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I guess you mean IP2,3 (Intercept points), right ?
If this is the case , as in an amplifier, you can use superposition to examine this relation :
If you draw the output terms seperately on a dB scale, you'll see that
a1*x(t) has a slope of 1 (10dB/dec) ,a2*x²(t) has slope=2(20 dB/dec) and a3*x³(t) has a slope=3 (30dB/dec).
You can solve the relation by equating : a1*x = a3*x³ ( which is the IP3 pt. )
or a2*x²=a1*x (which is IP2 pt. )
as far as I remember
a2 = Diode junction conductance (1/R_junction)
a3 = α(I_dc+I_sat), α=q/nKT
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