From quadratic form to vertex form

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SUMMARY

The quadratic expression $-x^2 + 4x - 1$ can be converted to vertex form using the method of completing the square. The correct transformation involves factoring out the negative sign from the quadratic term, resulting in $y = -(x - 2)^2 + 3$. This indicates that the vertex of the parabola is located at the point (2, 3). The initial attempts to complete the square were incorrect due to mismanagement of the negative sign.

PREREQUISITES
  • Understanding of quadratic functions and their properties
  • Familiarity with the method of completing the square
  • Basic algebraic manipulation skills
  • Knowledge of vertex form of a quadratic equation
NEXT STEPS
  • Practice converting various quadratic equations to vertex form
  • Explore the implications of vertex form on graphing parabolas
  • Learn about the effects of different coefficients in quadratic equations
  • Study the relationship between standard form and vertex form of quadratics
USEFUL FOR

Students studying algebra, educators teaching quadratic functions, and anyone looking to improve their understanding of polynomial transformations.

mathlearn
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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)
 
Last edited:
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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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