Question

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!!!

Answer 1

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
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