how do you find the real zeros and multiplicity of f(x)=-x^2(x^2-4)(x-5)? 
How do you determine whether the graph crosses or touches the x-axis at 
each x-intercept?

The real zeros of a function are the real values of x which do these two 
things:

1. Cause the expression -x²(x²-4)(x-5) to equal to zero.

2. Cause the graph of y = -x²(x²-4)(x-5) to be tangent to (just touch) the
x-axis when the value has an even multiplicity and cross the x-axis when it
has an odd multiplicity.

First we write -x²(x²-4)(x-5) in terms of linear factors, i.e. factors 
which only have terms in x¹, or just x (to the first power).

-x²(x²-4)(x-5) 

Write the x² as (x)(x) and (x²-4) as its factored form (x-2)(x+2)

-(x)(x)(x-2)(x+2)(x-5)

We set each of these equal to 0.

(We can ignore the negative sign since if any one of those factors are 0,
the annexation of a negative sign cannot affect the result of 0.)

We have 0 as a zero of multiplicity 2, since (x) appears 2 times and 2 is even,
so the graph will TOUCH but not CROSS THROUGH the x-axis where x is 0.

we have 2 as a zero of multiplicity 1, because (x-2) appears only 1 time and 1
is odd, so the graph will CROSS THROUGH the x-axis where x = 2.

we have -2 as a zero of multiplicity 1 , because (x+2) appears only 1 time and
1 is odd, so the graph will CROSS THROUGH the x-axis where x = -2.

we have 5 as a zero of multiplicity 1, because (x-5) appears only one time and
1 is odd, so the graph will CROSS THROUGH the x-axis where x = 5.

Here is the graph:



Edwin