Being that your graph has an oblique asymptote, an improper rational function should be the kind of function that comes to mind. By the looks of the graph, as we let x tend toward either \infty or -\infty, the function gets closer and closer in value to a linear function, so we should consider an improper rational function whose numerator is 1 greater in degree than its denominator. Also, a rational function is equal to 0 wherever its numerator, but not its denominator, is equal to 0. Since you have three x-intercepts, the numerator should be a polynomial that is the product of three factors, each one corresponding to a different x-intercept. A preliminary construction might look like the following:
f(x)=k\frac{(x+40)(x-.5)(x-277.4)}{?}
The numerator, when expanded, appears to be a polynomial whose dominant term is x^3. Therefore, we should consider a denominator that is a polynomial of degree 2. Also, the graph is undefined and has a vertical asymptote at only one value of x: -6. Notice also that the graph of f(x) tends toward -\infty as x approaches -6 from either side. Therefore, the denominator should be a quadratic with at least one (x+6) factor. To find the other factor, observe either that there is no sign change of the function across the vertical asymptote at x=-6, or that for any factor other than (x+6) to be present in the denominator, the graph would have to have another vertical asymptote that isn't at x=-6, so the denominator clearly has repeated (x+6) factors. Our construction now takes the following form:
f(x)=k\frac{(x+40)(x-.5)(x-277.4)}{(x+6)^2}=k\frac{(x+40)(\frac{2x-1}{2})(\frac{5x-1387}{5})}{(x+6)^2}=k\frac{(x+40)(2x-1)(5x-1387)}{10(x+6)^2}
=k\frac{(2x^2+79x-40)(5x-1387)}{10(x^2+12x+36)}=k\frac{10x^3-2379x^2-109773x+55480}{10x^2+600x+1800}
A function of this form is EXACTLY equal to 0 at all three of the intercepts shown on the graph provided. And, if we let k=-\frac{1}{5}, it has an oblique asymptote of -\frac{1}{5}x+\frac{2379}{50}, which has the same slope as the asymptote on the graph but NOT the same y-intercept. Also, the equation you came up with has the correct slope and y-intercept of the oblique asymptote, but it is NOT EXACTLY equal to 0 when x=277.4, so one of two things must be going on here: either the x-intercepts displayed on the graph are slightly off (rounded), or the actual equation of the oblique asymptote is slightly different than -\frac{1}{5}x+50.
