Question

How do I convert the equation `f(x)=x^2+6x+5` to vertex form?

Answer 1

The answer is `f(x)=(x+3)^2-4`.

To convert to vertex form, you must complete the square.

`f(x)=(x^2+6x+k)-k+5`

We add `k` to complete the square inside the brackets, but whatever we add inside, we must subtract outside the brackets to keep the equation balanced.

`k=(b/2)^2=(6/2)^2=9`

Substitute `k` to get your final solution:

`f(x)=(x^2+6x+9)-9+5`
`=(x+3)^2-4`

You can derive a formula algebraically:

`f(x)=x^2+bx+c`
`=(x^2+bx+(b/2)^2)+c-(b/2)^2`
`=(x+b/2)^2+c-(b/2)^2`

It is a slightly more complicated if `a!=1`:

`f(x)=ax^2+bx+c`

You must factor out `a` first, then continue as before:

`f(x)=a(x^2+b/ax+k)+c-ak`
`=a(x^2+b/ax+(b/(2a))^2)+c-a(b/(2a))^2`
`=a(x+b/(2a))^2+c-(ab^2)/(4a^2)`
`=a(x+b/(2a))^2+c-(b^2)/(4a)`

With a little rearrangement, it should look very familiar:

`=a(x+b/(2a))^2-(b^2-4ac)/(4a)`

If you can complete the derivation yourself, you should have no trouble completing the square with any question. When using the formula, be very careful when `a` is negative.

Here is another example:

How do I graph `y=x^2+4x-5`?