Posted by Rebekah on Wednesday, April 24, 2013 at 6:39am.
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
- Trig - Reiny, Wednesday, April 24, 2013 at 8:03am
let 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
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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