Parametric
x= t^4 + 4t^3 - 8t^2
y= 2t^2 - t
dy/dt = 4t - 1
dx/dt = 4t^3 + 12t^2 - 16t
Horizontal tangents occur when dy/dt = 0, provided dx/dt is not zero at that point
so for horizontal tangent:
4t - 1 = 0
=> t = 1/4
so a horizontal tangent occurs when t = 1/4
Vertical tangents occur when dx/dt = 0
so for vertical tangent:
4t^3 + 12t^2 - 16t = 0
=> 4t(t^2 + 3t - 4) = 0
=> 4t(t + 4)(t - 1) = 0
=> t = 0, t = -4, t = 1 .........the vertical tangents occur at these values of t.
You may want to see Tangents with Parametric Equations
to find the points that the tangents occur, just plug in the values we found for t into the original x and y equations, you will get coordinate points for each value of t that way