# Quadratic Equations - Finding the Coordinate of The VertexHello, I am given `y=5x^2-10x+7` and I must change it to the form of `y=a(x-h)^2+k` . So I've already done this and the answer is...

Quadratic Equations - Finding the Coordinate of The Vertex

Hello,

I am given `y=5x^2-10x+7` and I must change it to the form of

*`y=a(x-h)^2+k` **. So I've already done this and the answer is `5(x-1)^2+2` *

*For the coordinates I wrote -1,2 but the answer is 1,2. I don't understand this and how do you get the vertex coordinate? what do I need to look for? *

*Thanks,*

*MB*

*print*Print*list*Cite

### 2 Answers

There are two reasons for converting a quadratic function from standard form `y=ax^2+bx+c` to vertex form `y=a(x-h)^2+k` . Either you are trying to solve for x for a specific y-value (usually zero) or you are trying to find the vertex of the parabola (either to help with graphing or find a maximum or minimum).

In this case, you are trying to find the vertex.

If a parabola is in vertex form `y=a(x-h)^2+k` , then the vertex is `(h,k)` . Notice that the h-value is the opposite sign of what is inside the brackets.

This means that in your case,

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

has a vertex at (1,2). Take the number inside the brackets, and change the sign. That is, -1 changes to 1 for the x-value of the vertex. The y-value of the vertex is the last number of the vertex form, in this case 2.

**The vertex of the parabola is (1,2).**

The reason why it is 1,2 is because you need to set the parenthesis equal to 0, which you didn't seem to do.

x-1=0

+1 +1

x=1

You solved it right just the last part was a bit off