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

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Answer 1 · 2016-03-29T14:37:38

Vertex#->(x,y)->(-2,11)#

Axis of symmetry #->x_("vertex") = -2#

Standard form #y=ax^2+bx+c#

Write as #y=a(x^2+b/ax)+c#

#x_("vertex") = (-1/2)xx b/a#

So for your question

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

Substituting #x=-2# gives

#y_("vertex")=-2(-2)^2-8(-2)+3 = -8+16+3 = 11#