Quadratic ExtremaDate: 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! |
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