The document explains the process of finding a polynomial function based on given roots and their multiplicities. It details the steps of identifying zeros, determining behaviors based on multiplicity, adjusting for the y-intercept, and ensuring the correct end behavior of the polynomial. The final polynomial derived from the example is f(x) = -1/10(x + 5)(x + 1)(x - 2)².

1. Finding the polynomial from data
Tyler Murphy
September 2, 2014
The

2. rst thing to realise is that this is essentially the reverse process of sketching a graph.
Consider the following problem:
1 Example:
Find the polynomial function with f(5) = f(1) = f(2) = 0 where 2 has an even
multiplicity and 5 and -1 have an odd multiplicity. Additionally, f(0) = 2, f(x) 0
for all x 5 and f(x) 0 for all x 2.
1.1 solution:
1.1.1 Step 1: The Zeroes
First consider what your zeroes are and what that looks like in terms of a polynomial
function.
In this case, the zeroes are -5, -1,and 2.
But what does this mean? Consider the function f(x) = x 5: What would the zero
be here? What about the function f(x) = (x 3)(x 5) What are the zeroes here?
This should give you an idea about the pattern involved here. If your zero is x = a
then your polynomial has a term (x a).
So this function has the terms (x 5); (x 1); and (x 2).
1.1.2 Using the Multiplicity
First, you have to recall what the multiplicity of a function tells us. If the multiplicity of a
zero is even, then the graph simply touches down at that point. If the multiplicity is odd,
then the graph passes through that zero.
1

3. So for this problem, we have an odd multiplicity at -5 and -1. So we know the graph
passes through these points. We also know that the function only touches at x = 2 instead
of passing through it because the multiplicity of 2 is even.
So we know that our polynomial now looks something like this:
f(x) = (x + 5)1(x + 1)1(x 2)2
Think about why this makes sense. Since -5 is the solution to x + 5 = 0 and it has an
odd multiplicity, we know that the polynomial must have a term that looks like (x + 5)k
where k is some odd number. This follows for the other terms.
But what if the multiplicity of -5 is some other odd number than 1? What if it's a
large odd number like 101? It very well could be for another polynomial that hits these
same points. Know that there are in

4. nitely many solutions to this problem. But in this
class, you'll only work with the smallest exponent solutions. If you have a graph that
accompanies the data, this will become more apparent because you should intuitively be
able to recognize the degree of the function simply by looking at the graph.
In this case, we have created a 4th degree polynomial, so we know the graph is shaped
something like a stretched out w. We will see how this

5. ts what we know.
1.1.3 Using the Y-intercept
It may be tempting to think you are done at this point. However, if you simply turned
this in, you'd be wrong. The y-intercept for the function we have so far is :
f(0) = (0 + 5)(0 + 1)(0 2)2 = (5)(1)(4) = 20
If you recall, the problem states that our y-intercept is -2, not 20.
So how do we correct this? The simple answer is to simply multiply the function by a
constant that will give us -2 instead of 20. So what is that constant and how do we

6. nd
it?
20 = 1
10 .
We simply need to solve the equation c 20 = 2. Clearly we see that c = 2
So now we have a polynomial:
f(x) =
1
10
(x + 5)(x + 1)(x 2)2
Let's see if this works. It's pretty obvious that our zeroes haven't changed by setting
the equation equal to zero and multiplying both sides by -10 (to get rid of the constant).
But what about the y-intercept? Is it now -2 like it should be? Let's check.
f(0) =
1
10
(0 + 5)(0 + 1)(0 2)2 =
1
10
(5)(4) =
20
10
= 2
So our y-intercept is now correct.
2

7. 1.1.4 Checking End Behavior
The last thing to do is to check that your polynomial function generates the correct ending
behavior. We have created a 4th degree polynomial. As we know with any even degree
polynomial, the left hand end behavior will match the right hand behavior. Odd degree
polynomials will be opposite. Consider the graphs of f(x) = x2 and f(x) = x3 to verify
this.
Our 4th degree polynomial has the same behavior on the left as the right. Since the
constant at the front of our polynomial is negative, we know our graph opens down. This

8. ts with what we are given: f(x) 0 for all x 5 and f(x) 0 for all x 2.
So, our polynomial has passed all requirements and is complete.
So the

9. nal solution is:
f(x) =
1
10
(x + 5)(x + 1)(x 2)2
Note that it is not necessary to multiply this all out or even to reduce the constant in
the front. A factored polynomial is completely acceptable and may be easier to work with
in later problems if you need it.
3