We have our function of y = 1+2x-x2 The gradient of a stationary (or turning) point is 0, in essence a peak or a trough. In order to determine the gradient, we need to differentiate our function with respect to x. This can be done by our usual method of bringing the power of x to the front and multiplying, and reducing the power as well (y' = 2 - 2x). We set this equal to 0 and solve for x ( x = 1). Now the nature is determine by the rate of change of the derivative (i.e. the second derivative), so we differentiate again ( y'' = -2). If this was a function of x, we would substitue in our value of x and we would get a value out. Here though, we simply have -2, and using the second derivative test ( max if y'' < 0 or min if y'' > 0) we can see that this graph has a local maximum.