Thread: Need help finding a formula from graph

How would we find the formula for problem 28b? I think the vertical asymptote is 3. And that I think the formula is in the format 1/x^2 but don't know how to get the formula. Please help me with this. Thank you so much!

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.

Great explanation! Thank you so much.

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.

brilliant idea !!,
thanks