Parametric Equation Problem The problem is, "to determine any differences between the curves of the parametric equations. Are the graphs the same? Are the orientations the same? Are the curves smooth? Explain."
(a) $x=t;\quad y=2t+1$
(b) $x=\cos\theta;\quad y=2\cos\theta +1$
(c) $x=e^{-t};\quad y=2e^{-t}+1$
(d) $x=e^t; \quad y=2e^t+1$
The only one I am having difficulty with is (b). The graph provided in the answer key is:

Why are the arrows pointing in both directions? Is there no orientation? Also, they describe the curve as not being smooth. Why is this?
EDIT:
I have another question; the criteria given in the problem above is the same for this problem.
The parametric equations are, (a) $x=\cos\theta$, $y=2\sin^2\theta$
(b) $x=\cos(-\theta)$, $y=2\sin^2(-\theta)$
where $0<\theta<\pi$.
With the information, $-1<\cos\theta<1 \implies -1<x<1$
and $0<\sin\theta<0\implies0<2\sin^2\theta<0\implies0<y<0$ 
In the answer key, however, they have $-1\le x\le1$ How did they get the "equal to" component in there?

Though,each of the parametric equation reduce to $y=2x+1,$ the ranges of the values of $x$ for the real values of $t,\theta $ will differ.
In $(a),x$  cam assume any real values.
But in $(b),-1\le x\le 1$
In $(c),(d),x>0$ for the real values of $t$.
Using this, each curve has orientation zero.
