Completing the Square: Solving -x^2 + 4x and Finding the Vertex

[Miike012]

Problem: -x^2 + 4x

- ( x^2 - 4x + (4/2)^2 ) - 4
= - (x - 2) - 4

Vertex is at (2, -4)

but the vertex is obviously at (2,4)...
what did I do wrong.

 

[Mark44]

Problem: -x^2 + 4x

- ( x^2 - 4x + (4/2)^2 ) - 4

The error is here. You have actually added -4 inside the parentheses to complete the square, so to keep the expression equal, you need to add + 4.

= - (x - 2) - 4

Vertex is at (2, -4)

but the vertex is obviously at (2,4)...
what did I do wrong.

 

[Miike012]

how did I add -4... then the expression would be x^2 - 4x - 4 which is not a perfect square...? Does it have something to do with x^2 being negative?

 

[Mark44]

Let's go through the steps.

-x² + 4x
= -(x² - 4x)
= -(x² - 4x + 4) + 4
I put in +4 inside the parentheses, but due to the minus sign out front, I have actually added -4, so to balance, I have to add + 4.

= -(x - 2)² + 4

 

[HallsofIvy]

Another way of looking at it is to do both addition and subtraction inside the parentheses:
$$-(x^2- 4x)= -(x^2- 4x+ 4- 4)= -(x^2- 4x+ 4)-(-4)= -(x- 2)^2+ 4$$

 

[PeterO]

Let's go through the steps.

-x² + 4x
= -(x² - 4x)
= -(x² - 4x + 4) + 4
I put in +4 inside the parentheses, but due to the minus sign out front, I have actually added -4, so to balance, I have to add + 4.

= -(x - 2)² + 4

or if you like

-x² + 4x
= -[x² - 4x]
= -[x² - 4x + 4 - 4]
= -[(x - 2)² - 4]
= -(x - 2)² + 4