Quadratic Extrema

Date: 07/20/2002 at 03:02:40
From: Yan
Subject: about maximum and minimum values

Dear Dr. Math,

Can you give me information about the minimum and maximum values of
quadratic functions? 

Yan

Date: 07/20/2002 at 08:14:56
From: Doctor Jerry
Subject: Re: about maximum and minimum values

Hi Yan,

A standard form for a quadratic function is

  y = a*x^2 + b*x + c,

where a, b, and c are constants and a is not zero.

The graph of this function is a parabola, either opening up (if a is 
positive) or opening down (if a is negative).  If the parabola opens 
up, then the quadratic function has a minimum, which occurs at the 
vertex of the parabola. If the parabola opens down, then the quadratic 
function  has a maximum, which occurs at the vertex of the parabola. 

Here's how to find the maximum or minimum. It is  much like the 
quadratic formula, which is used to find the places where the parabola 
crosses the x-axis.

  y = a*x^2 + b*x + c

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

Now add and subtract (b/(2a))^2 = b^2/(4a^2) within the parentheses:

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

Rearrange:

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

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

    = a(x+b/(2a))^2 - [b^2-4*a*c]/(4a)

Now, if a>0, we see that in the above expression, the term

  a(x+b/(2a))^2 

is either positive or zero.  If it is zero, then y is as small as it 
can be. Hence, the point 

  (-b/(2a),- [b^2-4*a*c]/(4a)) 

is the minimum point on the parabola.  

If a<0, a similar argument can be made, showing that the point 

  (-b/(2a),- [b^2-4*a*c]/(4a)) 

is the maximum point on the parabola.  

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 07/21/2002 at 11:43:25
From: Yan
Subject: Thank you (about maximum and minimum values)

Thanks for replying to my question! Your answer is excellent.
It was just what I wanted to know. Thanks a lot!