Effect of Change of Potential Difference applied to an X-Ray tube I am a high school student, so I know only the basics of X-Rays. I simply know about continuous X-rays, cutoff wavelength and threshold wavelength.
Now if I increase the potential applied to the X-ray tube, I am certain that the minimum wavelength of emitted radiation decreases. I am unsure about its intensity. How does that change?
 A: An X-ray tube makes X-rays by Bremsstrahlung (quickly decelerating the
electrons), but also by exciting the atoms of the anode.  Just as
an electric discharge makes a red glow in neon gas, so  spectrum features of
the anode metal are prominent in the X-rays produced.
So, to make a higher energy of X-ray, an effective procedure is to
change the anode material to something with higher atomic number (the more 
charge in the nucleus, the higher the binding energy of the inner electrons, thus
the higher the possible energy of fluorescence).
In order to stand up to high currents, an anode is usually a good  heat
conductor (copper) or resistant to high temperature (molybdenum, tungsten).
Copper's  highest fluorescence is 8.98 keV, molybdenum  20 keV, and tungsten 69 keV.   
So if you want a high intensity fluorescence, you need to provide voltages
higher than the fluorescence to excite that intense X-ray emission, or provide
very high voltages and currents and  hope the Bremstrahllung  (continuous
spectrum) is intense enough (and the anode doesn't melt).
Because the Bremstrahllung includes a lot of low energy radiation, it is
usual to use an X-ray tube window that acts to filter out the unwanted low
energy radiation.   Both Bremsstrahlung and fluorescence energies can be no greater than the energy per electron.
To convert from energy $E$ (keV) to frequency $F$ (Hz) use Planck's constant
$h = 4.1\times 10^{-18}\ \mathrm{keV\,sec}$,
$$F = E /h$$
A: For a typical x-ray tube used in medical imaging, the radiation output (as measured in mGy/mAs) changes roughly with the square of the ratio of the ${kV_p}$
$$
y_1/y_2 = ({kV_p}_1/{kV_p}_2)^2
$$
where $y_n$ is the radiation output in mGy/mAs at ${kV_p}_n$
So if you measure the output of an imaging tube at 70 kVp and then change to 100 kVp (maintaining all other parameters the same), you can expect the output of the tube to roughly double.
A typical medical x-ray tube normally has about 2-3 mm Al filtration at the output port of the x-ray tube which serves to remove most of the low energy x-rays in the beam that don't contribute to image formation.
The graph below comes from data I've collected over several years of testing radiography units

