How do you graph #f( x ) = x ^ { 3} - 5x ^ { 2} + 9x + 6#?

1 Answer
Dec 22, 2017

See explanation...

Explanation:

Given:

#f(x) = x^3-5x^2+9x+6#

#color(white)(f(x)) = (x-5/3)^3+2/3(x-5/3)+317/27#

Hence #f(x)# is rotationally symmetric about the point #(5/3, 317/27)#

#f'(x) = 3x^2-10x+9#

which has no real zeros

#f'(5/3) = 3(5/3)^2-10(5/3)+9 = 25/3-50/3+27/3 = 2/3#

So at the point of inflexion at #(5/3, 317/27)# the cubic curve has slope #2/3#.

The #y# intercept of #f(x)# is at #(0, 6)#, being #(0, f(0))#.

The #x# intercept is somewhere in the interval #(-1, 0)#, since:

#f(-1) = -1-5-9+6 = -9 < 0#

Linearly interpolating, the #x# intercept is approximately at #x = -0.4# being #-6/(9+6)#

Let's get a better approximation with one step of Newton's method:

#-0.4-(f(-0.4))/(f'(0.4)) = -0.4-1.536/13.48 ~~ -0.51#

Hence we find the graph looks somewhat like this...

graph{(y-(x^3-5x^2+9x+6)) = 0 [-10, 10, -50, 50]}