Match the parametric equations with the graphs labeledI-VI. Give reasons for you

Consider the parametric equation,

This equation's graph matches with that of (V).
As t goes to infinity, both x and y go to infinity. The graph must also contain the point ${\displaystyle (1,0)}$ achieved at ${\displaystyle t=0}$
This function should not have any oscillatory behavior.
The parametric equations match graph (V).

${\displaystyle x=t^2-2t,y=\sqrt{t}}$

The equation's graph matches with that of (I).
As f goes to infinity, both x and y go to infinity. The only graphs that have this behavior are (V), (I) and (IV). The graph must also contain the point ${\displaystyle (0,0)}$ achieved at ${\displaystyle t=0}$ Only graph (I) satisifies this requirement.

${\displaystyle x=\sin 2t,y=\sin (t+\sin 2t)}$

This equation's graph matches with that of (II).
The graph must contain the point ${\displaystyle (0,1)}$ achieved at ${\displaystyle t=\frac{\pi }{2}}$. Graph (II) is the only graph that satisifies this.

${\displaystyle x-\cos 5t,y=\sin 2t}$

This equation's graph matches with that of (IV).
The graph must contain the point ${\displaystyle (1,0)}$ achieved at ${\displaystyle t=0}$. Graph (IV) is the only graph that satisifies this.

${\displaystyle x=t+\sin 4t,y=t^2+\cos 3t}$

This equation;s graph matches with that of (IV).
As t goes to infinity, both x and y go to infinity. The the onli graph that have this behavior are (V), (I), and (IV). However, graph (IV) is the only one that shows any oscillatory behavior.

${\displaystyle x=\frac{\sin 2t}{4+t^2},y=\frac{\cos 2t}{4+t^2}}$

This equation's graph matches with that of (III).
The only graphs that exhibit oscillatory behavior are graphs (III), (II), (IV), and (VI). The graph must also satisfy the fact that the point ${\displaystyle (0,0)}$ cannot be a point on the graph. Graph (III) and (IV) satisfy this requirement.
For graph (IV), as t goes to infinity, both x and y go to infinity . This is not the case for these equations.