Need help with Parabola and formulas, please!?

Question

I need to put these formulas into standard form as well as vertex form. If you know how to do these please reply to this!
The problems are:
First:
y= -1x^2 -16x + 15

Second:
(2,5)(4,-3)(-1,2)

Third:
y=2g^2 -3
g=x-5

I really don't understand how to do this, please help me!

Answer 1

Standard form is y = ax^2 + bx + c
Vertex form is y = a(x-h)^2 + k where (h,k) is the vertex.

First is in standard form already; the x coordinate of the vertex is x = -b over 2a so x = 16 over -2 or -8.
That would make y = -64 + 128 + 15 = 79

So plug these in to get vertex form.

The second, without using a graphing calculator, would be solved by simultaneously solving the three equations you'd get when you plug in each x and y.

5 = a(2^2) + b(2) + c
-3 = a(4^2) + b(4) + c
2 = a(-1)^2 + b(-1) + c

Good luck, this takes too much writing to type in here!

The third you'd get by replacing the x in the first equation with (x-5)

y = 2(x-5)^2 - 3 (which is vertex form, see #1 above)
y = 2(x-5)(x-5) - 3
y = 2(x^2 - 10x + 25) - 3

Do this out, which will give the standard form.

[Edited: found multiply error in part 1]

Answer 2

The equation of a parabola can take two main forms. One is y = Ax^2 +Bx + C , or x = Ay^2 +By + C This is standard form in some text books.

The second form is (y-k)^2=2p(x-h), [axis parallel to x -axis]
or (x-h)^2 = 2p(y-k) [axis parallel to the y-axis. where (h,k) is the vertex of the parabola. This is the standard in other text books.

First, you already have it in the first form mentioned.
So y = -(x^2+16x -15)
y= -(x^2 +16x +64 -64-15) [Complete square]
y = -(x +8)^2 -79
-(x+8)^2 = y+79 which is the vertex form.

Second, you have been given three points through which the parabola passes. So plug each pair of x and y values in the ist form and you will get three simultaneous equations in A, B, and C. Solve thes simultaneous equations and then form equation using the values found for A, B, and C. Then change to vertex form bby completing the square.

Third, substitute x-5 for g getting y+3 = 2(x-5)^2 so you have the vertex form. Now solve for y and you will have the 1st standard form.

Answer 3

1)y= -(x^2+16x-15) = -(x-8)^2 +79 the vertex is(8,79) because it is the maximum value .(you are completing the square)
2)I don´t know what you must do.If you try to find the equ. of a parabola through these points
y=ax^2+bx+c. Put into this the coordinates of those points.You´ll
get a system of three equ.with three unknowns (a,b,c)
3) y=2(x-5)^2-3 Vertex (5,-3)