Given y = y1-y2, where Y1 is inversely proportional to 3x, Y2 is positively proportional to the square of X, and when x = 1, y = 6, when x = - 1, y = - 2 1) 2) when x = - 2, the value of Y

Given y = y1-y2, where Y1 is inversely proportional to 3x, Y2 is positively proportional to the square of X, and when x = 1, y = 6, when x = - 1, y = - 2 1) 2) when x = - 2, the value of Y

1) Let Y1 = K1 / (3x), y2 = K2 (x) ^ 2,
Then: y = y1-y2 = K1 / (3x) - K2 (x) ^ 2
When x = 1, y = 6,6 = (K1 / 3) - (K2)
When x = - 1, y = - 2, - 2 = (- K1 / 3) - (K2)
By solving the above equations, K1 = 12, K2 = - 2
Substituting y = K1 / (3x) - K2 (x) ^ 2, y = (4 / x) + 2x ^ 2 is obtained
2) When x = - 2, y = (- 4 / 2) + 2 (- 2) ^ 2 = 6