Question 1121879
y=11/(8x^3) is graphed.  If the problem is (11/8)x^3, that is a different matter
x can't be 0, or the equation is undefined. As x approaches 0 from the positive side, the equation goes to positive infinity.  As x approaches 0 from the negative side, the equation goes to negative infinity.
x=1 y=11/8
x=2 y=11/64
as x gets large, y goes to 0 from the positive side

x=-1, y=-11/8
x=-2, y=-11/64 and as x becomes more negative, y goes to 0 from the negative side

{{{graph(300,300,-10,10,-2,2,11/(8x^3))}}}