Question

Horizontal and vertical tangents for x=t^3-3t and y=t^3-3t^2?

Answer 1

even easier than the last one
horizontal \(\frac{dy}{dt}=0\) and vertical \(\frac{dx}{dt}=0\)

Answer 2

Oops, forgot to say I needed the points...

Answer 3

k set \(\frac{dy}{dt}=0\) that will give the value of \(t\) where the tangent is horizontal
once you have \(t\) you know \(x\) and \(y\)

Answer 4

I have the values for t but I'm not sure how to convert them into points...do I plug them into the original equations for the points?

Answer 5

yes

Answer 6

i get \(t=0, t=2\) for horizontal, but i didn't write it down
is that correct?