Question

How do you rewrite `f(x) = x^2 - 8x + 12` in standard form `f(x) = (x-h)^2 + k`?

Answer 1

`(x-2sqrt(2))^2+4`

Explanation:

The coefficient of the `x` term has to have a square root `(b)` as `(x +b)^2 = x^2 +b^2x +...` The constant is adjusted for afterwards. There is not an integer square root of 8 so you would use `sqrt(8)` instead giving the part solution:

`x^2 -8x +12 -> (x-sqrt(8))^2`

You must not use the equals sign yet as `(x-sqrt(8))^2`
does not have the same value as `x^2 -8x +12`. That is why I have used 'maps to' (has a relationship).

Ok! Now we see what we have so far will give us then make adjustments so that the source equation does indeed have the same value.

What we have is: `(x-sqrt(8))^2 = x^2 -8x +8` The constant is 8 less than the original constant of 12 so our 'build' so far is 12-8=4 less than it should be. So we put it back!

`(x-sqrt(8))^2+4`

By the way, it is good mathematical practice to change any number that you are taking the root of such that you factor it down into a prime number then take any squared numbers outside of the root. Let me show you what I mean!

`sqrt(8) = sqrt(2 times 4) = sqrt(2 times 2^2) = 2 sqrt(2)`

So the final answer is :
`(x-2sqrt(2))^2+4`