What is the axis of symmetry and vertex for the graph #y=2x^2 +8x-3#?

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Answer 1 · 2016-02-28T13:07:32

#x=-2" "#the axis of symmetry
#(-2,-11)" "# the Vertex

From the given
#y=2x^2+8x-3#
from the first two terms of the right side of the equation, factor out the number 2 then complete the square.

#y=2(x^2+4x+4-4)-3#

#y=2(x+2)^2-8-3#

#y=2(x--2)^2-11#

#2(x--2)^2=y+11#

#(x--2)^2=1/2(y--11)" " "#The vertex form

by inspection , we can see that #x=-2# is the axis of symmetry and the vertex is at #(h, k)=(-2, -11)#

graph{(x--2)^2=1/2(y--11)[-20,20,-12,10]}

God bless...I hope the explanation is useful.