Question

How do you find the vertex of `y = x^2-8x+7`?

Answer 1

Complete the square to reformulate the quadratic in vertex form:

`y = x^2-8x+7 = (x-4)^2+(-9)`

So the vertex is at `(4, -9)`

Explanation:

In the general case we can complete the square as follows:

`y = ax^2+bx+c = a(x+b/(2a))^2+(c - (b^2)/(4a))`

Notice especially the `b/(2a)` term, which gives us the `-4`. Normally you just observe the `a` and `b` terms and look at what you get when you multiply out `a(x-b/(2a))^2` rather than following the above formula. On this occasion, let's just use the formula...

In our case `a=1`, `b=-8` and `c=7`, so

`y = x^2-8x+7`

`= 1*(x+(-8)/(2*1))^2+(7-((-8)^2)/(4*1))`

`=(x-4)^2+(7-64/4)`

`=(x-4)^2+(7-16)`

`=(x-4)^2+(-9)`

Vertex form of a quadratic is:

`y = a(x-h)^2+k` where `(h,k)` is the vertex.

So in our case `(h, k) = (4, -9)`

Why is this the vertex?

Well `(x-h)^2 >= 0` will only be zero when `x-h = 0` - that is when `x = h`. When `x=h` then `y = 0+k = k`, hence `(h, k)`.