F(x)=3x/x^2-16
x = -16 x = -4 x = 0 x = 4 x = 16

Somehow you have to involve the pythagorean theorem. You need 3 sides that will obey

a^2 + b^2 = c^2
We will pick
JK = a
JL = b
KL = c The longest side may not be c. We'll just see what happend.
JK^2
JK^2 = (x2 - x1)^2 + (y2 - y1)^2
JK^2 = (2 - -1)^2 + (2 - 3)^2
JK^2 = (3)^2 + (1)^2
JK^2 = 9 + 1
JK^2 = 10

JL^2  
JL^2 = (2 - -2)^2 + (2 - - 1)^2
JL^2 = 4^2 + 3^2
JL^2 = 25

KL^2
KL^2 = (-1 - -2)^2 + (3 - - 1)^2
KL^2 = ( -1 + 2)^2 + 4^2
KL^2 = (1)^2 + 16
KL^2 = 17

17 + 10 ≠ 25  I may have made a mistake. Check my numbers. This is how it is done, however.