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

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Answer 1 · 2016-08-14T11:45:08

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

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

$y = a x^{2} + b x + c This is the standard form$ $y=ax^2+bx+c" This is the standard form"$

$The vertex form is expressed as$ $"The vertex form is expressed as "$

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

$F = (h, k)$ $F=(h,k)$

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

$Solution -1:$ $"Solution -1:"$

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

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

$h = - 3$ $h=-3$

$k = 1$ $k=1$

$Solution-2:$ $"Solution-2:"$

$h = - \frac{b}{2 a} = - \frac{6}{2 \cdot 1} = - 3$ $h=-b/(2a)=-6/(2*1)=-3$

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

$a = 1$ $a=1$

$The vertex form:$ $"The vertex form:"$

$y = 1 \cdot {(x + 3)}^{2} + 1$ $y=1*(x+3)^2+1$

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

How do you write y=x2+6x+10y = x^2 + 6x + 10 in vertex form?