# What is the vertex form of y=8x^2 − 6x + 128 ?

Jan 30, 2016

color(blue)(y_("vertex form")=8(x-3/8)^2+ 126 7/8

$\textcolor{b r o w n}{\text{ explanation given in detail}}$

#### Explanation:

Given: $\text{ } y = 8 {x}^{2} - 6 x + 128$ ..........(1)

Write as $\text{ } y = 8 \left({x}^{2} - \frac{6}{8} x\right) + 128$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Now we start to change things a step at a time.}}$

$\textcolor{g r e e n}{\text{Change the bracket so that this part becomes:}}$
$8 {\left\{x - \left(\frac{1}{2} \times \frac{6}{8}\right)\right\}}^{2}$

$\textcolor{g r e e n}{\text{Now put back the constant giving:}}$
$8 {\left\{x - \left(\frac{1}{2} \times \frac{6}{8}\right)\right\}}^{2} + 128$

$\textcolor{g r e e n}{\text{But this change has introduced an error so we can not yet equate it}}$ $\textcolor{g r e e n}{\text{to } y .}$

$y \ne 8 {\left\{x - \left(\frac{1}{2} \times \frac{6}{8}\right)\right\}}^{2} + 128$

$\textcolor{g r e e n}{\text{We fix that by adding another constant ( say k ) giving:}}$

$y = 8 {\left\{x - \left(\frac{1}{2} \times \frac{6}{8}\right)\right\}}^{2} + 128 + k$ .........................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{To find the value of } k}$

$\textcolor{g r e e n}{\text{Equate (2) to (1) through } y}$

$8 {\left\{x - \left(\frac{1}{2} \times \frac{6}{8}\right)\right\}}^{2} + 128 + k \text{ " =" } 8 {x}^{2} - 6 x + 128$

$8 \left({x}^{2} - \frac{3}{8} x - \frac{3}{8} x + \frac{9}{64}\right) + 128 + k \text{ "=" } 8 {x}^{2} - 6 x + 128$

$\cancel{8 {x}^{2}} - \cancel{6 x} + \frac{9}{8} + \cancel{128} + k \text{ "=" } \cancel{8 {x}^{2}} - \cancel{6 x} + \cancel{128}$

$k = - \frac{9}{8}$ ....................................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute (3) into (2)

color(blue)(y_("vertex form")=8(x-3/8)^2+ 126 7/8

Note$\text{ } \frac{3}{8} = 0.375$

So
color(blue)(" "x_("vertex") = (-1)xx(-3/8) = + 0.375)
$\textcolor{b l u e}{\text{ "y_("vertex}} = 126 \frac{7}{8} = 126.875$ 