Question

How do I convert the equation `f(x)=x^2-8x+15` to vertex form?

Answer 1

To convert this equation into vertex form, you want to use a method called completing the square. I will walk you through the steps to do this:

First, group the first two terms together in brackets for now, and common factor the coefficient beside `x^2` out if necessary. There isn't a coefficient there so this is not necessary.

`f(x)=(x^2−8x)+15`

Next, take the x term (-8x), divide it by 2x, square it, then add and subtract that number inside the brackets so as to not change the meaning of the equation (16-16=0, therefore no change) like so:

`((−8x)/(2x))^2=16`

`f(x)=(x^2−8x+16−16)+15`

Now, take the subtracted term (-16) and multiply it by the coefficient outside of the brackets (there is none so skip this), and move it outside of the brackets:

`f(x)=(x^2−8x+16)+15−16`

Now, simplify the outside of the brackets, then change the `x^2` to just `x`, and divide the -8x by 2x. Also, drop the +16 inside the brackets, then square the bracket like so:

`(-8x)/(2x) = -4`

`f(x)=(x−4)^2−1`

And that's it! The original equation is now in vertex form! In case you were wondering or didn't know, the vertex `(h,k)` or `(x,y)` would be (4, -1). To find this, the axis of symmetry (h or x) is the number (with the opposite sign) inside the brackets and the optimal value (k or y) is the number (same sign) on the outside of the brackets!

Hopefully I was of some help and hopefully you've understood this! :)

Answer 2

The vertex form is `y = (x - h)^2 +k`
or, by expansion: `y = (x^2 - 2hx +h^2) + k`

The only term with `x` is `2hx` from the vertex form and `8x` from the standard form ( `x^2 - 8x + 15` )

Therefore, `2hx = 8x` `rarr` `h=4`

Substituting `h=4` into `y = (x^2 - 2hx +h^2) + k`

We have `y = x^2 - 8x + 16 + k`
For this to be equivalent to the standard form:
`y = x^2 - 8x + 15`

`16 + k = 15` `rarr` `k = -1`

With `h = 4` and `k = -1`, the vertex form is
`y = (x - 4)^2 -1`