Question

Convert the equation to the standard form for a parabola by completing the square on x or y as appropriate.
y^2 + 2y - 2x - 3 = 0

Answer 1

since we have y^2, lets assume this is asking for a y solution

Answer 2

do you recall how to complete a square?

Answer 3

with numbers not variables

Answer 4

lets see if we can decompose this

a complete square is also a perfect square; 4 is a perfect square, 9 is a perfect square, 16 is a perfect square etc...

we want a poly such that:

(y+n)^2 = y^2 + 2yn + n^2

Answer 5

and we are given that 2yn = 2y from the setup

Answer 6

okay so the n is the square

Answer 7

n is the value we have to add and subtract to "complete" the square inthe original setup

Answer 8

y^2 + 2y - 2x - 3 = 0

(y^2 + 2y) - 2x - 3 = 0

(y^2 + 2y +n - n ) - 2x - 3 = 0

(y^2 + 2y +n) - n - 2x - 3 = 0

(y+n)^2 - n - 2x - 3 = 0

Answer 9

so n would be 3

Answer 10

n=1

2yn = 2y when n=1

and i got a few typos in that last post

y^2 + 2y - 2x - 3 = 0

(y^2 + 2y) - 2x - 3 = 0

(y^2 + 2y +n^2 - n^2 ) - 2x - 3 = 0 ; n^2 is the add/subtract

(y^2 + 2y +n^2) - n^2 - 2x - 3 = 0

(y+n)^2 - n^2 - 2x - 3 = 0

Answer 11

i forget what standard form is; thats the one where it has at most 3 terms right?

Answer 12

(y+1)^2 = 2(x+2)
this is an example of one of my choices

Answer 13

hmm, so this might be using hte geometric definition of a parabola ...

Answer 14

lets use n=1

(y+1)^2 - 1 - 2x - 3 = 0

(y+1)^2 = 1 + 2x + 3

(y+1)^2 = 2x + 4

(y+1)^2 = 2(x+2) does fit, im just unsure about the whole "standard" part :)

Answer 15

okay thanks I will try that because the other choices are different and they don't make any since because they start with
(y-1)^2

Answer 16

as long as you posted it correctly above .... thats all i got to go by ;)

Answer 17

yeah the problem is right I just didn't know if the right side was + or -

Answer 18

thank you again