Fist we find the roots of the equation, ie the values of #x# st #y=0#:
# y=0 => -(x-2)(x+2)(x-4)=0 #
# :. (x-2)(x+2)(x-4)=0 #
# :. x=2, x=-2, x=4 #
Next we find the critical points, ie the values for which #y'=0#, we will need to multiply out as a cubic in order to this:
# y=-(x-2)(x+2)(x-4) #
# :. y=-(x-2)(x^2-4x+2x-8) #
# :. y=-(x-2)(x^2-2x-8) #
# :. y=-(x^3-2x^2-8x-2x^2+4x+16) #
# :. y=-(x^3-4x^2-4x+16) #
# :. y=-x^3+4x^2+4x-16 #
So, differentiating wrt #x# gives:
# :. y'=-3x^2+8x+4 #
# y'=0 => -3x^2+8x+4 = 0#
# :. 3x^2-8x-4 = 0#
# :. 3x^2-8x-4 = 0 #
This quadratic does not factorise, so we use the quadratic formula:
# x=(-(-8)+-sqrt((-8)^2-4(3)(-4)))/(2(3)) #
# :. x=(8+-sqrt(64+48))/(6) #
# :. x=(8+-sqrt(112))/(6) #
# :. x=-0.43, 3.10 # (2dp)
, or equally we can solve by completing the square:
# 3x^2-8x-4 = 0 #
# :. 3(x^2-8/3x-4/3) = 0 #
# :. (x^2-8/3x-4/3) = 0 #
# :. (x-8/6)^2 -(8/6)^2-4/3 = 0 #
# :. (x-8/6)^2 -64/36-4/3 = 0 #
# :. (x-8/6)^2 = 64/36+48/36 #
# :. (x-8/6)^2 = 112/36 #
# :. (x-8/6) = +-sqrt(112/36) #
# :. x=8/6 +-sqrt(112/36) #
again leading to #x=-0.43, 3.10 # (2dp)
The coordinates of these points are then given by using the original formula for #y# to give:
# x=-0.43 => y=-16.9 # (1dp)
# x=3.10 => y=5.0 # (1dp)
These values of #x# correspond to the critical points (or turning points). we now use the second derivative to determine the nature (max, min or point of inflection) of these points.
from earlier, # y'=-3x^2+8x+4 #
# :. y''=-6x+8 #
# x=-0.43 => y''=10.5 > 0 # (1dp)
# x=3.10 => y''=-10.6<0 # (1dp)
So we can now determine the nature of the turning points which are:
# (-0.43,-16.9) # minimum
# (3.10, 5.0) # maximum
We now have enough to sketch the curve:
graph{-(x-2)(x+2)(x-4) [-10, 10, -20, 20]}
