How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of #y=x^2+8x+12#?

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Answer 1 · 2018-06-21T15:59:03

Minimum (-4,-4)
Axis of symmetry #x=-4#
y intercept (0,12)
x intercept (-2,0) and (-6,0)
Domain #(-oo, oo)#
Range #[-4, 00)

#x=[-b]/[2a]# gives the x coordinate for the vertex

#x=[-8]/2=-4#

When #x=-4# #y=[-4]^2+8xx-4+12#

#y=16-32+12=-4#

The vertex is (-4,-4)

If we factor the equation #y=(x+2)(x+6)#

So the x intercepts are when #y=0 => (x+2)(x+6)=0#

#x=-2 or x=-6#