Conference rusure::math

1425.0. "Help finding extreme of a parabola?" by WFOVX8::SPORBERT (All the little ghoulies are calling your name) Thu Apr 18 1991 17:45

	 Hi All!  

I have a question concerning solving a quadratic formula.  
How do you solve the graph of a quadratic function using the 
y intercept and an extreme point.  Actually I just need to 
know how to find the extreme point.  My school book says 
something like this:

y = ax^2 + bx + c

Factor an a out of the two terms containing x.

y = a(x^2 + b/ax) + c

then complete the square of the terms within the parenthesis. 
(this is where I get lost...)

y = a(x^2 +b/ax + b^2/4a^2) + c - b^2/4a

  = a(x + b/2a)^2 +c = b^2/4a

HUH???

Is there an easier way to do this???? 8-)

                           Thanks - Ed