How do you write $y = x^2 + 6x + 10$ in vertex form?

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Answer 1 · 2016-08-14T11:45:08

$color(green)(y=(x+3)^2+1)$

$y=x^2+6x+10$

$y=ax^2+bx+c" This is the standard form"$

$"The vertex form is expressed as "$

$y=a(x-h)^2-k$

$F=(h,k)$

$F=(h,k) "represents the focus coordinates"$

$"Solution -1:"$

$y=x^2+6x+9+1$

$color(green)(y=(x+3)^2+1)$

$h=-3$

$k=1$

$"Solution-2:"$

$h=-b/(2a)=-6/(2*1)=-3$

$k=(-3)^2+6*(-3)+10=9-18+10=1$

$a=1$

$"The vertex form:"$

$y=1*(x+3)^2+1$

$color(green)(y=(x+3)^2+1)$

How do you write y = x^2 + 6x + 10y = x^2 + 6x + 10 in vertex form?