Find the points on the curve where tangent is horizontal
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Take derivative of both and set them equal to zero:
x= t^4–4t
dx/dt = 4t^3–4=0
4t^3 = 4
t^3=1
t=1
Horizontal tangent plug t back into original equations
x= 1–4= -3
y = 1–2 = -1
(x,y) = (-3,-1)
y= t^3–2
dy/dt = 3t^2 -0=0
3t^2=0
t=0
x= t^4–4t
dx/dt = 4t^3–4=0
4t^3 = 4
t^3=1
t=1
Horizontal tangent plug t back into original equations
x= 1–4= -3
y = 1–2 = -1
(x,y) = (-3,-1)
y= t^3–2
dy/dt = 3t^2 -0=0
3t^2=0
t=0
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