# What is the the vertex of y = (x - 2)^2 + 5x+4 ?

Sep 4, 2016

Vertex$\to \left(x , y\right) = \left(- \frac{1}{2} , \textcolor{w h i t e}{.} \frac{31}{4}\right)$

#### Explanation:

Square the brackets giving:

$y = {x}^{2} - 4 x + 4 + 5 x + 4$

$y = {x}^{2} + x + 8$
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Using part of the process of completing the square ( a sort of cheat method, but permitted).

Consider standard form $y = a {x}^{2} + b x + c$

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

In this case $a = 1$

In that we have $1 {x}^{2}$ (not normally written this way).

Thus $y = a \left({x}^{2} + \frac{b}{a} x\right) + c \text{ "->" } y = \left({x}^{2} + x\right) + 8$

color(blue)(x_("vertex")->(-1/2)xx(b/a) " "->" "(-1/2)xx1 = -1/2)
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Determine ${y}_{\text{vertex}}$ by substitution for $x$

y=x^2+x+8" "->" "color(blue)(y_("vertex")=(-1/2)^2-1/2+8 = 31/4)