# What is the vertex form of y=(x-6)(x-3)?

Apr 4, 2016

$\textcolor{b l u e}{y = {\left(x - \frac{9}{2}\right)}^{2} - \frac{9}{4}}$

#### Explanation:

given:$y = \textcolor{b l u e}{\left(x - 6\right) \textcolor{b r o w n}{\left(x - 3\right)}}$

Multiply out the brackets giving

$y = \textcolor{b r o w n}{\textcolor{b l u e}{x} \left(x - 3\right) \textcolor{b l u e}{- 6} \left(x - 3\right)}$

$y = {x}^{2} - 3 x - 6 x + 18$

$y = {x}^{2} - 9 x + 18$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compare to standard form $y = a {x}^{2} + b x + c$

Where $a = 1 \text{ ; "b=-9" ; } c = 18$

The standard for the vertex form of this equation is:

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - \left[{\left(\frac{b}{2}\right)}^{2}\right]$

So for your equation we have

$y = {\left(x - \frac{9}{2}\right)}^{2} + 18 - \left[- \frac{81}{4}\right]$

$\textcolor{b l u e}{y = {\left(x - \frac{9}{2}\right)}^{2} - \frac{9}{4}}$