The equation requires that u have to give at least an energy equal to the wave function of the metal to eject an electron,
if that condition is not met then nothing will happen,
so the work function is the EXTRAPOLATED INTERCEPT of the graph,
as it can be clearly seen that when f=0 , KE = -w , which is impossibe as KE cannot be negative,
so the graph never intersects the X axis, its only the extended or extrapolated graph(on which professors show with a dotted line) which intersects it.
But the stopping potential does depend on the kinetic energy of the electrons. The stopping potential is defined as the potential necessary to stop any electron (or, in other words, to stop even the electron with the most kinetic energy) from 'reaching the other side'.
As you already stated, the maximum kinetic energy is given by
$$K_\text{max}=h\nu-e\phi_\text{em}$$
In order to stop an electron with this amount of kinetic energy, you have to impose an electric field such that it will lose exactly this amount of energy while traversing it, so that it stops just slightly before reaching the other end of the setup shown in your picture.
The energy gained or lost by a charged object traversing this static electric field is given by the simple formula
$$\Delta K=q(V_\text{final}-V_\text{initial})$$
In the case of an electron, $q=e$, while the difference in potential can be denoted by $\Delta V\equiv V_\text{final}-V_\text{initial}$
Now, if we want to stop the most energetic electrons, but only barely, we have to make sure that $|\Delta K|=K_\text{max}$. Let us denote the corresponding potential difference, the stopping potential, by $V_0$. Then, we obtain
$$\Delta K=K_\text{max}=eV_0=h\nu-e\phi_{em} $$
As you see, we are only considering the most energetic electrons when we want to make sure all of them are stopped. This explains why the velocity does not appear as a variable.
Best Answer
You must see the curves in the light of photoelectric equation;
Photo electron Kinetic Energy (max) = h. frequency - W (work function)
From above equation, it is clear that the number of photons incident should have no effect on the energy of the photoelectrons.
However, If the intensity of light is changed (increased or decreased) the number of photoelectrons that are released gets affected .
As each photon is capable of releasing only one electron, then if there are more photons, then there is release of more electrons.
So for three intensities the photo current curve takes different paths depending upon the available electrons.
However as the photo active plate is of same material having unique work function the stopping potential remains same as defined by the guiding equation.