How do you graph y=-x^3-x^2+5x?

1 Answer
Mar 21, 2017

y=f(x)=-x^3-x^2+5xy=f(x)=x3x2+5x

This is a cubic function with a negative aa value, so its end behavior is:
lim_(x->-oo)f(x)=oo
lim_(x->oo)f(x)=-oo

y=-(x)(x^2+x-5)
From the factored form above, there is a zero at x=0, so f(0)=0

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

f'(x)=-3x^2-2x+5=-(x-1)(3x+5)
f'(x) changes sign at x=-5/3 and at x=1. It changes from negative to positive at x=-5/3 and from positive to negative at x=1. This shows how the slope of f(x) changes.