How do I find the image of the functions $y=2$ and $y = 2x - 6$? The domain of the following function $$y=2$$ is just 2? And the image of it?
I don't think I quiet understand what the image of a function means. The domain is all values that it can assume, correct?
Could you please try to define the image of this equation too: $$y = 2x - 6$$ so I can try to understand it?
 A: This really depends on the context of the question.  I'm going to make the assumption here that you're discussing functions from the real numbers to the real numbers.  Strictly speaking, the domain of the function should be given explicitly when you define the function.  However, often one defines a function and then says that the domain is the set of all real numbers for which the function is defined.  In this interpretation, the domain of your function $y=f(x)=2$ is all of $\mathbb{R}$ and its image is $\{2\}$.  Similarly, the domain of $y=f(x)=2x-6$ is all of $\mathbb{R}$ and its image is $\mathbb{R}$.
To give you a different example, the domain of the function $y=f(x)=\frac{1}{x-1}$ is $\mathbb{R} \setminus \{1\}$ and the image is $\mathbb{R} \setminus \{0\}$.
A: The image of a function is simply the set of all possible values the function can take. If you have a function $f(x)$, and evaluate it for all possible values of $x$ (i.e., for all $x$ in the domain of f), then the resulting set of values is called the image of $f$.
You can also find a image of a certain subset of the domain - i.e., instead of evalulating $f(x)$ for all $x$ in the domain, you only evaluate for a certain subset.

Formally, the image $f(U)$ of some subset $U$ of the domain of $f$ is
  $$
  f(U) = \left\{f(x) \,:\, x \in U\right\} \text{.}
$$

To check whether $y$ lies in the image of $f$, you thus have to ask

Is there an $x$ such that $f(x)=y$?

To check whether $y$ lies in the image of some subset $U$ of the domain of $f$, you similarly ask

Is there an $x \in U$ such that $f(x)=y$?

The image all sets $U$ which aren't empty under $f(x)=2$ is thus simply $\{2\}$, because $f$ never takes any value other than 2. 
A: When you define a function you should always provide a domain (i.e. a start set) and a co-domain (i.e. a target set). Writing
$$
  f:A\rightarrow B
$$
means that $A$ is the domain and $B$ the co-domain. $f$ is a function if it "associates" every single element $a\in A$ with a unique element $b\in B$. We call such $b$ the image of $a$ and denote it by $f(a)$. Then the image of $f$ is the set of all images.
When talking about real valued functions such as $f(x)=2x-6$, it is implied that the domain is the largest subset of $\mathbb R$ for which $f$ is defined (e.g. the domain of $f(x)=\sqrt x$ is $\{x\geq 0\}$), and that the co-domain is $\mathbb R$.
Now, regarding your examples, $y=2$ is short for the (real valued) function $f(x)$ defined as
$$
\begin{array}{crcl}
  f: & \mathbb R & \longrightarrow & \mathbb R \\
     &  x & \longmapsto & 2
\end{array}
$$
(note that it is well defined for every $x\in\mathbb R$, so that by the convention above the domain of $f$ is the whole $\mathbb R$). The image of $f$ is just the set $\{2\}$, since no other values of the co-domain are "reached" by $f$.
Concerning $f(x)=2x-6$, since no specification is given, as before we assume
$$
\begin{array}{crcl}
  f: & \mathbb R & \longrightarrow & \mathbb R \\
     &  x & \longmapsto & 2x-6
\end{array}
$$
i.e. that the domain is the largest subset of $\mathbb R$ for which the function is defined (so, again the whole $\mathbb R$) and that the co-domain is the whole $\mathbb R$. Now, for example the image of the element 5 is 4, since $f(5)=4$. It is easy to show that the image of $f$ is the whole co-domain $\mathbb R$ (in which case we say that $f$ is surjective, which is not the opposite of injective!). In general, you can find the image of $f$ by "turning $y=f(x)$ in something like $x=\ldots$" where the rhs is an expression that does not contain the variable $x$, and then see for which values of $y$ the equation has sense and the rhs is part of the domain. In this example,
$$
y=2x -6 \Rightarrow 2x=y+6 \Rightarrow x= \frac{y+6}2
$$
and the last expression makes sense (i.e. can be computed) for every $y\in\mathbb R$. We then conclude that the image of $f$ is the whole $\mathbb R$.
If we "restricted" the function as follows:
$$
\begin{array}{crcl}
  f: & [0,+\infty) & \longrightarrow & \mathbb R \\
     &  x & \longmapsto & 2x-6
\end{array}
$$
then this time finding its image would result in
$$
x= \frac{y+6}2 \quad\text{and}\quad x\geq 0
$$
since now the domain in which $x$ lies is $[0,+\infty)$, i.e.
$$
\frac{y+6}2\geq 0 \Leftrightarrow y\geq -6
$$
Therefore, in this case the image of $f$ would be $[-6,+\infty)$.
I hope this example makes it clear.
A: An image is a subset of the co-domain with respect to a certain  pre-image, which is a subset of the domain. For the function $y=2x-6$, for example, given a pre-image of $[2,10]$, the image is $[-2,14]$. For the function $y=2$, since any input value in the domain will result in $y=2$, besides the null set, the only possible pre-image is ${2}$, and the only possible image is $2$.
A: The value of f when applied to x is the image of x under f. y is alternatively known as the output of f for argument x.
Now Y=2 is a constant function. The image of Y is always 2 for any value of x.
So the Image of function Y=2 is the set containing element 2.
Image of 2, under Y=2x−6, is the set containing the element 2(2)-6=4-6=-2. 
