function

[md2]

if we have function like this y = ( x - 2 ) ^ 2 + 1, what we will get if we do : y = - ( ( x - 2) ^ 2 + 1 ) , and more to find the top of function ( highest value ) doesnt matter the ( x - 2) ^ 2 ..... for x-2

beacause theres allways is 0 to find the top of function

 

[Greg H.]

First just to clarify, do you mean these functions?

 

[math]y = (x - 2)^2 + 1[/math]

 

and

 

[math]y = -((x - 2)^2 + 1)[/math]

 

Second, I don't know what you're trying to determine. Can you elaborate?

[CaptainPanic]

!

Moderator Note

 

md2,
You must write posts that we can understand. What do you want to ask or discuss? If you do not write posts that other people can understand, you cannot post on our forum. I hope you understand this warning.

So far, nobody understood any of your posts.

[md2]

and this part x ^ 2 also doeasnt matter, bacause the x ^ 2 after set the x = 0 will be 0, so in this function only matter + 1, to find the top of function. and b value must be > 0 , or < 0, its clear

i am trying to find the maximum and minimum of this function. and what else use this equal y = b / x has sense ?

and in equal y = (x-3)^2 + 2, line provide by minimum of this function is y = 3 / 2 x . with equal (x-6)^2 + 1 the line provide by minimum of this function is y = 1/6 x

Edited June 4, 2014 by md2

[md2]

and what more (b) cant be more ( for x^2 + b ) next values

for example :

y = x^2 + b, x = 1, x = 2, maximum is only 1 point

and the

b < x^2 + b

Edited June 4, 2014 by md2

[md2]

y = (x - 2)^2 / b , for maximum of function y = (x - 2 )^2 + b , for maximum y = 0 , when x = 2

 

 

y

y = ( x - 3 ) ^ 2 + b

b < ( x - 3 ) ^ 2

Edited June 5, 2014 by md2

[md2]

b = ( x - 3 ) ^ 2 + b

Edited June 6, 2014 by md2