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

Apr 9, 2017

$y = 3 {\left(x - 5\right)}^{2} + 7$

#### Explanation:

First we need to complete the square .

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

WAIT We added a $3 \cdot 25$ 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 $y - 82 + 3 \cdot 25 = 3 \left({x}^{2} - 10 x + 25\right)$

${x}^{2} - 10 x + 25$ is a perfect square now, so let's simplify it to $\left(x - 5\right) \left(x - 5\right)$ or just ${\left(x - 5\right)}^{2}$.

Now our equaation is starting to look like vertex form!
$y - 82 + 75 = 3 \left({\left(x - 5\right)}^{2}\right)$
$y - 7 = 3 \left(x - {5}^{2}\right)$
$y = 3 {\left(x - 5\right)}^{2} + 7$

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.