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`x=t^2-t+2 , y=t^3-3t` Find all points (if any) of horizontal and vertical tangency to the curve.

Parametric curve (x(t),y(t)) has a horizontal tangent when its slope `dy/dx` is zero, i.e. `dy/dt=0` and `dx/dt!=0` .


Curve has a vertical tangent if its slope approaches infinity i.e. `dx/dt=0` and `dy/dt!=0`


Given equations of the parametric curve are:


`x=t^2-t+2`


`y=t^3-3t`


`dx/dt=2t-1`


`dy/dt=3t^2-3`


For horizontal tangents:


`dy/dt=0`


`3t^2-3=0`


`=>3t^2=3`


`=>t^2=1`


`=>t=+-1`


Corresponding points on the curve can be found by plugging the values of t in the parametric equation,


For t=1,


`x_1=1^2-1+2=2` 


`y_1=1^3-3(1)=-2`


For t=-1,


`x_2=2^2-2+2=4`


`y_2=2^3-3(2)=2`


Horizontal tangents are at the points (2,-2) and (4,2)


For vertical tangents,


`dx/dt=0`


`2t-1=0`


`=>t=1/2`


Corresponding points on the curve for `t=1/2` are,


`x=(1/2)^2-1/2+2`


`x=1/4-1/2+2`


`x=(1-2+8)/4`


`x=7/4`


`y=(1/2)^3-3(1/2)`


`y=1/8-3/2`


`y=(1-12)/8`


`y=-11/8`


Vertical tangent is at the point `(7/4,-11/8)`

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