How does finding the $y$-intercept of a graph reduce error compared to finding the average of the data set? In my particular case I have the equation
$$
\frac{1}{d_i} = -\frac{1}{d_o}+\frac{1}{f} 
$$
and I'm plotting $\frac{1}{d_i}$ against $\frac{1}{d_o}$ to give an intercept of (in theory) -1 and a $y$-intercept of $\frac{1}{f}$. How does this reduce the error versus just taking the average of all my values for $f$? I can see how finding the slope accounts for systematic errors, as adding a constant to every $x$ or $y$ value would only increase the intercept, but I don't see how finding the intercept mitigates errors.
 A: If you find the intercept, your result will be more heavily weighted by the points close to the Y axis (where $d_0$ is large). By contrast, if you take the average of all the $\frac{1}{f}$ values, each value will be equally weighted.
Is there a reason to believe the values of $f$ will change with $d$? Or alternatively, if there is less error in the value of $\frac{1}{d}$ for large $d$, then you will get a more accurate result.
Without knowing how the error of either quantity depends on its magnitude it's hard to make a definitive assertion that one method is better than the other.
A: This is quite a difficult experiment to analyse because as the percentage error in one quantity $\frac {1}{d_\rm o}$ increases the percentage error in the other quantity $\frac {1}{d_\rm i}$ decreases and vice versa and so it is not just a case of points being more accurate at one end of the graph as compared with the other.  
Think of it this way.
Assume the focal length of the lens is approximately $10\,\rm cm$ and the error in a measurement is $0.5 \, \rm cm$.
When ${d_\rm o} \approx 20 \,\rm cm$ and ${d_\rm i} \approx 20 \,\rm cm$ just adding errors gives an error in the focal length of approximately $2.5 \%$ whereas if the object is at infinity the measured focal length is in error by approximately $5 \%$.  
The find the focal length as an average of many values one would have to use a weighted average with the weighting being $\dfrac{1}{\rm error^2}$ which would be quite laborious by hand or even if one set up a spread sheet to do the calculations.  
Drawing a graph with appropriate error bars is not necessarily more accurate but you are in effect doing the averaging of your results with less effort.
At the same time you can use your physics knowledge to compare theory and experiment as you know that the graph should be symmetrical with the intercepts on the two axes the same $(=\frac 1f)$.  
