Question

How do you write `F(x) = 3x^2 − 30x + 82` into vertex form?

Answer 1

`y=3(x-5)^2+7`

Explanation:

First we need to complete the square .

`y=3x^2-30x+82`
`y-82=3x^2-30x`
Now let's find the constant that will make `3x^2-30x` a perfect square.
First we need to factor out that `3`
`y=82=3(x^2-10x)`
Now we need to find that constant. To do that, we halve `10` and then square that number, like so `(1/2*10)^2`, which is 25. That gives us `y-82=3(x^2-10x+25)`

WAIT We added a `3*25` on one side of the equation, but we need to keep both sides equal. What we do to one side we must do to the other. Our real equation is `y-82+3*25=3(x^2-10x+25)`

`x^2-10x+25` is a perfect square now, so let's simplify it to `(x-5)(x-5)` or just `(x-5)^2`.

Now our equaation is starting to look like vertex form!
`y-82+75=3((x-5)^2)`
`y-7=3(x-5^2)`
`y=3(x-5)^2+7`

Just to double check our work, we can graph both equations. If we did this right, they should look identical

graph{y=3x^2-30x+82}
graph{y=3(x-5)^2+7}

You have to zoom out a fair bit, but the graph are the same. Good job! We were right.