changing equation to standard form

[renob]

1. The problem statement, all variables and given/known data

Sketch the region in the xy-plane that is bounded between the graphs of the given functions. Find the points of intersection of the graphs.

1) y=x^2+2x+2

2)y=-x^2-2x+2


3. The attempt at a solution

I already completed the square for equation 1):
y=(x+1)^2+1

Im having trouble completing the square for the second equation because of the negative values. I tried factoring out the negative sign:

-(x^2+2x-2)

but that just makes the 2 negative.

How should I go about converting the second equation?

[renob]

nevermind, i got it

[semc]

Oh well from what i can see,-x^2-2x+2 = -(x^2+2x+2) +4. so....-x^2-2x+2=-(y)+4. If you can sketch the graph for the 1st one the second one shouldn't be a problem:biggrin:

[HallsofIvy]

1. The problem statement, all variables and given/known data

Sketch the region in the xy-plane that is bounded between the graphs of the given functions. Find the points of intersection of the graphs.

1) y=x^2+2x+2

2)y=-x^2-2x+2


3. The attempt at a solution

I already completed the square for equation 1):
y=(x+1)^2+1

Im having trouble completing the square for the second equation because of the negative values. I tried factoring out the negative sign:

-(x^2+2x-2)
No, -(x^2+ 2x- 2)= -x^2- 2+ 2, not x^2- 2x+ 2.
x^2- 2x+ 2= (x- 1)^2+ 1.

but that just makes the 2 negative.

How should I go about converting the second equation?

[semc]

hmm did i make a mistake somewhere? Where did the x^2- 2x+ 2 come from haha::rolleyes:

[renob]

it comes out to be -(x+1)^2+3

[renob]

No, -(x^2+ 2x- 2)= -x^2- 2+ 2, not x^2- 2x+ 2.
x^2- 2x+ 2= (x- 1)^2+ 1.

I think you misread the second equation