# How do you write y=3x^2-18x+5 in vertex form?

Mar 27, 2015

Vertex form: $y = a {\left(x - h\right)}^{2} + k$

(That's what vertex form is in the textbook here. Other forms are possible.)

Method 1

Your teacher probably wants you to complete the square. (It is an important technique, used for other things also.)

This method relies on the fact that: ${\left(x \pm a\right)}^{2} = {x}^{2} \pm \left(2 a\right) x + {a}^{2}$

$y = 3 {x}^{2} - 18 x + 5 = 3 \left({x}^{2} - 6 x \text{ --------}\right) + 5$ (Leave yourself some apce inside the parentheses)

If the stuff in parentheses is going to be a perfect square ${\left(x - a\right)}^{2}$, then it must have $2 a = 6$ which tells me that $a = \left(\frac{1}{2}\right) \left(6\right) = 3$.

So the stuff inside parentheses is not a perfect square becuase it is missing ${a}^{2}$ which we would like to be $9$.

Now, (within limits) we are in charge here! We will add $9$ inside the parentheses to make it a perfect square

$\textcolor{red}{\text{PROBLEM}}$ $3 \left({x}^{2} - 6 x + 9\right)$ is $3 {\left(x - 3\right)}^{2}$, but it is not equal to what we started with.

$\textcolor{g r e e n}{\text{Solution}}$ We will add and subtract the $9$ and regroup.

It looks like this:

$y = 3 {x}^{2} - 18 x + 5 = 3 \left({x}^{2} - 6 x \text{ --------}\right) + 5$

$= 3 \left({x}^{2} - 6 x + 9 - 9\right) + 5$

Now regroup, keeping the perfect square were we want it to be. (We're in charge (within limits).)

$y = 3 \left({x}^{2} - 6 x + 9\right) - 3 \left(9\right) + 5$
(Convince yourself that this really is equal to what we started with. Do the algebra to simplify it.)

Now we'll write it this way:

$y = 3 {\left(x - 3\right)}^{2} - 27 + 5$

$y = 3 {\left(x - 3\right)}^{2} - 22$

(If your class uses a different vertex for, it's probably:
$\left(y - k\right) = a {\left(x - h\right)}^{2}$ so the answer would be $\left(y + 22\right) = 3 {\left(x - 3\right)}^{2}$)

Method 2

Depends on your knowing the vertex formula (which was 'discovered' by completing the square).

$y = a {x}^{2} + b x + c$ has vertex with $x$-coordinate $\frac{- b}{2 a}$

(Why> Well, start with $y = a {x}^{2} + b x + c$ and complete the square to get $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + \text{I'll leave it to you to find this}$

So $y = 3 {x}^{2} - 18 x + 5$ has vertex at $\frac{- \left(- 18\right)}{2 \left(3\right)} = 3$

And when $x = 3$,
we have $y = 3 {\left(3\right)}^{2} - 18 \left(3\right) + 5 = 3 \left(9\right) - 6 \left(9\right) + 5 = - 3 \left(9\right) + 5 = - 27 + 5 = - 22$
(I now, I do arithmetic kinda weird.)
So $a = 3$, $h = 3$ and $k = - 22$

$y = 3 {\left(x - 3\right)}^{2} - 22$