Math Integral grade 11

Question

Please help answer question 3 with steps, much appreciated!

3 . A curve is such that dx/dx = 6/x^2 - 5x and the point (3,5.5) lies on the curve. Find the equation of the curve.

Answer 1

y' = dy/dx = 6/x^2 -5x

y = integral of 6/x^2 -5x

y' = 6x^-2 -5x

y =-6x^-1 - 5x^2/2 + C

5.5 = -6/3 - 5(3)^2/2 + C

C = 11/2 + 2 + 45/2 = 60/2 = 30

f(x) = y = -6/x - (5/2)x^2 + 30

check the answer, plug in the point

5.5 =-6/3 - (5/2)(3)^2 + 30

5.5 =-2 -45/2 +30= 11/2=5.5