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No, the graph depicts the smallest limit where the transistor just starts to operate. Current is very low, less than 70 uA. I think the bias is very low too.
Saturation occurs when the transistor is biased strongly so that C-E current is maximum. In other words, C-E is down to its lowest 'On' resistance. Further increase in bias current will have no effect. This is typically at a bias of several mA.
If we're talking about a mosfet (as I believe the graph portrays) then the bias is in volts, nevertheless the same concepts apply.
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Your graph would be the area inside the green region below.
Coming back to the original question regarding the slope (gradient) of the Ids-Vs graph: The slope of this curve dI/dV is identical to the dynamical output conductance of the device, which is rather small (equivalent to a
large output resistance). This property is the reason to treat transistors as (non-ideal) current sources.