First, we must find the value of x that corresponds to the stationary point, i.e. where the gradient (dy/dx) is zero. This will be in terms of 'a' (if we change 'a,' we change the shape of the function by a transformation that is not linear, which means we will change the x-coordinate and the y-coordinate corresponding to the stationary point). Differentiating (and remembering that a/x = ax-1):
dy/dx = 8x -ax-2 = 0
8x = ax-2
8x3 = a
x = a1/3/2
As mentioned, plugging in this value of x will always allow us to find the x-coordinate of the stationary point.Now, we must find the value of 'a' which makes the stationary point correspond to a y-value of 32 (using some quite involved algebra!):
32 = 4[(a2/3)/4] + a/(a1/3/2) + 5
27 = a2/3+ 2a2/3
9 = a2/3
a = 93/2 = 27