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

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Answer 1 · 2015-06-06T17:02:44

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

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

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

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

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

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

In general, if $y = a x^{2} + b x + c$ $y = ax^2+bx+c$ then

$y = a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $y = a(x+b/(2a))^2 + (c - b^2/(4a))$

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

and axis of symmetry $x = - \frac{b}{2 a}$ $x = -b/(2a)$

How do you find the axis of symmetry for a quadratic equation y=x2+8x+12y=x^2+8x+12?