What does a second power graph look like anyone have any examples?

A second power graph would be a parabola. Do you know what a parabola looks like? You see them all the time in arches, etc. They are almost like a U, but the ends go outward from each other and not parallel. Graph y= x^2, which is the most basic parabola and you will see.

You mention y=Ax^2, and the "A" determines if the parabola goes up or down (positive, it's smiling; negative, it's frowning). It also determines how wide or narrow it is. The higher the absolute value of A is, the more narrow it will be, the lower - the wider it will be.

hope this helps!

here is an example i had one like this

The subject in my textbook is "parabolas translated from the origin"

and standard equations. I was zipping right along until I came to

the following question:

State the co-ordinates of the vertex, and sketch the graph

of the following: y^2+8y+8x-8=0

I know I have to make this equation look like the stardard -

either (x-h)^2 = 4p(y-k) or (y-k)^2 = 4p(x-h) -

Hello = I'm glad im able to help.

You have taken the right first step in approaching this problem -

you want to make the equation given to you fit into one of the two

general equations, either the one with x to the second power and

y to the first power or the one with x to the first power and

y to the second power. Since the given equation, y^2 + 8y + 8x - 8 = 0,

has x to the first power and y to the second power, we want to make

it look like the general equation, (y-k)^2 = 4p(x-h), since that is

the general equation that also has x to the first power and y to

the second power. Does that make sense?

So, now we need only find out what h and k are. First, let's get the

x's and y's on different sides of the equation. Put the constant

with the x (you can put it with the y if you want, but things will be

easier if you put it with the x). So, your rewritten equation is:

y^2 + 8y = -8x + 8

Now this is looking much closer to the general form, right?

The left-hand side of the equation needs to be changed, though, and

we change it by using a process called "completing the square."

When you complete the square of an equation, say y^2 + ay (for any

constant, a), you take the coefficient of the y term, divide by 2,

and square it. If you add this constant, (a/2)^2, to the original

equation, you get y^2 + ay + (a/2)^2. This factors into the equation,

(y + a/2)^2. This will make more sense in context, so let's complete

the square on this problem.

To get the above equation so that we have something that looks like

(y-k)^2 on the left, we complete the square on the left. We don't

want to change the equation's meaning, so anything we do on the left

we must also do on the right. So to complete the square, we take

the coefficient of the y term, 8, divide by 2 (so we get 4), and square

it, so we get 16. Now we add 16 to both sides of the equation. Our

new equation is:

y^2 + 8y + 16 = -8x + 24

The left-hand side of the equation factors:

y^2 + 8y + 16 = (y + 4)^2

Now your equation is:

(y+4)^2 = -8x + 24

With only a few more steps that I'll let you do, you can put this

equation into the general form, and then you will know what the vertex is.

I hope this makes sense to you. This method of completing the square is

just a methodical way of finding out what number you should add to each

side of the equation so the side with the variable to the second power

factors into a square. At the stage where you have the equation,

y^2 + 8y = -8x +8, you know you have to add a constant to both sides

of the equation so the left side will become a perfect square.

Completing the square is just a nice method for figuring out what that

constant should be.

I hope this helps. Feel free to write back with any other questions.

Good luck!