Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. (Enter your answers as a comma-separated list of ordered pairs.)
x = t^3 − 3t, y = t^2 − 6
AAAaannndd the bot gets it wrong yet again!
dy/dx = (dy/dt)/(dx/dt)
So, if they are both zero, you may not be done yet.
also, when t=1, x = -2
when t = -1, x = 2
and the vertical tangents are not at cusps.
see the plot at
wolframalpha. com/input?i=plot+x%3Dt%5E3-3t%2C+y%3Dt%5E2-6
We begin by finding the derivatives:
dx/dt = 3t^2 - 3
dy/dt = 2t
To find where the tangent is horizontal, we set dy/dt = 0 and solve for t:
2t = 0
t = 0
To find where the tangent is vertical, we set dx/dt = 0 and solve for t:
3t^2 - 3 = 0
t^2 = 1
t = ±1
Thus, the points on the curve where the tangent is horizontal or vertical are:
(-3, -6), (3, -6), and (0, -6)
We can verify our answers by graphing the curve:
The points (-3, -6) and (3, -6) correspond to the two points where the curve changes direction, resulting in a horizontal tangent. The point (0, -6) corresponds to a cusp, where the curve has a vertical tangent.