Answer:
There is a horizontal tangent at (0,-4)
The tangent is vertical at (-2,-3) and (2,-3).
Step-by-step explanation:
The given function is defined parametrically by the equations:
[tex]x=t^3-3t[/tex]
and
[tex]y=t^2-4[/tex]
The tangent function is given by:
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
[tex]\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}[/tex]
The tangent is vertical at when [tex]\frac{dx}{dt}=0[/tex]
[tex]\implies \frac{3t^2-3}{2t}=0[/tex]
[tex]\implies 3t^2-3=0[/tex]
[tex]\implies 3t^2=3[/tex]
[tex]\implies t^2=1[/tex]
[tex]\implies t=\pm1[/tex]
When t=1,
[tex]x=1^3-3(1)=-2[/tex] and [tex]y=1^2-4=-3[/tex]
When t=-1,
[tex]x=(-1)^3-3(-1)=2[/tex] and [tex]y=(-1)^2-4=-3[/tex]
The tangent is vertical at (-2,-3) and (2,-3).
The tangent is horizontal, when [tex]\frac{dy}{dx}=0[/tex] or [tex]\frac{dy}{dt}=0[/tex]
[tex]\implies 2t=0[/tex]
[tex]\implies t=0[/tex]
When t=0,
[tex]x=0^3-3(0)=0[/tex] and [tex]y=0^2-4=-4[/tex]
There is a horizontal tangent at (0,-4)
