How do you find the maximum value of # y = -x^2 + 8x - 4#?

1 Answer
Steve M · Shwetank Mauria
Oct 20, 2016

12

Explanation:

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

# :. y' = -2x + 8 #

and #y''=-2# (See Note below)

As such we have maximum (as #y''<0#) at #y'=0 => -2x + 8 = 0 #

# :. 2x = 8 #
# :. x = 4 #

When #x=4 => y=-4^2+8(4)-4 = -16+32-4 =12 #

Note: As this is a quadratic and the coefficient of #x^2 <0# it should be obvious that the critical point corresponds to a maximum without the need to examine the second derivative.

Incidentally, this can also be fond without calculus by completing the square:
# y = -x^2 + 8x -4 #
# y = -(x^2 - 8x +4) #
# y = -((x-4)^2 -4^2+4) #
# y = -((x-4)^2 -16+4) #
# y = -((x-4)^2 -12) #
# y = -(x-4)^2 +12 #

So maximum of #y=12# occurs when #x-4=0=>x=4#
graph{-x^2 + 8x -4 [-16.17, 23.83, -4.8, 15.2]}

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