find the equation of the parabola passing through the origin with the turning point (2;-8).help mi.

Answer 1

the equation must look like:
y = a(x-2)^2 - 8
and it must pass through (0,0)
0 = a(-2)^2 - 8
8 = 4a
a = 2

equation: y = 2(x-2)^2 - 8

Answer 2

or, you know that one root is at (0,0) and since the vertex is at x=2, the other root must be at (4,0).

So, y = ax(x-4)
y(2) = -8, so
-8 = a*2(-2)
a = 2

y = 2x(x-4)
This agrees with Reiny's solution above.