How do you find the axis of symmetry for a quadratic equation $y=x^2+8x+12$ ?

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Answer 1 · 2015-06-06T17:02:44

$y = x^2+8x+12$

$=(x+4)^2-4^2+12$

$=(x+4)^2-16+12$

$=(x+4)^2-4$

So the vertex is given by $x = -4$ , $y = -4$

... the axis of symmetry being the vertical line $x = -4$

In general, if $y = ax^2+bx+c$ then

$y = a(x+b/(2a))^2 + (c - b^2/(4a))$

which has vertex at $(-b/2a, c - b^2/(4a))$

and axis of symmetry $x = -b/(2a)$

How do you find the axis of symmetry for a quadratic equation y=x^2+8x+12y=x^2+8x+12?