The equation of a quadratic curve is y=x^2+ax+b. The points (6,-4) and (4,-6) lie on this curve. Find the co-ordinates of the turning point of the curve.

If y=x^2 + ax + b is the equation for the curve, then the points which lie on the curve must satisfy this equation.Inputing the points into the equation gives:-4 =6^2+6a+b -6=4^2+4a+bNow solve simultaneously to find a and b. Eliminate b by subtracting 2. from 1.1. - 2. : 2 = 20 +2a. Rearranging gives a = -9Substituting a =-9 into any of the equations gives b = =14. So the equation of the curve is y=x^2-9x+14. Completing the square to find the turning point gives: (x-(9/2))^2-(81/4)+14 which simplifies to (x-(9/2))^2-25/4. So co-ordinates of turning point is: (9/2,-25/4)