Does $y = (-1)^x$ where $x∈ℝ$, change exponentially? Is $y = (-1)^x$ an exponential curve, or just a sinusoidal one, can it be said to change exponentially as with positive exponents?

I'm sure W/A showed this as being sinusoidal with an integer period. But now shows the answer as -1.
From the answers I understand that their is no definitive definition but their is an equivalent with complex numbers, where instead of the real number -1, the equivalent complex number {-1, 0} is used, and that is sinusoidal as described above.
 A: If you are talking about $-1^x$, then no: $-1^x = -1$.
The expression $(-1)^x$ on the other hand is not defined for most $x\in \mathbb{R}$.
If you mean the complex-valued function $e^{i\pi x}$ by $(-1)^x$, then you get:
$$\Re(e^{i\pi x}) = \cos(\pi x)$$
$$\Im(e^{i\pi x}) = \sin(\pi x)$$
That's probably, what Wolfram Alpha shows you.
A: Here is a fun bit of insanity that occurs if we allow formal manipulations of negative numbers raised to a power: 
Let $x>0$.  Then $x$ can be approximated by a sequence of rationals $\left\{\frac{2a_n}{2b_n+1}\right\}_{n=1}^{\infty}$, where $a_n$ and $b_n$ are positive integers defined for $n \in \{1, 2, 3\ldots\}$, so that: 
$$ x = \lim_{n\rightarrow\infty} \left[\frac{2a_n}{2b_n+1} \right]\: \: \: \mbox{(Equation 1)} $$
See paragraph at bottom for a formal proof of this. 
Now for each $n \in \{1, 2, 3, \ldots\}$ we have: 
$$ (-1)^{\frac{2a_n}{2b_n+1}} = ((-1)^{2a_n})^{\frac{1}{2b_n+1}} = (1)^{\frac{1}{2b_n+1}} = 1 $$
where the second equality uses the fact that $(-1)^{even} = 1$, and the last equality uses the fact that any odd root of 1 is 1.  Assuming continuity, we get:
$$ (-1)^x = \lim_{n\rightarrow\infty} (-1)^{\frac{2a_n}{2b_n+1}} = 1 $$
This holds for all $x>0$, including $x=3$, and so $(-1)^3 = 1$. But of course $(-1)^3=-1$.  So $1=-1$.  

To understand why (Equation 1) holds, first note that $x$ can be approximated by a sequence of rationals $\{p_n/q_n\}_{n=1}^{\infty}$ such that $p_n$ and $q_n$ are positive numbers for all $n \in \{1, 2, 3, \ldots\}$, and $\lim_{n\rightarrow\infty} p_n/q_n=x$, so it follows that: 
$$\lim_{n\rightarrow\infty} \frac{2^np_n}{2^nq_n+1} =\lim_{n\rightarrow\infty} \frac{p_n}{q_n}=x$$ 
We can simply define $a_n = 2^{n-1}p_n$ and $b_n = 2^{n-1}q_n$, so that (Equation 1) holds. 
