The equation of a curve is y=px+ h/x^3 ,(x not equal to 0) where h and k are constants. The
gradient of the curve y at the point (2,22) is −60. Hence, find the turning points of the curve and determine the nature of the turning points.
You must have done some work on it -- too bad you didn't see fit to post what you tried ...
y = px + h/x^3
y' = p - 3h/x^4
so, now we know that
2p + h/8 = 22
p - 3h/16 = -60
Now just solve for p and h.
y = -27/4 x + 284/x^3
y' = -(27/4 + 852/x^4)
since y' is never zero, there are no turning points.
The graph looks something like a hyperbola, with a slope asymptote (y' approaches -27/4 as x gets large)