Graphing

[mercuryv8]

The title for this post is inappropriate sorry, I hit post when I should have hit preview.

 

I'm trying to re-arrange the first equation in the series below to the format

[math]

y=a(x-p)^2+q

[/math]

Using the completing the square method, now I have a process down where I get the right answers. But I don't understand what happens to the "8x" them in the last step...I just leave it out and take the sqroot of the "16" and the "x^2" to get the (x-4) in brackets?

 

I don't fully understand why one-half of the coefficien of the x-term is added and subtracted either?

 

 

 

[math]

F(x) = 3x^2-24x+40

[/math]

[math]

F(x) = 3(x^2-8x)+40

[/math]

[math]

F(x) = 3(x^2-8x+16-16)+40

[/math]

[math]

F(x) = 3(x^2-8x+16)-48+40

[/math]

[math]

F(x) = 3(x^2-8x+16)-8

[/math]

[math]

F(x) = 3(x-4)^2-8

[/math]

 

 

any help in explaining this to me is greatly appreciated.

 

Thanks in advance

 

Nic

[timo]

I don't understand what happens to the "8x" them in the last step...I just leave it out and take the sqroot of the "16" and the "x^2" to get the (x-4) in brackets?

You don´t "just leave it out". It´s part of the (x-4)²:

(x-4)² = (x-4)(x-4) = x² -4x -4x + 16 = x² -8x +16

If you read that equation from the right to the left (equations can always be read both ways round), then that´s exactly what you´ve done/used in the last step.

 

I don't fully understand why one-half of the coefficien of the x-term is added and subtracted either?

I don´t fully understand the question. You´re asking why 16 (the square of one-half of the coefficient of the x-term) is added and substracted?

[mercuryv8]

 

 

I don´t fully understand the question. You´re asking why 16 (the square of one-half of the coefficient of the x-term) is added and substracted?

 

 

Yeah?

 

And thanks for your help with the factoring of the equation...It makes more sense now.

 

Nic