
Originally Posted by
spacefire5458
Is it y=4x^2?
y = 4x^2 is certainly a parabola.
The general form for a parabola is:
y = Ax^2 + Bx + C.
Here if A >0, the parabola opens upward, and if A is negative it opens downward. For the example you gave y = 4x^2, the value of A is 4, B is 0 and C is 0. So this parabola opens upward.
If you can convert this to the form y-k =a(x-h)^2, then the vertex is at the point (h,k). For the example case of y = 4x^2, this can convert to:
y-0 = 4(x-0)^2, so this tells you the vertex is at (0,0).