Trig
posted by Rebekah .
A parabola has its vertex on the graph of the line y=3x+1 and passes through (1,10). If it is the same size,?
shape, and direction as the graph of y=3x^2, find the equation (s) of all possible parabolas
Thanks!!!

let the parabola have equation
y = 3(xp)^2 + q
(1,10) lies on it, so
10 = 3(1p)^2 + q
also the vertex (p,q) lies on y = 3x+1
thus:
q = 3p + 1
sub back into equation above
10 = 3(1p)^2 + 3p + 1
10 = 3  6p + 3p^2 + 3p + 1
3p^2 3p 6 = 0
p^2  p  2 = 0
(p2)(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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