What is the the vertex of $y = (x + 2)^2 + 3x+4$ ?

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Answer 1 · 2017-03-22T21:04:39

$"Vertex{-3.5" , "-4.25}$

$y=(x+2)^2+3x+4$

$y=x^2+4x+4+3x+4$

$y=x^2+7x+8" (1)"$

$(d y)/(d x)=0$

$(d y)/(d x)=2x+7=0$

$2x+7=0$

$2x=-7$

$x=-7/2=-3.5$

$"use (1)"$

$y=(-7/2)^2-7(7/2)+8$

$y=49/4-49/2+8$

$y=(49-98+32)/4$

$y=-17/4=-4.25$

$"Vertex{-3.5" , "-4.25}$

What is the the vertex of y = (x + 2)^2 + 3x+4 y = (x + 2)^2 + 3x+4 ?