# What is the the vertex of y=(x - 8)^2 + 20x+70 ?

Nov 25, 2015

$\text{the vertex is at: } \left(x , y\right) \to \left(- 2 , \textcolor{w h i t e}{.} 131\right)$

#### Explanation:

Expanding the bracket:
color(blue)(y=color(brown)((x^2-16x+64 ))+20x+70

Collecting like termes

$y = {x}^{2} + 4 x + 135. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$

Consider the $= 4 x$ term

$\textcolor{g r e e n}{{x}_{\text{vertex}} = \textcolor{b l a c k}{\left(- \frac{1}{2}\right) \times \left(+ 4\right) =} - 2} \ldots \ldots \ldots \ldots \ldots \left(2\right)$

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$\textcolor{b r o w n}{\text{Note that the equation must be in the form}}$
$\textcolor{b r o w n}{y = a \left({x}^{2} + \frac{b}{a} x\right) + c \text{. In your case } a = 1}$

color(brown)("so you end up with "color(green)( x_("vertex")= (-1/2)xx(b/a)))
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute (2) into (1) giving:

$y = {\left(- 2\right)}^{2} + 4 \left(- 2\right) + 135$

$y = 4 - 8 + 135$

$y = 131$

$\textcolor{b l u e}{\text{So the vertex is at: } \left(x , y\right) \to \left(- 2 , 131\right)}$