Solution

Einstein's photoelectric equation: $h\nu = \phi + KE_{max}$
$\Rightarrow KE_{max} = h\nu - \phi = h\nu - h\nu_0$, where $\nu_0$ is the threshold frequency.
This is a straight line of the form $y = mx + c$ with slope $h$ (positive, same for all metals) and intercept $-\phi$ on the KE axis.
For $\nu \lt \nu_0$ no photoelectrons are emitted, so the kinetic energy is zero (the graph does not exist below $\nu_0$).
KE becomes zero exactly at $\nu = \nu_0$ and then increases linearly with $\nu$.
So the graph is a straight line with positive slope that starts from a non-zero frequency (the threshold frequency) on the frequency axis.

Concept KE versus frequency graph

Einstein's photoelectric equation in past papers

10 questions from this chapter have appeared across 8 exam years.