# How would you find the vertex form of the parabola graph in the image attached?

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The vertex form of a parabola is given by,

`y = j(x-h)^2+k`

But we can write the equation as a a general quadratic equation and then convert to vertex form.

`y = ax^2+bx + c`

We have to find, a, b and c.

1. at `x = 0, y = 5`

`5 = axx0^2+bxx0+c`

`c = 5`

2. at `x = -3, y = 2`

`2 = axx(-3)^2+bxx(-3)+5`

`2 = 9a-3b+5`

`9a-3b=-3`

`3a -b = -1`

3. at `x = 3, y =2`

`2 = axx3^2+bxx3+5`

`2=9a+3b+5`

`9a+3b=-3`

`3a+b=-1`

We can solve for a and b,

by adding, `6a = -2 and a=-1/3`

by substracting, `2b=0 and b =0`

Therefore the equation is,

`y = -1/3x^2+5`

This is also equivalent to the vertex form,

where `j = -1/3` ` , h = 0 and k = 5`

The vertex form is,

`y = -1/3(x-0)^2+5`

Vertex form of a parabola is

`y = j(x - h)^2 + k`

Where, the vertex is located at (h,k)

Our goal is to figure out what j, h, and k are from looking at the graph.

We can look at the graph and see that the vertex is located at (0,5). So that means that h = 0, and k = 5.

We can rewrite our equation with this new information.

`y = j(x-0)^2 + 5`

Now we just need to find j.

That dot on the right of the graph tells us that when x = 3, y = 2.

So we can plug x = 3, y = 2 into our equation, this will let us solve for the last variable in our equation, j.

`2 = j(3-0)^2 + 5`

`2 = j*9 + 5`

`-3 = j*9`

`j = -3/9 = -1/3`

Now we have found what j, h, and k. We can write our equation in vertex form.

`y = -1/3(x-0)^2 + 5`