anyone who knows ALG II, please help?

Question

oh god i have a whole bunch of worksheets given to me yesterday, and it's due today at 1:00! If anyone knows how to solve this, please explain!
Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions:
Vertex (0, -2), passing through (3, 25)
thank you!

Answer 1

Suppose the equation is y = ax^2 + bx + c. Since the point (0,-2) is on the parabola, we use x = 0 and y = -2 in the equation: -2 = a*0^2 + b*0 + c, so c = -2, and we know the equation is y = ax^2 + bx - 2.

Now the vertex is at (0,-2) and the axis is vertical tells us that the axis is the y-axis. That says that the curve is symmetric with respect to the y-axis, so that if (x,y) is on the parabola, then (-x,y) is also on the parabola. In particular, since (3,25) is on our curve, so is (-3,25). Using these two points, we get these equations:
25 = 9a + 3b - 2 and 25 = 9a - 3b -2. Adding the equations, we have 50 = 18a - 4, or 54 = 18a, so a = 3.

Finally, using a = 3, x = 3, y = 25, and c = -2 we have
25 = 3*9 + 3b - 2, so b = 0. Our equation is y = 3x^2 - 2.

Answer 2

The vertex form of the equation of a parabola is
y=a(x-h)^2 +k, where the vertex is (h,k).

This is a good form to use, because you know the vertex: (0,-2).
Replace h and k with 0 and -2, to get

y = a(x-0)^2 + (-2)
y = ax^2 - 2

To find "a", use your other point, (3,25). Replace x with 3 and y with 25:

25 = a(3^2) - 2
...and so on.

Answer 3

First off, SOLVE YOUR OWN PROBLEMS!! what youre doing is unethical and should be done.
You said you had a lot of worksheets?
If you had enough time to post this, you have enough time to answer the questions.