Find the coordinates of the turning point of y=x^2-5x+4

There are various methods of doing this, one by the symmetry of graph, one by completing the square and one by differentiation and setting f prime equal to zero. Depending on the specification, some will/won't be required. Symmetry of the graph:Factorise the quadratic and set it to zero to find the roots of the equation. (x-4)(x-1)=0, so x=4 and x=1 are the roots of the equationNext find the halfway x value between the two routes: this is the x coordinate of the turning point, and put this value through the equation to find the corresponding value of y. x=(4+1)/2=2.5, y=(2.5)²-5(2.5)+4=-2.25So coordinates of turning point: (2.5,-2.25)Completing the square:(x-2.5)²-2.25=0At the minimum value of y (x-2.5)² cannot be negative and so the smallest value it can be is zero, hence y=0-2.25The minimum value of y is -2.25.For x coordinate, which value of x must we put in to make (x-2.5)²=0? x=2.5So coordinates of turning point: (2.5,-2.25)Differentiation (assuming knowledge of the down and decrease rule and principles of differentiation):At turning points, the gradient is zero. Hence, differentiating the equation and setting this equal to zero allows us to work out the x coordinate:y'=2x-5, 2x-5=0, x=5/2=2.5Now put this value through the original equation to find the corresponding value of y. x=(4+1)/2=2.5, y=(2.5)²-5(2.5)+4=-2.25So coordinates of turning point: (2.5,-2.25)