40413

Functions of the form ((x)(x+a)(x-a))/((x+b)(x-b)) where a>b seem to obey the general shape of the displayed graph. In order to work this out, I first considered the asymptote's on the graph. The function seemed to have two vertical asymptotes, evenly spaced from 0 which I originally thought were at x=1 and x=-1. So I knew that the denominator of my rational function should have roots at 1 and -1. Furthermore I considered the limits as x approached 0,1,-1 and infinity of the function shown. I knew that the limit as x approached positive infinity of the function was an oblique asymptote (that looked like the line y=x). The limit as x approached 1 from the right handside was negative infinity, and so I knew that the numerator must be negative for x close to 1 from the RHS. Symmetrically I knew that my numerator must be positive for x close to-1 from the LHS. Keeping this in mind, I then considered the roots of my function. They were 0, a number greater than 1 (greater than my positive vertical asymptote) and less than -1, (less than my negative vertical asymptote). So the roots of my numerator were 0,2,-2. So I initially wrote down x(x+2)(x-2) on my numerator. This numerator even satisfied my previously imposed conditions, because if I pick an x value close to 1 from the RHS, I will get a negative value on my numerator. And if I pick an x value close to -1 from the LHS, I will get a poitive value in my numerator, this is because (x-2)(x) would always yield negative, and a negative * a negative is a positive. So in general, the form of such a function would be ((x)(x+a)(x-a))/((x+b)(x-b)). Moreover, on taking the limit to negative or positive infinity of the expression chosen above, as the numerator is a polynomial of degree one greater than the polynomial below, the function gets closer and closer to x^3/x^2 = x. Which yields the oblique asymptote. Thank you for reading my solution!