# What is the vertex form of y= 6x^2-9x+3 ?

Mar 2, 2016

( x $-$ 3/4 )^2 = 1/6 ( y + 3/8 ).
The vertex of the parabola is ( 3/4, $-$3/8).

#### Explanation:

The axis of the parabola is in the positive y-direction.
The focus is at ( 3/4, $-$3/8 $-$1/24.) = ( 3/4, $-$5/12.), below the vertex on the axis x = 3/4.

Mar 2, 2016

$y = 6 {\left(x - \frac{3}{4}\right)}^{2} + \frac{39}{16}$

The solution method has been shown in a lot of detail

#### Explanation:

Given:$\text{ } y = 6 {x}^{2} - 9 x + 3$ ................................(1)

This process introduces an error. This error is dealt with by introducing a correction constant.

Let the correction constant be $\textcolor{g r e e n}{k}$

$\textcolor{b l u e}{\text{Step 1}}$

Write as:$\text{ } y = 6 \left({x}^{2} - \frac{9}{6} x\right) + 3$

Note that$\text{ } 6 \times \left(- \frac{9}{6}\right) x = - 9 x$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2}}$

$\textcolor{b r o w n}{\text{Move the index of 2 from "x^2" to outside the brackets.}}$

We have now changed the value of the RHS so unable at this stage to equate it to $y$

Write as:$\text{ "6(x^(color(magenta)(2))-9/6x)+3" " ->" } 6 {\left(x - \frac{9}{6} x\right)}^{\textcolor{m a \ge n t a}{2}} + 3$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3}}$

$\textcolor{b r o w n}{\text{Remove the right hand side "x" from inside the brackets.}}$

$6 {\left(x - \frac{9}{6} \textcolor{m a \ge n t a}{x}\right)}^{2} + 3 \text{ "->" } 6 {\left(x - \frac{9}{6}\right)}^{2} + 3$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4}}$

$\textcolor{b r o w n}{\text{Add correction constant of } \textcolor{g r e e n}{k}}$

$6 {\left(x - \frac{9}{6}\right)}^{2} + 3 \text{ " ->" } 6 {\left(x - \frac{9}{6}\right)}^{2} + 3 + \textcolor{g r e e n}{k}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 5}}$

$\textcolor{b r o w n}{\text{Halve the "9/6" inside the brackets}}$

$\implies y = 6 {\left(x - \frac{9}{12}\right)}^{2} + 3 + k$.........................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for } k}$

Expand the brackets of equation (2)

$y = 6 {x}^{2} - 9 x + \frac{81}{144} + 3 + k$

The $\frac{81}{144}$ is the error that $k$ is correcting

So to get rid of $\frac{81}{144}$ we make k $- \frac{81}{144}$

so our constant become: $\text{ } 3 - \frac{81}{144} = 2 \frac{7}{16} = \frac{39}{16}$

Equation (2) becomes: $\text{ "y=6(x-9/12)^2+39/16" } \ldots . . \left({2}_{a}\right)$

But $\frac{9}{12} = \frac{3}{4}$ so we now have

color(magenta)(" "y=6(x-3/4)^2+39/16)" ".....(2_b)