It cuts x axis at three places, so we should have THREE values of x at which y is 0, one negative, one near or 0 itself and one positive..
Check the choices..
(A) \(y=x^3-x^2=x^2(x-1)\), so x=0 and 1..Only two values..Eliminate
(B) \(y=-x^3+x^2=x^2(1-x)\), so x=0 and 1..Only two values..Eliminate
(C) \(y=-x^3+x=x(1-x^2)=x(1-x)(1+x)\), so x=-1, 0 and 1..three values..possible
(D) \(y=x^3-x=x(x^2-1)=x(x-1)(1+x)\), so x=-1, 0 and 1..three values..possible
(E) \(y=x^3+x\).. Negative value NOT possible as x will never be 0

Between C and D, look at the coefficient of x^3..
If it is positive, it will sloping downwards on left and sloping upwards on right and vice versa..
So, here it should be positive..so x^3-x
You can also substitute values and see, say x= -2, then y=(-2)^3-(-2)=-6..True

D
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Why did we assume y to be negative. please help

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