V max divided by two doesn't change. And so if we look at the corresponding Y coordinate point, at the corresponding x coordinate point for this Y coordinate, in the case of no inhibitor, present it's here. In the case of the inhibitor, present it's farther along that x axis, and so the Km is greater. So competitive inhibitors increase the parent Km value. So since most of the active sites are occupied by the inhibitor, a larger amount of substrate needs to be present to actually overcome and displace that inhibitor to reach that same enzyme rate. And this means a higher amount of substrate is actually needed to reach that rate of VMAX divided by two.