Question

Investigate the short-run behavior of the the third degree polynomial u(x)=x^3-x^2-6x.

Describe the graph of u(x). Where does it cross the x-axis? Where does it cross the y-axis? Where is u(x) positive? Where is u(x) negative?

Answer 1

The graph is an s shape because the leading power of the polynomial is 3. Geometrically, the graph "crosses" the x axis wherever the value of y is zero. Since u(x) is just another way to write "y", we can write this as \[y=x^3-x^2-6x\]. Now fully factor the polynomial and set it equal to zero (y is zero). Solve, and you will have where the graph crosses the x axis. so....\[0 = x(x^2-x-6) = x(x-3)(x+2)\].....which is true when x = 0, 3, or -2. Therefore, the graph crosses the x axis at these x coordinates. The graph crosses the y axis when x is zero, so evaluate u(0), which is simply equal to zero. Therefore the graph crosses the y axis at (0,0). The graph is positive when \[u(x)>0...x(x-3)(x+2)>0\], so now you have a polynomial inequality. To solve this, divide the number line up where the function is equal to zero and test points in those regions. Doing so you will see that the solutions are...\[[x|x \epsilon(-2,0)\cup(3,\infty)]\], solve in a similar manner to see where it is decreasing.

Answer 2

so solve \[x(x-3)(x+2) < 0\]

Answer 3

to see where the function is decreasing

Answer 4

This could more easily be done with calculus, which I am assuming that this is not a calculus class.

Answer 5

A plot is attached.