How do you find the turning point of a quadratic equation?
We factorise the equation by completing the square.Take y=x2 + ax + bWhen we complete the square we have y= (x+1/2a)2-a2+b which is y= (x+1/2a)2 + c where c = -a2+bThe turning point would be at (-1/2a, c). Remember we always set (x+1/2a) = 0 to find x and y.For example, find the turning point of the equation y=x2+4x-12Firstly, we complete the square: y = (x+2)2-4-12 and this simplifies to y = (x+2)2 -16Setting (x+2)=0 we have x=-2. When x= -2 we have y= -16. Hence our turning point is (-2,-16).Another way of thinking about the graph by y= x2 and the transformation the graph will have with y= (x+1/2a)2 + c where c = -a2+bTake f(x) = x2We know when f(x)= (x+a)2. We shift the graph to the left by a, on the x axis and when f(x) = x2 + a. We shift the graph up by a, on the y axis.
