Solution

40464

First name
Louis d M
School
EMS
Age
16

There are two vertical asymptotes. Since we are allowed to choose what we want for these, we can choose an easy number, such as x=1. The other one is the same line reflected in x=0, so that asymptote would have the equation x=-1. Thus our equation will have to divide by 0 at -1 and 1:
y=(something)/(x-1)(x+1)
= (something)/(x^2 - 1)
The curve also has an asymptote to some line y=mx. Again, we can make m=1 to make our equation easier. To make out curve get closer to y=x for numbers with a large magnitude, since there is an x^2 term on the bottom of the fraction, we want an x^3 term on the top. The curve crosses just next to where the asymptotes cross the x-axis, so we can call this value something slightly larger in magnitude than 1 e.g. 1.1. Since the curve has to cross at x=-1.1, x=0 and x=1.1, the cubic has the equation y=x(x-1.1)(x+1.1), which expands like this:
= x(x^2-1.21)
= x^3 - 1.21x
So, the equation of our curve is:
y = (x^3 - 1.21x)/(x^2 - 1)
Attached is a picture of the graph in Desmos, with the curve in red and the asymptotes in blue.