Originally Posted by
ebaines To find a possible function, consider that this function becoms undefined at x = 3, similar to how the graph 1/x^2 becomes undefines at x = 0. So your first guess might be the function is 1/x^2 shifted to the right by 3. Recall that the graph of a function function f(x-a) is the same as f(x) but shifed a units to the right. So to shift 1/x^2 by 3 units to the right you have to change the x to x-3:
y = 1/(x-3)^2.
But if you do the math you'll see that this is consistently too low by one. So you need to add one to the equation:
y = 1/(x-3)^2 + 1.