some math formula collide in my head

[Vastor]

ok, it's said

 

The Quadratic Equation f(x) = 3x^2 - 6x + 11 can be expressed in the form f(x) = 3(x - h)^2 + k. Find the value of h and of k

 

f(x) = 3x^2 - 6x + 11

 

i use the completing the square method to convert it into vertex form

 

= 3x^2 - 6x + (6/2)^2 + 11 - (6/2)^2

 

= 3x^2 - 6x + 9 + 2

 

at this part start to puzzle me, what should i do with 3, which is a = 3

this number appear to be the same at ax^2 + bx + c = a(x - h) + k

 

then, should i remove the 3 from the factorisation?

 

which mean

 

x^2 - 6x +9 +2 = (x - 3)^2 + 2

 

and then put a to the equation?

 

3(x - 3)^2 + 2?

 

so, that when i check again,

 

3(x-3)(x-3) +2 = 3x^2 - 18x + 27 + 2, this is fail.

 

anyone can help?

Edited May 17, 2011 by Vastor

[DrRocket]

ok, it's said

 

The Quadratic Equation f(x) = 3x^2 - 6x + 11 can be expressed in the form f(x) = 3(x - h)^2 + k. Find the value of h and of k

 

f(x) = 3x^2 - 6x + 11

 

i use the completing the square method to convert it into vertex form

 

= 3x^2 - 6x + (6/2)^2 + 11 - (6/2)^2

 

= 3x^2 - 6x + 9 + 2

 

 

This is true, but has nothing to do with the given problem.

 

Think about what was asked. Factor accordingly.

[Vastor]

ha3, nvm, i find after surf on yahoo answer for a while.

 

need confirmation thought.

 

quadratic equation, f(x) = 0

 

0 = 3x^2 - 6x + 11

 

-11 = 3(x^2 - 2x)

 

-11 + [3(2/2)^2] = 3[x^2 - 2x + (2/2)^2]

 

- 8 = 3(x-1)^2

 

f(x)= 3(x-1)^2 + 8

 

right? then, i check it...

 

3(x^2 - 2x + 1) + 8 = 3x^2 - 6 + 11

 

tadaa..

 

urgh, another problem,

 

ok, it's said.

 

(h, k) is the turning point of the curve y = a(x -2)^2 - 8, another info. y-intercept = -6,

 

determine the value of a, h and k, where (h, k) is the min point

 

so,

 

i got h = 2, k = -8

 

so, how to determine the a ?

Edited May 17, 2011 by Vastor

[CaptainBlood]

You can't just arbitrarily remove numbers (3 in your case) in the factored part of the equation and plugging them into another equation.   Just think that if you remove a number from factored part of the equation, you are affecting the whole factored part of the equation with that number, in other words that number has an effect on all parts of the factored part of the equation.

Edited May 18, 2011 by CaptainBlood

[Vastor]

You can't just arbitrarily remove numbers (3 in your case) in the factored part of the equation and plugging them into another equation. Just think that if you remove a number from factored part of the equation, you are affecting the whole factored part of the equation with that number, in other words that number has an effect on all parts of the factored part of the equation.

 

arbitrarily remove numbers? ermm, u respond to which post actually?

[imatfaal]

Hey Vastor - you are not being so lucky with people telling you that you have made an error! Again I think you maths is ok - if difficult to follow. To explain to the pirate Captain

 

-11 = 3(x^2 - 2x)

-11 + [3(2/2)^2] = 3[x^2 - 2x + (2/2)^2]

Perhaps if I put in an intermediate stage or two

-11 = 3(x^2 - 2x)

-11 + 3(2/2)^2 = 3(x^2 - 2x) + 3(2/2)^2

-11 + 3(2/2)^2 = 3[(x^2 - 2x) + (2/2)^2]

-11 + 3 = 3[(x^2 - 2x) + 1]

-8 = 3(x^2-2x +1)

It becomes clear you have just added 3 to each side of the equation - quite legitimate. If I was marking this I would require at least the first intermediate stage and I would need to see how you came up with the formulation (2/2)^2.

Have a bash at the other problem and I am sure that someone will step up to help

[Vastor]

hmm, i think i should 'slow down' on my calculation... fortunately, i already learn the Latex, so no more confusing-looking math...

 

 

and about [math](\frac{2}{2})^2[/math], it's the so-called Completing The Square method, which is actualy mean to be [math](\frac{b}{2})^2[/math]

 

thanks for ur reply btw..

[imatfaal]

Oh Yeah - I understood what you where doing, I was saying you should make it clear that you do. At the level I learnt up to , and the level you are at, much of the credit given by teachers is for you demonstrating knowledge of methods and understanding the implementation and the logic as to why they work. If you leave this out then you are risking the marker assuming that you do not understand the methodology especially if you make an arithmetical error and your final answer is awry.

[physics confusion]

I think i understand the problem you are having with the 'a' variable. The equation 'y=a(x-h)^2+k' consists of the variables a, h and k with the x value. in the equation, 'a' is the dilation factor; 'h' is the x point; and k is the y point, so the vertex is at the points (x,y). The 'a' variable has no effect on the vertex, it effects the shape (dilation) of the graph.

 

Hope it helps