# how do I demonstrate algebraically that the equaitons: (X-2)^2+ 1 and

(-x-2)^2 +1 are equivalent?

well my guess is that if i expand it it will equal the same maybe?

so this is what i have done so far:
(x-2)^2+1
(X^2+4X+4)+1
now collect all like terms os would it be as follows:
X^2+4X+5

now for
(-x-2)^2+1
(x^2+4x+4)+1
collect like terms
X^2+4X+5

hey wait they are the same did i just prove they were equivalent?

yes, you proved it.

No, I'm sorry, but they aren't equivalent. If you graph them you'll see that the second one is the first one shifted 4 points to the left. I'll prove it.

Here's what was originally posted;

how do I demonstrate algebraically that the equaitons: (X-2)^2+ 1 and
(-x-2)^2 +1 are equivalent?

well my guess is that if i expand it it will equal the same maybe?

so this is what i have done so far:
(x-2)^2+1
(X^2+4X+4)+1
now collect all like terms os would it be as follows:
X^2+4X+5

now for
(-x-2)^2+1
(x^2+4x+4)+1
collect like terms
X^2+4X+5

hey wait they are the same did i just prove they were equivalent?

The first one is (X-2)^2+ 1.
When you expand it you get
x^2-4x+4+1 = x^2-4x+5.
You made a mistake expanding this; you had 4x, not -4x which it should be.
The second is (-x-2)^2 +1.
When we expand this we get
x^2+4x+4+1 = x^2+4x+5
The second one can also be see to be
((-1)(x+2))^2+1=(x+2)^2+1=((x-2)+4)^2+1
This is the first parabola shifted 4 units to the left. You can also see that the lowest point for the first one is 2 and the lowest for the second is -2)
If you meant to say that they are equivalent in some geometric sense, but simply shifted, then you'd be correct. But as they are written they do not identify the same parabola.

whatis the answer for 3y-6y+5-8-3y

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