# What is the vertex form of y=-1/7(6x-3)(x/3+5) ?

Jan 4, 2018

$\textcolor{b r o w n}{\text{reworking the solution}}$

#### Explanation:

This is a link to a step by step guide to my shortcut approach. When applied properly it should only take about 4 to 5 lines all depending on the complexity of the question. https://socratic.org/s/aMg2gXQm

The objective is to have the format $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c + k$

Where $k$ is a correction making $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c \textcolor{w h i t e}{\text{d}}$ have the same overall values as $y = a {x}^{2} + b x + c$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Answering the question - the more formal approach}}$

$\textcolor{b r o w n}{\text{This is one of those situations where you just have to}}$$\textcolor{b r o w n}{\text{remember the standard form steps}}$

Lets multiply out the brackets

$y = - \frac{1}{7} \left(6 x - 3\right) \left(\frac{x}{3} + 5\right) \text{ } \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(1\right)$

$y = - \frac{1}{7} \left(2 {x}^{2} + 30 x - x - 15\right)$

$y = - \frac{1}{7} \left(2 {x}^{2} + 29 x - 15\right)$

Factor out the 2 from $2 {x}^{2}$. We do not want any coefficient in front of the ${x}^{2}$

$y = - \frac{2}{7} \left({x}^{2} + \frac{29}{2} x - \frac{15}{2}\right)$

Just for ease of reference set $g = {x}^{2} + \frac{29}{2} x - \frac{15}{2}$ giving:

$y = - \frac{2}{7} g \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left({1}_{a}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The x-intercepts are at $y = 0$ giving

${y}_{\text{x-intercept}} = 0 = - \frac{2}{7} g$

Thus it must be true that for this condition $g = 0$ thus we have:

$g = 0 = {x}^{2} + \frac{29}{2} x - \frac{15}{2}$

Add $\frac{15}{2}$ to both sides

$\frac{15}{2} = {x}^{2} \textcolor{red}{+ \frac{29}{2}} x$

To make the right hand side into a perfect square we need to add ${\left(\frac{1}{2} \times \textcolor{red}{\frac{29}{2}}\right)}^{2} \to {\left(\frac{29}{4}\right)}^{2}$ so add $\frac{841}{16}$ to both sides giving:

$\frac{15}{2} + \frac{841}{16} \textcolor{w h i t e}{\text{d")=color(white)("d}} {x}^{2} + \frac{29}{2} x + \frac{841}{16}$

$\frac{15}{2} + \frac{841}{16} \textcolor{w h i t e}{\text{d")=color(white)("d}} {\left(x + \frac{29}{4}\right)}^{2}$

$\frac{961}{16} \textcolor{w h i t e}{\text{d")=color(white)("d}} {\left(x + \frac{29}{4}\right)}^{2}$

$g = 0 = {\left(x + \frac{29}{4}\right)}^{2} - \frac{961}{16}$

But from $E q u a t i o n \left({1}_{a}\right) \text{ } y = - \frac{2}{7} g$ giving

$y = 0 = - \frac{2}{7} \left[{\left(x + \frac{29}{4}\right)}^{2} - \frac{961}{16}\right]$

$\textcolor{m a \ge n t a}{\text{I will let you finish this off.}}$ 