The example below shows a quadratric function transitioning using "completing the square" into the "Vertex" form, so that (h,k) are easily visible....
f(x) = 2x^2 − 4x + 1
= 2(x^2 − 2x) + 1
= 2(x^2 − 2x + 1) + 1 − 2 ------Where did this -2 come from?
= 2(x − 1)^2 − 1
Since completing the square in this case means adding (b/2)^2 to the trinomial inside the parentheses and to the term outside the parentheses, why would you subtract by 2?
I know this is correctly solved, but I'm wondering why...
THANKS!
f(x) = 2x^2 − 4x + 1
= 2(x^2 − 2x) + 1
= 2(x^2 − 2x + 1) + 1 − 2 ------Where did this -2 come from?
= 2(x − 1)^2 − 1
Since completing the square in this case means adding (b/2)^2 to the trinomial inside the parentheses and to the term outside the parentheses, why would you subtract by 2?
I know this is correctly solved, but I'm wondering why...
THANKS!
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In the third line you added a +1 inside the parentheses to create the square.
since there's a 2 multiplier outside the parenthesis, you actually added 2
You have to subtract it so the equation is unchanged
since there's a 2 multiplier outside the parenthesis, you actually added 2
You have to subtract it so the equation is unchanged
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You need to subtract 2 to balance out the completing square.
Compare the expression,
2(x^2 − 2x) + 1
to
2(x^2 − 2x + 1) + 1
= 2x^2 - 2x + 2 + 1
Note that the second expression is 2 more than the first expression. By completing the square, we've added this 2 to the expression. To cancel it out and make the expressions the same, we subtract 2.
Compare the expression,
2(x^2 − 2x) + 1
to
2(x^2 − 2x + 1) + 1
= 2x^2 - 2x + 2 + 1
Note that the second expression is 2 more than the first expression. By completing the square, we've added this 2 to the expression. To cancel it out and make the expressions the same, we subtract 2.