Question 468073
f(x) = -3x^3 - 6x^2 + 10x + 9

f(-1) = -3(-1)^3 - 6(-1)^2 + 10(-1) + 9 = 3 - 6 -10 + 9 = 12 - 16 = -4

F(0) = 0 - 0 + 0 + 9

since the equation is continuous and it is negative at x = -1, and it is positive at x = 0, the equation must cross the x-axis somewhere in between.

a graph of this equation is shown below:

{{{graph(600,600,-2,2,-20,20,-3x^3 - 6x^2 + 10x + 9)}}}

you can see from the graph that it definitely crosses the x-axis somewhere in that interval.

the intermediate value theorem states that, if the function is continuous, and it is negative at one point in an interval and positive in another point in that interval, then it must be 0 somewhere in between.