# How do you write the quadratic function y=x^2-8x+10 in vertex form?

Sep 8, 2017

For an in depth explanation example see my solution:
https://socratic.org/s/aHWWFAP6

$y = {\left(x - 4\right)}^{2} - 6$

#### Explanation:

$\textcolor{b l u e}{\text{Preamble:}}$

I will be jumping steps to give you the answer. Working through my example should help with understanding what is going on.

Basically you may turn any equation into any form you wish as long as you incorporate something that takes it back to its original form.

Example

$y = 2 x + 4$ change this to $y = 2 x - 5$ which is not the same but
$y = 2 x - 5 + 9$ is the same thing but looks different.
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$\textcolor{b l u e}{\text{Answering the question - reduced explanation}}$

$y = {x}^{2} - 8 x + 10$

$y = {x}^{2} - 8 x + 10 + k \textcolor{w h i t e}{\text{cc}}$ at this stage $k = 0$

Take the square outside the brackets

$y = {\left(x - 8 x\right)}^{2} + 10 + k$

$y = {\left(x - \frac{8}{2}\right)}^{2} + 10 + k$

$y = {\left(x - 4\right)}^{2} - 6$
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Check

${\left(x - 4\right)}^{2} = {x}^{2} - 8 x + 16$

Putting it all together

$y = {x}^{2} - 8 x + 16 - 6$

$y = {x}^{2} - 8 x + 10 \textcolor{red}{\leftarrow \text{ Matches so true}}$