Discovering the Maximum Value of a Quadratic Equation

[3.141592654]

Homework Statement

Maximizing the function f(X)=6X-1/2X^2+82. That's it, so 6X-1/2X^2+82 = 82-1/2(X^2-12X) = 82-1/2(X-6)^2+18. So f(X)=100-1/2(X-6)^2. Since (X-6)^2 is positive for any X not =6, than the max is x=6. The part I don't understand is when rearranging the function, how do you get from -1/2(X^2-12X) to -1/2(X-6)^2+18?

Homework Equations

The focus of the problem assumes my question is already common knowledge, so I don't think there are any relevant equations.

The Attempt at a Solution

Not sure what to do, at first I worked backwards trying to expand (x-6)^2+18, which is X^2-12X+54, but I'm at a dead end. This isn't actually for a class, but thanks for your help in advance!

 

[GregA]

Expand $$-\frac{1}{2}(x-6)^{2}$$ and tell us how much smaller it is than $$-\frac{1}{2}(x^{2}-12x)$$

 

[3.141592654]

alright, so -1/2(X-6)^2 expanded =-1/2X^2+6X-18. 18 smaller than -1/2(X^2-12X). Very helpful, thank you. That answers my question, but this brings me to another question. If given -1/2(X^2-12X), is there a way to use calculation to find that it equals -1/2(X-6)^2+18? I guess what I am asking is how can I look at this function, -1/2(X^2-12X) and then determine that it is equal to -1/2(X-6)^2+18? Do you factor an x and expand or is there any way to do this? Thanks again!

 

[Defennder]

This involves a method known as completing the square. Have you learned that?

 

[GregA]

alright, so -1/2(X-6)^2 expanded =-1/2X^2+6X-18. 18 smaller than -1/2(X^2-12X). Very helpful, thank you. That answers my question, but this brings me to another question. If given -1/2(X^2-12X), is there a way to use calculation to find that it equals -1/2(X-6)^2+18? I guess what I am asking is how can I look at this function, -1/2(X^2-12X) and then determine that it is equal to -1/2(X-6)^2+18? Do you factor an x and expand or is there any way to do this? Thanks again!

It is known as completing the square
If I have some quadratic of the form say...ax^2 + bx + c then first I can factor out an a to get
$$a(x^{2} +\frac{b}{a}x + \frac{c}{a})$$
Then I can add and subtract $$(\frac{b}{2a})^{2}$$ to get
$$a(x^{2} +\frac{b}{a}x + \frac{b^{2}}{4a^{2}} - \frac{b^{2}}{4a^{2}} + \frac{c}{a})$$
Which can then be converted to
$$a((x +\frac{b}{2a})^{2}) + d)$$ where $$d = \frac{c}{a} - \frac{b^{2}}{4a^{2}}$$
You can of course pull d out of the brackets where it would then be equal to $$c - \frac{b^{2}}{4a}$$

 

[Gib Z]

Is it part of your knowledge that the axis of symetrry of a parabola of the form y= ax^2 + bx + c is the line x=-b/2a. They introduced that to us in Australia before applying Completing the square for maxima/minima. If you do, and say the parabola is concave down - where does the maximum value always occur?