What's the exact process to convert y=-2x²+x+1 from the standard form y=ax²+bx+c to the vertex form y=a(x-h)²+k [Quadratic Equations]

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Koalanet21

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I have a problem in my process.
I can do these steps :

Step0: y=-2x²+x+1
Step1: y=-2(x²-(1/2)x)+1
Step2: (1/2)²=1/4
Step3: y=-2(x²-(1/2)x+1/4)+1-(1/4)(-2)
And then I cannot find a number which multiplied gives 1/4 and added gives 1/2.
It could have been 1/2 if in Step3, there had been -x instead of - (1/2) x. But it's not the case and I can't find an error at Step 0, 1 or 2.

Please, can someone help me?
 
The "process" is "completing the square".

I presume that you know that \(\displaystyle (x+ a)^2= x^2+ 2ax+ a^2\). Compare that with \(\displaystyle -2x^2+ x= -2(x^2- x/2)\). \(\displaystyle 2ax= -(1/2)x\) so \(\displaystyle a= -1/4\) and \(\displaystyle a^2= 1/16\).

So we can make \(\displaystyle -2(x^2- x/2)\) a "perfect square by adding 1/16 inside the parentheses. Of course, in order not to change the value you must also subtract it: \(\displaystyle -2(x^2- x/2)= -2(x^2- x/2+ 1/16- 1/16)= -2(x^2- x/2+ 1/16)+ 1/8=-2(x- 1/4)^2+ 1/8\)

So \(\displaystyle -2x^2+ x+ 1= -2(x- 1/4)^2+ 1/8+ 1= -2(x- 1/4)^2+ 9/8\).
 
Koalanet21 said:
I have a problem in my process.
I can do these steps :

Step0: y=-2x²+x+1
Step1: y=-2(x²-(1/2)x)+1
Step2: (1/2)²=1/4
Step3: y=-2(x²-(1/2)x+1/4)+1-(1/4)(-2)
And then I cannot find a number which multiplied gives 1/4 and added gives 1/2.
It could have been 1/2 if in Step3, there had been -x instead of - (1/2) x. But it's not the case and I can't find an error at Step 0, 1 or 2.

Please, can someone help me?
Click to expand...
You forgot that the number you square to get the constant you add is half of the coefficient of x: not 1/2 itself, but 1/4. So the constant you should have added is 1/16, not 1/4. Your step 2 was wrong.
 
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