x=t+1 , y=t^2+3t Find all points (if any) of horizontal and vertical tangency to the curve.

Parametric curve (x(t),y(t)) has a horizontal tangent if its slope dy/dx is zero i.e. when dy/dt=0 and dx/dt!=0
It has a vertical tangent, if its slope approaches infinity i.e. dx/dt=0 and dy/dt!=0
Given parametric equations are:
x=t+1
y=t^2+3t
dx/dt=1
dy/dt=2t+3
For Horizontal tangents,
dy/dt=0
2t+3=0
=>2t=-3
=>t=-3/2
Corresponding point on the curve can be found by plugging in the value of t in the equations,
x=-3/2+1=-1/2
y=(-3/2)^2+3(-3/2)
y=9/4-9/2
y=(9-18)/4=-9/4
Horizontal tangent is at the point (-1/2,-9/4)
For vertical tangents,
dx/dt=0
However dx/dt=1!=0
So there are no vertical tangents.