# What is the vertex form of y=(x + 21)(x + 1) ?

Mar 23, 2016

color(blue)("Vertex "-> (x,y)->(-11,-100)

#### Explanation:

For a more detailed explanation of method see the example of
http://socratic.org/s/asZq2L8h .Different values but the method is sound.

Given:$\text{ } y = \left(x + 21\right) \left(x + 1\right)$

Let $k$ be the error correcting constant

Multiply out giving

$\text{ } y = {x}^{2} + 22 x + 21$

y=(x^(color(magenta)(2))+22x)+21+k" "color(brown)("No error yet so k=0 at this stage")

Move the power to outside the bracket

y=(x+22color(green)( x))^(color(magenta)(2)) +21+k" "color(brown)("Now we have the error "->k!=0 )

Remove the $x$ from $22 \textcolor{g r e e n}{x}$

$\text{ } y = {\left(x + \textcolor{red}{22}\right)}^{2} + 21 + k$

Multiply $\textcolor{red}{22} \text{ by } \left(\frac{1}{2}\right) = \textcolor{b l u e}{11}$

$\text{ } \textcolor{g r e e n}{y = {\left(x + \textcolor{red}{22}\right)}^{2} + 21 + k}$

$\text{changes to } \textcolor{g r e e n}{y = {\left(x + \textcolor{b l u e}{11}\right)}^{2} + 21 + k}$

The error introduced is ${\left(a \times \frac{b}{2}\right)}^{2} \to {\left(1 \times \frac{22}{2}\right)}^{2} = + 121$

So $k = - 121$ to 'get rid' of the error
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So
$\text{ } \textcolor{m a \ge n t a}{y = {\left(x + \textcolor{b l u e}{11}\right)}^{2} + 21} - 121$

$\text{ } y = {\left(x + 11\right)}^{2} - 100$

$\text{ "color(blue)("Vertex } \to \left(x , y\right) \to \left(- 11 , - 100\right)$