SOLUTION: I am trying to find the y-intercepts of {{{ x+2=(y+1)^2 }}} all I have been able to figure out is that the vertex is (-2,-1), if you could also show me how it is done that would be

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I am trying to find the y-intercepts of {{{ x+2=(y+1)^2 }}} all I have been able to figure out is that the vertex is (-2,-1), if you could also show me how it is done that would be      Log On


   

Question 387929: I am trying to find the y-intercepts of +x%2B2=%28y%2B1%29%5E2+ all I have been able to figure out is that the vertex is (-2,-1), if you could also show me how it is done that would be great. Thank you.
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I am trying to find the y-intercepts of +x%2B2=%28y%2B1%29%5E2+ all I have been able to figure out is that the vertex is (-2,-1), if you could also show me how it is done that would be great. Thank you.
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x+2=(y+1)^2
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Let x = 0,then solve for "Y":
(y+1)^2 = 2
y+1 = +sqrt(2) or y+1 = -sqrt(2)
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y = -1+sqrt(2) or y = -1-sqrt(2)
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Those are the y-intercepts of the parabola.
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Cheers,
Stan H.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!



Hi,         

+x%2B2=%28y%2B1%29%5E2+  OR y = ± sqrt(x+2) - 1

+x=%28y%2B1%29%5E2+-+2 

 y-intercepts are when x= 0

 +0+=%28y%2B1%29%5E2+-+2

 +0+=+y%5E2+%2B+2y+%2B+1+-+2+

  +0+=+y%5E2+%2B+2y+-1+

  y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

  y+=+%28-2+%2B-+sqrt%288%29%29%2F%282%29+

   y+=+%28-2+%2B-+sqrt%284%2A2%29%29%2F%282%29+

   y+=+%28-2+%2B-+2sqrt%282%29%29%2F%282%29+

   y+=+%28-1+%2B-+sqrt%282%29%29

      y = -2.414

      y = .414