How do you graph #y=-2x^2-4x+6#?

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Answer 1 · 2017-04-07T02:23:45

#y# is a parabola with an absolute maximum of #8# at #x=-1#
#y# has zeros at #x=-3# and #x=1#
#y =6# at #x=0#

#y=-2x^2-4x+6# is the equation of a parabola

Since the coefficient of #x^2<0# #y# will have an absolute maximum

#y = 0 -> -2x^2-4x+6 = 0#

#x^2+2x-3=0#

#(x+3)(x-1) = 0#

Hence #y# has zeros at #x=-3# and #x=1#

#y(0) = 0 + 0 +6 = 6#

#y# will have its absolute maximum value where #y' =0#

Hence where #y' = -4x-4 =0#

#x=-1#

#:.y_max = y(-1) = -2(-1)^2 -4(-1) +6#

#= -2+4+6=8#

Hence #y# has an absolute maximum of #8# at #x=-1#

There points can be seen on the graph of #y# below:

graph{-2x^2-4x+6 [-11.12, 11.38, -2.275, 8.975]}