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December 20, 2014

December 20, 2014

Posted by **Rebekah** on Wednesday, April 24, 2013 at 6:39am.

shape, and direction as the graph of y=3x^2, find the equation (s) of all possible parabolas

Thanks!!!

- Trig -
**Reiny**, Wednesday, April 24, 2013 at 8:03amlet the parabola have equation

y = 3(x-p)^2 + q

(1,10) lies on it, so

10 = 3(1-p)^2 + q

also the vertex (p,q) lies on y = 3x+1

thus:

q = 3p + 1

sub back into equation above

10 = 3(1-p)^2 + 3p + 1

10 = 3 - 6p + 3p^2 + 3p + 1

3p^2 -3p -6 = 0

p^2 - p - 2 = 0

(p-2)(p+1) = 0

p = 2 or p = -1

if p=2, then q = 3(2) + 1 = 7

and the parabola is y = 3(x - 2)^2 + 7

if p = -1, then q = 3(-1) + 1 = -2

and the parabola is y = 3(x+1)^2 - 2

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