Finding a Polynomial from a Graph (3 Important Steps) | jdeducation (2023)

polynomials. Algebra students spend countless hours studying polynomials. well maybe notinnumerablehrs. Your polynomial training probably started in high school when you were learning about linear functions. This happened in the days when math went from lots of numbers to lots of letters!

In this article, we'll look at how to write the equation of a polynomial function given its graph. There are many things to consider in this process. Let us begin!

What is a polynomial? (checking the polynomials)

First we need to reiterate a few things about polynomials. What is a polynomial? If we know anything about language, the word "poly" means "many" and the word "nomio" means "concepts".

A polynomial is an expression with many terms. A monomial is an expression, but for our purposes we will consider it a polynomial.

All of the following expressions arePolynome:

• 5x⁴- 2x³+x – 10
• 10x³+ 7x²– 5x + 8
• -2x²
• (1/2)x -1
• 5

The following expressions are NOT polynomials:

A nice property of polynomial graphs is that they are smooth. Think of the graph of a parabola or the graph of a cubic function.

There are no sharp turns or corners on the map. Now let's look at one type of problem that we will solve in this lesson.

Example 1 - Graph of a 3rd degree polynomial

The graph of a 3rd degree polynomial is shown. Write the functional equation.

Before we solve the above problem, let's check the definition ofdegree of a polynomial.

For example the polynomialF(X) = 5X⁷+ 2X³– 10 is one7th gradePolynom.

Now let's write a function for the given graph. We're going to make great use of an important theorem of algebra:The factor set.

The set of factors helps us enormously when working with polynomials:if we know a zero of the function, we can find a factor.

Coming back to our example problem, there are several key points in the graph: the three zeros and the y-intercept. Let's identify these points:

Notice that the graph goes straight through the x-axis three times. Imagine zooming in on each x-intersection.

Although the function isn't linear, if you zoom in on any of the intersection points, the graph will increase in sizelinear aspect. For example, if you zoom to zero (-1, 0), the polynomial graph looks like this:

Please note: this is the graph of a curve, but it looks like a straight line! By the factor theorem we know that if -1 is zero then (X+ 1) is a factor.

Since -9 and 4 are also zeros, (X+ 9) y (X– 4) are also factors. Therefore we can write our polynomial as follows:

• F(X) =A(X+ 1)(X+ 9)(X– 4)

Now we can calculate the value of the constantA. We can do this using another point on the chart.

Usually a point that is easy to find on a chart is thisj-intersection which we have already discovered as point (0. -4). If hejsection is not at the intersection of the grid lines of the chart, it may not be easy to determine definitely from the chart.

In this case, sometimes a relative high or low on the chart can be easy to read. Or find a point on the chart that coincides with the intersection of two grid lines.

calculationA, replace the values ​​of (0, -4) with (X,j) into the equation:

• -4 =A(0 + 1)(0 + 9)(0 – 4)
• -4 =A(1)(9)(-4)
• -4 = -36A
• A= 1/9

So the polynomial function is:

• F(X) = (1/9)(X+ 1)(X+ 9)(X– 4)

If we wanted to put that in standard form, we'd have to multiply it. Let's not bother this time! 🙂

How can we find the degree of a polynomial from its graph?

Let's analyze the degree of a polynomial in a little more detail. Suppose we get the graph of a polynomial but not the degree. How can we find the degree of the polynomial? The fundamental theorem of algebra can help us with this.

For example, a linear equation (degree 1) has a root. A quadratic equation (degree 2) has exactly two roots. A cubic equation (degree 3) has three roots. Etc. When counting the number of roots, we include both complex roots and multiple roots.For our purposes in this article, we will only considerrealestate. That means:

Memory:The real zeros of a polynomial correspond to thisX-Intersections of the chart. The roots of a polynomial are the solutions of the equationF(X) = 0.

AndPage(X) = 2(X- 3)²(X+ 5)³(X- 1). The zeros are 3, -5 and 1.

We say that zero 3 has multiplicity 2, -5 has multiplicity 3, and 1 has multiplicity 1. When a polynomial is factored, the multiplicity is the power of each factor.

The graph goes straight through theX-Axis.

How does this help us in our search for the degree of a polynomial from its graph? First, let's look at some polynomials of different degrees to create a pattern.

1st Class

j= 2X+ 3

1 real zero.

Note 2

j=X²

1 real zero with a multiplicity of 2.

The graph bounces off theX-Axis.

Note 2

j= (X- 3)²+ 1

No real zeros.

The graph does not touch or crossX-Axis.

This is a situation for another day!

Note 2

y = ½ (X– 5)(X+ 2)

2 real zeros.

In eachX-intersection, the graph runs straight through theX-Axis. The graph looks roughly linear at every zero. Each zero has a multiplicity of one. Each zero is a single zero.

3rd grade

j=X³

1 real zero.

The graphic is borderingX-axis and crosses to the other side.

The zero that occurs inX= 0 has multiplicity 3.

3rd grade

j= 0,10 (X-1)(X+ 4)(X– 5)

3 real zeros.

In eachXsection that cuts right through the graphX-Axis. The graph looks roughly linear at every zero. Each zero has a multiplicity of 1.

