How do you find the axis of symmetry of the quadratic equation #y = 2x^2 + 24x + 62#?

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Answer 1 · 2016-06-12T07:47:23

Complete the square to find the axis of symmetry is #x=-6#

Complete the square to get into vertex form:

#y = 2x^2+24x+62#

#=2(x^2+12x+31)#

#=2(x^2+12x+36-5)#

#=2(x+6)^2-10#

#=2(x-(-6))^2+(-10)#

The equation:

#y = 2(x-(-6))^2+(-10)#

is in standard vertex form:

#y = a(x-h)^2+k#

where #a=2# is a constant multiplier determining the steepness of the quadratic and #(h, k) = (-6, -10)# is the vertex.

The (vertical) axis of symmetry passes through the vertex, so has equation:

#x = -6#