Example curve: y = x2 + 4x + 5The first step is to differentiate the equation to give us the gradient at a general point. As a quadratic equation is an example of a polynomial, the solution is as follows:dy/dx = 2x2-1 + 14x1-1 + 0*5 = 2x + 4Since we know that stationary points are defined as points where the tangent line is horizontal (i.e. that the gradient is zero), the next step is simply to equate our previous equation with 0, and rearrange. This gives us x = -2. The final step step is to plug our value of x back into the original equation, to give us the corresponding y value, and hence the complete result: (-2, 1)