What is the axis of symmetry and vertex for the graph #y=x^2+8x+12#?

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Answer 1 · 2018-05-10T02:28:58

The axis of symmetry is #x = -4#, the vertex is #(4, -4)#

We can find the axis of symmetry in the form of a line, #x#

To find the value of line #x#, we use this equation:
#x = (-b)/(2a)#.

In our given graph, #a = 1#, and #b = 8#

After inserting values:

#x = (-8)/(2*1) = (-8)/2 = -4#

The equation for our axis of symmetry is #x = -4#

To find the vertex, we must convert our parabola's equation to vertex form: #y = a(x - h)^2 + k#. We can do this by completing the square.

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

#y - 12 = x^2 + 8x#

#y - 12 + 16= x^2 + 8x + 16#

#y + 4= (x + 4)^2#

#y = (x+4)^2 - 4#

#h = -4, k = -4#

Our vertex is h,k, and thus, (-4,-4).