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

Question

Paisley Johnson · Accepted Answer

12

# 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]}

Cameron Alexander · Answer

undefined To find the maximum value of the function y = -x^2 + 8x - 4, you can follow these steps:

1. Determine the derivative of the function y' = -2x + 8.
2. Set the derivative equal to zero to find critical points: -2x + 8 = 0.
3. Solve for x: -2x = -8, x = 4.
4. Plug the critical point(s) back into the original function to find the corresponding y-value(s): y = -4^2 + 8(4) - 4 = 8.
5. Therefore, the maximum value of the function is y = 8 at x = 4.