I'm convinced with the graph except for when x=0.
When $x=0$, the collector is at 0 potential. So photoelectrons that are emitted from the plate are not influenced by any electric field.
Since there is no electric field, the photoelectrons are equally likely to go in a particular direction (towards collector plate) regardless of their energy.
So, since the same amount of electrons are being emitted from the plate by source of different frequencies and since the electrons are equally likely to go towards the collector plate,
I say that at $x=0$, the photocurrent should be the same for all three curves.
Please tell me whether what I have said is right or wrong. And if it's wrong, why is it wrong?
Experiment diagram: 
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Best Answer
Electric current law in terms of current density $j$: $$ j = \rho\,u$$ where $\rho$ is charge density and $u$ - electron drift velocity. Expressing equation in terms of electron drift kinetic energy : $$ j = \rho\sqrt{2\frac{E_k}{m_e}}$$ where $m_e$ is electron mass. Finally, incident photon does work by freeing electron from a surface and accelerating it, so : $$ j = \rho\sqrt{2\frac{h\nu-\psi}{m_e}} $$ where $\psi$ is work function - energy needed to free electron from a surface. Now you clearly see that increasing incident photon frequency, makes photo-current density bigger, because photon transfers more energy to electron drift velocity. When you apply additionally external electric field, depending on voltage sign - it either stars stopping electrons, by reducing their kinetic energy or in reverse - by accelerating them even to a greater speeds until saturation is reached. Hope that helps.
EDIT:
This equation is for no external electric field applied ($V_{\textrm{external}}=0$).