I need help with this 2 dimensional kinematics equation

A particle's trajectory is described by x=(1/2t^3-2t^2) m and y=(1/2t^2-2t) m, where t is in s.

So what is the the trajectory's direction at t=0 and at t=4? Measure it as an angle from the x-axis

x=(1/2t^3-2t^2) and y=(1/2t^2-2t)

trajectory's direction = dy/dx = dy/dt * dt/dx
y=(1/2t^2-2t)
dy/dt = t -2
dx/dt = 3/2 t^2 - 4t

trajectory's direction = dy/dx = dy/dt * dt/dx
= (t-2) / ( 3/2 t^2 - 4t)

at t=0
trajectory's direction = (t-2) / ( 3/2 t^2 - 4t)
= -2 / 0 =infinity = tan-1 A
A = 90 deg
answer

at t=4
trajectory's direction = (t-2) / ( 3/2 t^2 - 4t)
= 2 / (24 - 16) = 0.25
A= tan -1 0.25 = 14 deg
answer