Question

What is the vertex form of `y= (x-1)(x – 6)`?

Answer 1

`y=(x-7/2)^2 -25/4`

Explanation:

Let's convert this into standard form. Then we can "complete the square" to solve for the vertex form.

`y=(x-1)(x-6)`

`y=x^2-6x-x+6`

`y=x^2-7x+6`

Now let's complete the square. To do that, we need to find a value that make `x^2-7x` a perfect square. To find that value, we take the middle term, `-7`, and we divide it by `2`. That gives us `-7/2`. Now we square the fraction: `49/4`

Now we have the value that makes the equation true. But!! we cannot introduce a new value! Not without immediately subtracting it, which would make the final value `0`.

`y=x^2-7x+6 + 49/4 - 49/4`

So, we added `49/4` and then `-49/4`. Now let's rearrange it so we have a perfect square.... and other stuff:

`y=x^2-7x+49/4+6-49/4`

Let's rewrite `x^2-7x+49/4` as a perfect square: `(x-7/2)^2`

Now our equation is `y=(x-7/2)^2 +6-49/4`

combine like-terms

`y=(x-7/2)^2 -25/4`