# `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`

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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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