What is the the vertex of $y = x^2+4x+20$ ?

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Answer 1 · 2017-02-26T13:09:23

Vertex $->(x,y)=(-2,16)$

The format of the question is already as:

$y=ax^2+bx+c" "->" "y=a(x^2+b/ax)+c$

as $a=1$

$x_("vertex")=(-1/2)xxb/a" "->" "=(-1/2)xx4 = -2$

So by substitution

$y_("vertex")=(-2)^2+4(-2)+20 = 16$

Vertex $->(x,y)=(-2,16)$

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What is the the vertex of y = x^2+4x+20 y = x^2+4x+20 ?