What symmetry does #f(x) = x^3-5x^2# have ?

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Answer 1 · 2017-12-22T23:35:01

#f(x)# has rotational symmetry of order #2# about the point #(5/3, -250/7)#

Given:

#f(x) = x^3-5x^2#

Note that this function is neither odd nor even. So it has no reflective symmetry about #x=0#, nor rotational symmetry about #(0, 0)#.

However, note also that:

#f(x) = x^3-5x^2#

#color(white)(f(x)) = (x-5/3)^3-25/3(x-5/3)-250/27#

So:

#f(x+5/3)+250/27# is an odd function and we can deduce that:

#f(x)# has rotational symmetry of order #2# about the point #(5/3, -250/7)#

graph{(y-x^3+5x^2)((x-5/3)^2+0.02(y+250/27)^2-0.01) = 0 [-10, 10, -40, 25]}