Graphing the function $(-2)^x$ When I wanted to graph $y=(-2)^x$ many graphing calculator apps refused to plot it. TI-Nspire CAS plotted it as shown in the first picture. I think the plot is not correct as only the envelopes should be there with no values between the envelopes as shown in the second picture and the $(-2)^x$ should look like $2^x$ and $-(2^x)$ plotted on the same graph but of course with many discontinuities as explained in my analysis below. Am I right? 
$(-2)^x$ as graphed by the TI-Nspire
this is what I think it should be
Here’s my analysis of the function: 
f(x)=(-2)^x 
First: when x >=0 


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*if x is an integer, x>=0 


(-2)^0=1 
(-2)^1=-2
(-2)^2=4
(-2)^3=-8
(-2)^4=16 
(-2)^x oscillates back and forth 
When x is an even integer (-2)^x is positive. 
When x is an odd integer (-2)^x is negative. 


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*when x is a rational number, x>0 
let x=p/q    , p>0, q>0 


(-2)^x=(-2)^(p/q)=((-2)^p)^(1/q) 
if p is even and q is odd, (-2)^x is a positive real value 
Example: (-2)^(100/51)= 3.8927
if p is odd and q is odd, (-2)^x is a negative real value. 
Example: (-2)^(99/51)=-3.8402
if p is odd and q is even, (-2)^x is an imaginary value (not defined in the set of real numbers) 
Example: (-2)^(99/50)=i 3.9449 
(Also in all the above cases, if you extend your analysis to include complex, we get q complex roots.) 
In the domain of rational numbers, (-2)^x oscillate or is undefined (imaginary or complex) 


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*When x is irrational, x>0 
(-2)^x has no real value. It has an infinite number of complex roots. 


Second: when x<0 
(-2)^x=1/(-2)^|x| 
Use the same approach above to analyze the behavior of the function. 
 A: The graph of $f(x)=(-2)^x$ is problematic for real numbers $x$. Think about what happens when $x=\frac{1}{2}$. Then $f(x)=(-2)^{\frac 1 2}=\sqrt{-2}$. Can you see why this is a problem to graph?
The graph of $f(x)=(-2)^x$ only makes sense for integer values of $x$. Also, as zwim pointed out in the comments, your second graph is not the graph of a function, as it is multivalued, i.e. one input of $x$ gives two outputs of $f(x)$.
A: $(-1)^x$ as defined in the real domain is only making sense for $x\in\mathbb Q_\text{odd}$
$=\{\frac pq\mid p\in\mathbb Z,q\in\mathbb N^*,\gcd(p,q)=1,\text{ and }q\text{ is odd }\}$
Have a look at this post for instance :
Exponential Equation with base -1, 0, and 1
Thus the graph of $(-2)^x$ has holes on all points of ${\mathbb Q_\text{odd}}^\complement$.
More than a continuous graph with infinity of holes it is in reality more of a collection of isolated points in the plane. But I agree that thet are so close from each other that visually it looks like the curve you represented.
But if you could zoom up with infinite factor you would see a dashed line.
