# What is the vertex form of y=1/5x^2 -3/7x -16?

Mar 7, 2016

$\textcolor{b l u e}{\text{Thus vertex form} \to y = \frac{1}{5} {\left(x - \frac{15}{14}\right)}^{2} - \frac{3181}{196}}$

#### Explanation:

You can very easily go wrong on this one. There is a small detail that can easily be over looked.

Let $k$ be a constant yet to be determined

Given:$\text{ } y = \frac{1}{5} {x}^{2} - \frac{3}{7} x - 16$.......(1)

$\textcolor{b l u e}{\text{Build the vertex form equation}}$

Write as:$\text{ } y = \frac{1}{5} \left({x}^{2} - \textcolor{g r e e n}{\frac{15}{7}} x\right) - 16$..........(2)

$\textcolor{b r o w n}{\text{Note that } \frac{15}{7} \times \frac{1}{5} = \frac{3}{7}}$

Consider the $\frac{15}{7} \text{from } \frac{15}{7} x$

Apply$\frac{1}{2} \times \frac{15}{7} = \textcolor{red}{\frac{15}{14}}$

At this point the right hand side will not be equal to y. This will be corrected later

In (2) substitute $\textcolor{red}{\frac{15}{14}} \text{ for } \textcolor{g r e e n}{\frac{15}{7}}$

$\frac{1}{5} \left({x}^{2} - \textcolor{red}{\frac{15}{14}} x\right) - 16 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . . \left({2}_{a}\right)$

Remove the $x$ from $\frac{15}{14} x$

$\frac{1}{5} \left({x}^{\textcolor{m a \ge n t a}{2}} - \frac{15}{14}\right) - 16$

Take the power (index) of $\textcolor{m a \ge n t a}{2}$ outside the bracket

$\frac{1}{5} {\left(x - \frac{15}{14}\right)}^{\textcolor{m a \ge n t a}{2}} - 16 \text{ "color(brown)("Note that an error comes from the } \frac{15}{14}$

$\textcolor{b r o w n}{\text{This is still not yet equal to y}}$
Add the constant value of $\textcolor{red}{k}$

$\frac{1}{5} {\left(x - \frac{15}{14}\right)}^{\textcolor{m a \ge n t a}{2}} - 16 + \textcolor{red}{k}$

$\textcolor{g r e e n}{\text{Now it is equal to } y}$

$y = \frac{1}{5} {\left(x - \frac{15}{14}\right)}^{2} - 16 + \textcolor{red}{k}$.........(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To determine the value of } k}$

If we were to expand the bracket and multiply by the $\frac{1}{5}$ we would have the extra value of $\frac{1}{5} \times {\left(- \frac{15}{14}\right)}^{2}$. The constant $k$ is to counter this by removing it.

$\textcolor{b r o w n}{\text{Let me show you what I mean. Compare equation (1) to (3)}}$

$\frac{1}{5} {x}^{2} - \frac{3}{7} x - 16 \text{ " =" "y" "=" } \frac{1}{5} {\left(x - \frac{15}{14}\right)}^{2} - 16 + k$

$\frac{1}{5} {x}^{2} - \frac{3}{7} x - 16 \text{ " =" } \frac{1}{5} \left({x}^{2} - \frac{15}{7} x + {\left(\frac{15}{14}\right)}^{2}\right) - 16 + k$

$\frac{1}{5} {x}^{2} - \frac{3}{7} x - 16 \text{ } = \frac{1}{5} {x}^{2} - \frac{3}{7} x + \left[\frac{1}{5} \times {\left(\frac{15}{14}\right)}^{2}\right] - 16 + k$

$\frac{1}{5} {x}^{2} - \frac{3}{7} x - 16 \text{ } = \frac{1}{5} {x}^{2} - \frac{3}{7} x + \left[\frac{45}{196}\right] - 16 + k$

$\cancel{\frac{1}{5} {x}^{2}} - \cancel{\frac{3}{7} x} - \cancel{16} \text{ } = \cancel{\frac{1}{5} {x}^{2}} - \cancel{\frac{3}{7} x} + \left[\frac{45}{196}\right] - \cancel{16} + k$

$\implies 0 = \frac{45}{196} + k$

$\implies \textcolor{red}{k = - \frac{45}{196}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So equation (3) becomes:

$y = \frac{1}{5} {\left(x - \frac{15}{14}\right)}^{2} - 16 \textcolor{red}{- \frac{45}{196}}$.........(3)

$y = \frac{1}{5} {\left(x - \frac{15}{14}\right)}^{2} - \frac{3181}{196}$

$\textcolor{b l u e}{\text{Thus vertex form} \to y = \frac{1}{5} {\left(x - \frac{15}{14}\right)}^{2} - \frac{3181}{196}}$