hello

 

I'm asked to write the equation y= 2x^2-4x+6 in the form y= a(x-h)^2

 

this is what i have been doing

 

first I divide by 2

 

x^2-2x+3

then I try to complete the square

 

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

(x+1)^2

 

now i don't know what else to do

 

Should I multiply that one times 2 because I divided by two and then add it to 3 to get 5? and so the answer would be

 

(x+1)^2+5

 

 

 

But I know that's wrong because I graphed both equations and they aren't the same graph

 

please help me.

well the correct square root would be (x-1)^2 +2=y/2

 

so then you multiply by 2 and your done.

Completing the square is when you half the coefficient of the term x, and then add it to the both sides.

 

i.e. [math]2x^2-4x+6=0[/math].

 

You factor it by 2.

[math]2(x^2-2x+3)=0[/math]

[math]2(x^2-2x)=-3[/math]

 

You half the second term by -2 and add it to both sides.

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

 

So you get: [math]2(x-1)^2+2[/math].

That's the answer. I have checked it, and it's right.

[math] 2(x-1)^2 +2 = 2x^2 - 4x + 2 +2 = 2x^2 - 4x + 4 \neq 2x^2 - 4x + 6 = y [/math]

 

Evon: You consistently forget factors, e.g. from 2(x² -2x +3)=0 => 2(x² -2x)=-6, not 2(x²-2x)=-3.

 

howareyou: You should try carrying around the "y=" part of the equation all the time. I think it would make things a lot easier for you.

A general solution would be this:

 

[math]ax^2 + bx + c \to a(x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} + c[/math]

 

Applying this to the original equations gives:

 

[math]2(x-1)^2 + 4[/math]

 

howareyou: I advise that after you complete the square that you expand the brackets, and see what you're missing. Although this relies on you knowing the first part of the general solution, which is the a(x + b/2a)² then once you enter that you can expand that and work out what to add on the end.