t^3 - 3t^2 I got that the graph has a vertical tangent when t is = to postive or negative one. And it is horizontal at t = 2. However, this implies that at the point (x,y) = (2, -4) the graph has both a vertical and horizontal tangent. How is this possible? Thanks
When t = 2, x = 2 and y = -4, just as you say. And dy/dx = 0 when t = 2, so the tangent is horizontal at (2, -4). Why do you think that the tangent is vertical when t = 2?

My friend from math class actually explained it to me, but thank you for your help. At the point (2, -4), the graph crosses itself, so as a result there are two tangents at that spot.

Mark44
Mentor
My friend from math class actually explained it to me, but thank you for your help. At the point (2, -4), the graph crosses itself, so as a result there are two tangents at that spot.
Right, but they occur for two different values of t.

When t = -1, (x, y) = (2, -4) and the tangent is vertical.
When t = 2 you get (2, -4) again, but this time with a horizontal tangent.

To find the values of t, I solved the equation t3 - 3t = 2, or t3 - 3t - 2 = 0, which in factored form is (t + 1)2(t - 2) = 0.

Thank you so much, I understand now