What is the axis of symmetry and vertex for the graph #y = –2x^2 – 32x – 126#?

3 general conceptual options.

1: Determine the x-intercepts and the vertex is #1/2# way between. Then use substitution to determine Vertex.

2: Complete the square and almost directly read of the vertex coordinates.

3: Start the 1st step of completing the square and use that to determine #x_("vertex")#. Then by substitution determine #y_("vertex")#
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Given: #y=-2x^2-32x-126#

#color(blue)("Option 1:")#

Try to factorise #-> -2(x^2+16x+63)=0#

Note that #9xx7=63 and 9+7=16#

#-2(x+7)(x+9)=0#

#x=-7 and x=-9#

#x_("vertex")=(-16)/2=-8#
By substitution you can determine #y_("vertex")#

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#color(blue)("Option 2:")#

Given: #y=-2x^2-32x-126#

#y=-2(x^2+16x)+k-126 larr" At this stage "k=0#

Halve the 16, remove the #x# from #16x# and move the squared.

#y=-2(x+8)^2+k-126 larr" "k" now has a value"#

Set #-2(8)^2+k=0 => k = 128#

#y=-2(x+8)^2+128-126#

#y=2(xcolor(red)(+8))^2color(green)(+2)#

#x_("vertex")=(-1)xxcolor(red)(8)=color(magenta)(-8)#
Vertex #->(x,y)=(color(magenta)(-8),color(green)(2))#
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#color(blue)("Option 3:")#
Given: #y=-2x^2-32x-126#

#y=-2(x^2+16x)+k-126#

#x_("vertex")=(-1/2)xx16 = -8#

By substitution determine #y_("vertex")#