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November 28, 2014

November 28, 2014

Posted by **Hokoslavia** on Sunday, March 9, 2008 at 11:36am.

- algebra -
**drwls**, Sunday, March 9, 2008 at 12:24pmThe P and Q coordinates in your first statement do not agree with the P and Q in your last statement. I would need to see the graph to know what this problem is all about.

- algebra -
**Damon**, Sunday, March 9, 2008 at 2:11pmI will try to reverse engineer this mess.

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

if the vertex is at (-1,3) and x=-1 is the axis

put in P at (-1,7) impossible because y = 3 when x = -1

so I will try the second pair of points

P at (0,4)

(4-3) = k (1)^2

so k = 1

so

y - 3 = (x+1)^2

try Q at (-3,7)

4 = (-2)^2

YES

so the parabola with axis of symmetry at x = -1 and vertex at (-1,3) can contain the SECOND pair of points P and Q

rewrite in standard form

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

y = x^2 + 2 x + 4

I have no idea what the two original points P and Q were. They are on some other curve.

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