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

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The discussion centers on completing the square for the quadratic equation -x^2 + 4x. The correct process reveals that when adding 4 inside the parentheses, the negative sign outside requires balancing by adding another +4 to maintain equality. This leads to the correct expression of -(x - 2)^2 + 4, indicating that the vertex of the parabola is at (2, 4). The initial confusion stemmed from misapplying the negative sign, which resulted in an incorrect vertex calculation. Ultimately, the vertex is confirmed to be at (2, 4).
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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.
 
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Miike012 said:
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.
Miike012 said:
= - (x - 2) - 4

Vertex is at (2, -4)

but the vertex is obviously at (2,4)...
what did I do wrong.
 
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?
 
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
 
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
 
Mark44 said:
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
 

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