MHB From quadratic form to vertex form

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$-x^2+4x-1$ should be converted to the vertex form of $y=k-(x-h)^2$

How can this be solved by factoring or any other method ?

My attempt to solve this problem , I will be using the completing the square method,

$\left(-x^2+4x+\frac{-b}{2a}\right)=1+\frac{-b}{2a}$

Here $\frac{-4}{-2}=2$

$\left(-x^2+4x+2\right)=1+2$

$\left(-x+2\right)^2=1+2$

$\left(-x+2\right)^2+3$

It's incorrect

Many Thanks (Happy)
 
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mathlearn said:
$-x^2+4x-1$ should be converted to the vertex form of $y=k-(x-h)^2$

How can this be solved by factoring or any other method ?

My attempt to solve this problem , I will be using the completing the square method,

$\left(-x^2+4x+\frac{-b}{2a}\right)=1+\frac{-b}{2a}$

Here $\frac{-4}{-2}=2$

$\left(-x^2+4x+2\right)=1+2$

$\left(-x+2\right)^2=1+2$

$\left(-x+2\right)^2+3$

It's incorrect

Many Thanks (Happy)
It's the negative in front of the x^2 term that's causing you problems. I'd factor it out at the beginning:
[math]y = -x^2 + 4x - 1 = -(x^2 - 4x + 1) = \text{ ... }[/math]

I get [math]y = -(x - 2)^2 + 3[/math].

-Dan
 
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