How do you graph #y=-2x^2+8x#?

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Answer 1 · 2016-08-13T06:23:44

A parabola with zeros at #x=0# and #x=4#; a maximum value of #8# at #x=2# (See graph below).

#y=-2x^2+8x#
#y = -2x(x-4)#

Therefore #y=0# at #x =0 or 4#

#y' = -4x+8#

#y'=0 # at #x=2#

Since the coefficient of x^2 is negative, #y# has a maximum value at #x=2#

Therefore, #y_"max" = -2*2^2+8*2 = -8+16 = 8#

graph{-2x^2+8x [-12.6, 15.89, -4.86, 9.38]}