Question 107150
Crosses the x axis at -4,0,1, in other words f(x)=0. 
It must have a form like at least like this, 
f(x)=g(x)*(x+4)*x*(x-1)
because
at x=-4, x+4=0, f(x)=0
at x=0, x=0, f(x)=0
at x=1, x-1=0, f(x)=0
We'll get to g(x) later.
OK, now we satisfy the crossing conditions. 
Let's look at what happens between (-4,0) and (0,1).
Between (-4,0), x+4>0, x<0, and x-1<0.
If you multiply the product of those you would have positive times negative times negative which would give me positive, that is, lies above the x axis. The function g(x) would have to be positive in this range also. 
Between (0,1), x+4>0, x>0, and x-1<0.
If you multiply the product of those you would have positive times positive times negative which would give me negative, that is, lies above the x axis. The function g(x) would have to be positive in this range also. 
Since you satisfy the conditions for above and below the x axis in the given ranges and you cross the x axis at the right points, we're almost done. 
Now for g(x). You can make it as complicated as I want as long as it stays positive in the given ranges(Example:g(x)=x*x). 
You also want to make the degree of the polynomial as small as possible. 
So let's make g(x) as simple as can be. 
Let's make g(x)=1.
Therefore,
{{{f(x)=(x+4)*x*(x-1)}}}
satisfies all of your criteria. 
If you want to expand it, it would be,
{{{f(x)=x^3+3x^2-4x}}}
{{{ graph( 300, 300, -5, 5, -10, 10, x^3+3x^2-4x)}}}