Well, we've seen polynomials of degree 1, 2, and 3. It seems we have situations where the graph goes right through theX-axis, the graph bounces off theX-axis, or the graph will rub against theX- intercept while driving through.

A closer examination of polynomials of degree higher than 3 will allow us to summarize our results. Hold:

Note, for polynomials of even degreej=X²,j=X⁴, jj=X⁶, as the power of the variable increases, the parabola flattens out near zero.

For odd degree polynomials:j=X³,j=X⁵, jj=X⁷, the graphic borders on theX-Axis each crossing theX-y-axis also flattens as variable power increases.

We now summarize our results:

These results will help us in the task of determining the degree of a polynomial from its graph. Sometimes we are unable to tell the exact power of the factor, only that it is odd or even.

(You can learnmore about straight functions here, jmore about odd functions here).

When this is the case, the problem is often written as "write the lowest degree polynomial that the function could represent". So if we know that a factor is not linear but has an odd degree, we would choose the power of 3. Let's consider another problem.

Example 2: Finding a polynomial from a graph

Given the graph below withj-Intersection 1.2, write a least degree polynomial that can represent the graph.

Solution

First we identify the zeros and their multiplicities using the information obtained so far.

Starting from the left side of the graph, we see that -5 is a zero, so (X+ 5) is a factor of the polynomial. When the graph bounces off theXaxis, -5 has a multiplicity of 2.

2 is then a zero (X– 2) is a factor. The graphic is borderingX-Axis. 2 has a multiple of 3.

6 is then a zero (X– 6) is a factor. The graph goes straight through theX-Axis. 6 has a multiple of 1. It is a single zero.

If we add the multiplicities 2 + 3 + 1 = 6, we can see that we have a sixth degree polynomial of the form:

• F(X) =A(X+ 5)²(X– 2)³(X– 6)

Use thej-Section(0,1,2) to solve for the constantA. PlugX= 0 Jj= 1,2.

• 1,2 =A(0 + 5)²(0 – 2)³(0 – 6)
• 1,2 =A(25)(-8)(-6)
• 1,2 = 1200A
• A= 0.001 or 1/1000

Hence our polynomial equation isF(X) = 0,001 (X+ 5)²(X– 2)³(X– 6).

Now let's change things up a bit. Suppose we have the function and want to plot the graph. How do we do that? Let's look at an example.

Draw a polynomial graph of a function

Example 3: Draw a polynomial graph from a function

Draw the polynomialPage(X) = (1/4)(X– 2)²(x + 3)(X– 5)

Solution

Let's find those firstX- Sections of the polynomial. Set the equation equal to zero and solve:

• (1/4)(X– 2)²(x + 3)(X– 5) = 0

This is easily solved by setting each factor to 0. We can see that we have 3 different zeros: 2 (multiplicity 2), -3 and 5. So we already have 3 points to plot on our graph.

Another easy to find item is thej-intercept. LeaveX= 0 and solve:

• j= ¼ (0 – 2)²(0 + 3)(0 – 5)
• j= ¼ (4)(3)(-5)
• j= -15

Now let's graph our four points:

Let's think a little more about how we're going to graph this function.

Since 2 has a multiplicity of 2, we know the graph will bounce off ofXAxis for points close to 2.

Since -3 and 5 each have a multiplicity of 1, the graph goes straight through theX-axis at these points.

What do we do now? How do we know if the graph from above goes through -3X-axis or from below theX-Axis?

A quick review of the final behavior will help us with this. The ultimate behavior of a function describes what the graph doesXapproaches ∞ or -∞.

Checking the final behavior of polynomials

Imagine multiplying our polynomial: the leading coefficient is 1/4, which is positive, and the degree of the polynomial is 4. So the function starts high and ends high.

Now we could draw the graph, but for more precision we can just fill in some values ​​forXand calculate the values ​​ofj.

By plotting these points on the graph, and drawing in arrows to show the final behavior, we can get a pretty good idea of ​​what the graph looks like!

Now we can draw the graph!

How to find a polynomial given points

Let's look at a different kind of problem. Suppose we have a set of points and we want to determine the polynomial function.

How many points do we need to write a unique polynomial? If we think about it a little, the answer will be obvious.

We know that two points uniquely determine a line. And it should make sense that three points can determine a parabola. That brings us to an important idea.

Example 4: Find a degree 4 polynomial from 5 points

a polynomialPage(X) of degree 4 has single zeros at -7, -3, 4, and 8. It also goes through the point (9, 30). Find the polynomial.

Solution

First notice that we have 5 points given so we can uniquely determine a quadratic polynomial from these points. Using the factor set, we can write our polynomial as

• Page(X) =A(X+7)(X+3)(X– 4)(X– 8)

Substitute the point (9, 30) to solve the constantA.

• 30 =A(9 +7)(9 +3)(9 - 4)(9 - 8)
• 30 =A(16)(12)(5)(1)
• 30 = 960A
• A= 1/32

Hence our polynomial Page(X) = (1/32)(X+7)(X+3)(X– 4)(X– 8).

Polynomials are a big part of algebra and beyond. Hopefully today's lesson gave you more tools to use when working with polynomials!

About the author:
Jean-Marie Gard is a freelance math teacher and tutor based in Massachusetts. You can contact Jean-Marie athttps://testpreptoday.com/.