Finding the Vertex bycompletingng the square"
Step 1. Insure the coefficient of the x^2 (the "a" term) is 1.
In this case it is otherwise divide thru by the "non one" coefficient. Step 2 Take the coefficient of the x term (b term) If the equations was modified to get the coefficient of the a term a one, use the modified coefficient of be, and divide by 2, (5/2) square this term (25/4) and add it to the equation.x2−5x+25/4+3Since you added a value to the equation you must balance the equation by subtracting it:x2−5x+25/4+3−25/4Step 3. Thus far we have formed a "perfect square" with the(x2−5x+25/4)is a perfec square(x−5/2)2Substitute the "squared" version getting:(x−5/2)2+3−25/4perform the addition and subtraction regroup the terms getting:f(x)=(x−5/2)2−13/4 now in standard form. Now from the standard form it is apparent that the vertex occurs at the point (5/2, -13/4)
