How does shift and scaling inside of a function affect its Fourier Transform?
For one-dimensional Fourier transforms,
$$\begin{align*}
x(t) &amp;\leftrightarrow X(f)\\
x(t-a) &amp;\leftrightarrow X(f)\exp(-j2\pi fa)\\
x(bt) &amp;\leftrightarrow \frac{1}{|b|}X\left(\frac{f}{b}\right)
\end{align*}$$
all of which are proved by change of variables in the definition of the
Fourier transform
$$X(f) = \int_{-\infty}^{\infty} x(t)\exp(-j2\pi ft) \mathrm dt.$$
It is good exercise to work out the details of each of these results 
at least once in one's life, and once one has done so, the two-dimensional case
is as easy as rolling off a log.  If you always look up the answers in a table
of Fourier transform properties, the right result will never come to mind just when
you need it most.

More generally, it is a good idea to remember these results in graphic
terms.  If $x(t)$ is a signal with Fourier transform $X(f)$, and $b &gt; 1$,
say $b = 2$,
then $y(t) = x(2t)$ is a time-compressed form of $x(t)$: while $y(0) = x(0)$,
$y(3) = x(6)$, that is, $6$ seconds worth of $x(t)$ have been squeezed down 
into $3$ seconds worth of $y(t)$.  Thus, the frequencies have doubled, and so
$Y(f)$ should be something like $X(f/2)$ (but see below).  More generally,


  Compressing the time axis by a factor of $b$ expands 
  the frequency axis by a factor of $b$.


But that is not the complete answer.  One of the more boring results in Fourier
transform that everyone tends to ignore is
$$y(0) = \int_{-\infty}^{\infty} Y(f) \mathrm df.$$
But, 
$$y(0) = x(0) = \int_{-\infty}^{\infty} X(f) \mathrm df$$
and since we have stretched the horizontal axis by a factor of $b$
in going from $X(f)$ to $X(f/b)$, we better reduce the magnitude of each
frequency component by a factor of $b$ so that the two integrals (= area
under the curve) work out to have the same value, that is,
$$Y(f) = \frac{1}{b}X\left(\frac{f}{b}\right), ~~ b &gt; 0.$$
The result for $b &lt; 0$ can be worked out similarly.