# What is the vertex form of y= x^2 +8x +16?

Apr 26, 2016

$\textcolor{b l u e}{y = {\left(x + 4\right)}^{2}}$

#### Explanation:

Consider the standard for $\text{ } y = a {x}^{2} + b x + c$

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color(blue)("Scenario 1:" -> a=1)" " (as in your question)

Write as
$y = \left({x}^{2} + b x\right) + c$

Take the square outside the bracket.
Add a correction constant k ( or any letter you so chose)

$y = {\left(x + b x\right)}^{2} + c + k$

Remove the $x$ from $b x$

$y = {\left(x + b\right)}^{2} + c + k$

Halve $b$

$y = {\left(x + \frac{b}{2}\right)}^{2} + c + k$

Set the value of $k = \left(- 1\right) \times {\left(\frac{b}{2}\right)}^{2}$

$y = {\left(x + \frac{b}{2}\right)}^{2} + c - {\left(\frac{b}{2}\right)}^{2}$

Substituting the value gives:

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

$\textcolor{b l u e}{y = {\left(x + 4\right)}^{2}}$
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By changing the brackets content so that it has $\frac{b}{2}$ and then squaring $\frac{b}{2}$ you introduce a value that was not in the original equation. So you remove this using $k$ and thus returning the whole to its original inherent value.
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$\textcolor{b l u e}{\text{Scenario 2:} \to a \ne 1}$

Write as
$y = a \left({x}^{2} + \frac{b}{2 a} x\right) + c + k$

and you end up with

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c + k$

In this case $k = \left(- 1\right) \times {\left(\frac{a b}{2 a}\right)}^{2} = - {\left(\frac{b}{2}\right)}^{2}$

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - {\left(\frac{b}{2}\right)}^{2}$
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