The [leverage](http://en.wikipedia.org/wiki/Leverage_%28statistics%29#Definition) is $h_{ii}=\frac{1}{n}+\frac{(x_i-\bar{x})^2}{\sum (x_i-\bar{x})^2}\,$.

The term $\frac{1}{n}$ and the denominator of the second term $\sum (x_i-\bar{x})^2$ are the same for every $i$, so the point with the largest $(x_i-\bar{x})^2$ has the highest leverage.

This means that the point furthest from the mean has the highest leverage.

![enter image description here][1]

In the diagram, point 3 is the furthest from $\bar x$ in the x-direction, so it will have the largest leverage of the three points.

  [1]: https://i.sstatic.net/qYIL7.png