A curve has parametric equations: x = 3t +8, y = t^3 - 5t^2 + 7t. Find the co-ordinates of the stationary points.

First differentiate: dx/dt = 3,   dy/dt = 3t² - 10t + 7

Using the chain rule: dy/dx = dy/dt * dt/dx = (3t² - 10t + 7)/3 

At stationary points, the gradient is equal to zero: 3t² - 10t + 7 = 0

Solve for t using the quadratic formula and substituting these values: a = 3, b = -10, c = 7

Solutions: t = 7/3 and 1

For the co-ordinates, substitute the values of t into the parametric equations: (15, 49/27) and (11, 3)