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)#