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.
y = t^2-6
dx/dt = 3t^2 -3
By the chain rule, you get dy/dx = 2t/ (3t^2-3).
Horizontal tangent lines happen when dy/dx= 0, so when 2t=0, so t=0. plugging t= 0 back to the x and y equations, x=0, y= -6, so (0, -6)
Vertical tangent lines happen when dy/dx is undefined, so when 3t^2-3, so when t= -1 and when t=1. plugging back to get x and y, we get (2,-5) and (-2, -5)
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