How do you write the vertex form equation of the parabola y=3x^2-12x+9?

Apr 26, 2016

$y = 3 {\left(x - 2\right)}^{2} - 3$ with vertex at $\left(2 , - 3\right)$

Explanation:

The general vertex form is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given
$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{2} - 12 x + 9$
Extract the $\textcolor{g r e e n}{m}$ factor (see the general vertex form above)
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{3} \left({x}^{2} - 4 x\right) + 9$

Since in its general form a squared binomial
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$
If ${x}^{2} - 4 x$ are the first two terms of a squared binomial,
the third term needs to be $+ 4$

If we add a $+ 4$ to complete the square inside the parentheses we will need to subtract $3 \times \left(+ 4\right)$ outside the parentheses to keep the same value.
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{3} \left({x}^{2} - 4 x \textcolor{b r o w n}{+ 4}\right) + 9 \textcolor{b r o w n}{-} \left(\textcolor{g r e e n}{3}\right) \cdot \left(\textcolor{b r o w n}{+ 4}\right)$

Re-writing as a squared binomial and simplifying the constant term
$\textcolor{w h i t e}{\text{XXX")y=color(green)(3)(x-color(red)(2))^2+color(blue)(} \left(- 3\right)}$
which is the vertex form for a parabola with vertex at $\left(\textcolor{red}{2} , \textcolor{b l u e}{- 3}\right)$

Here's a graph of the original equation for verification purposes:
graph{3x^2-12x+9 [-1.637, 7.132, -4.24, 0.142]}