# What is the vertex form of y=7x^2-9x-32?

Jan 17, 2016

${y}_{\text{vertex form}} = 7 {\left(x - \frac{9}{14}\right)}^{2} - \frac{977}{28}$

#### Explanation:

Given: $y = 7 {x}^{2} - 9 x - 32$......................(1)

Write as:

$y = 7 \left({x}^{2} - \frac{9}{7} x\right) - 32$

Now write as

$y = 7 {\left(x - \left[\frac{1}{2} \times \frac{9}{7}\right]\right)}^{2} - 32 \textcolor{b l u e}{+ \text{correction}}$

$y = 7 {\left(x - \frac{9}{14}\right)}^{2} - 32 \textcolor{b l u e}{+ \text{correction}}$..........................(2)
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Consider $7 {\left(x - \frac{9}{14}\right)}^{2}$
This gives: $7 \left({x}^{2} - \frac{9}{7} x + \frac{81}{196}\right)$
We need the $7 \left({x}^{2} - \frac{9}{7} x\right)$ but the $7 \left(+ \frac{81}{196}\right)$ is an extra value we need to get rid of. This is why we have a correction. In this case the correction value is:$\textcolor{b l u e}{7 \left(- \frac{81}{196}\right) = - \frac{81}{28}}$
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So equation (2) becomes:

$y = 7 {\left(x - \frac{9}{14}\right)}^{2} - 32 \textcolor{b l u e}{+ \left(- \frac{81}{28}\right)}$..........................(2_a)

$y = 7 {\left(x - \frac{9}{14}\right)}^{2} - \frac{977}{28}$

("So " x_("vertex")=(-1)xx(-9/14)color(white)(..)=+9/14)