Graphing Polynomial Functions


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A polynomial function is a relation between some variable to another variable with a few restrictions to limit its behavior. In much simpler terms, a polynomial function is simply a subset of functions altogether. Polynomial functions are special in that they contain no discontinuities in their behavior, have distinct slopes and features, and have end behaviors that approach infinity. These functions are excellent for demonstrating real life situations, such as banking trends.

In this tutorial, we will define a polynomial function and learn how to graph them and study the shape of the graph.


Mathematically, a polynomial function is simply a relation between two variables with these rules in mind:

  • Variables cannot have fractional, negative, or imaginary exponents
  • Variables cannot exist in the denominator of any ratio
  • Variables cannot be exponents themselves
  • Variables cannot exist within absolute value delimiters
  • Variables cannot be applied to a trigonometric function

Polynomial functions may only contain constant exponents or coefficients. This seems very limiting, but there are still so many possible behaviors polynomials functions can exhibit.

The following are examples of true polynomial functions since they follow all the above restrictions.

\begin{align} y = x^2+2x-3 \\ y = 3x^3+2x^2-\frac{7}{2} \\ y = 2(x-3)(x+3)^2 \\ y = 8x+13 \\ \end{align}

The following do not follow the above restrictions and therefore cannot be considered polynomial functions.

\begin{align} y = 4x^{1/2}+x^{-2}-x^{1+i} \\ y = x^2+3x-{4 \over x+1} \\ y = 6x^5-3^x \\ y = 5+|-x+2|-\sqrt{2x^3} \\ y = \tan(4x^2+3x-1) \\ \end{align}


A polynomial can be written in different ways. Sometimes, it is written in descending degree order, and sometimes it is a representation of a bunch of factors. There are two important forms of polynomials that are extremely useful when extracting information out of them.

Standard Form:

\begin{equation} f(x) = a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n \end{equation}

Don't let its looks scare you. This simply means that you write each term according to the exponent the term has. For example, $y = 3x^2+4x-4$ is in standard form. So is $y = 6x^{16}-x^{11}+13x^8-92x^4+122x-1$. However, the polynomial $y = 4x+5x^2-3$ is close, but isn't in standard form. For it to be in standard form, the 5x2 term should be in front of the 4x term.

Factored Form:

\begin{equation} f(x) = a(x-r_1)(x-r_2)...(x-r_{n-1})(x-r_n) \end{equation}

This form is also very useful. This form directly exposes the prime factors of a particular polynomial and therefore makes it very easy to determine x-intercepts of the graph. Some examples include $y = (x+3)(x-2)$, $y = (x+4)(x^2+3x-2)$, and $y = 3(x-2)^2(x+7)$. The polynomial $y = (x+6)(x-4)+5x^2$ is not factored because a 5x2 is being added rather than multiplied. To be in factored form, the factors must be multiplied with one another.


The degree of a polynomial function is the highest exponential value of the independent variable in the function. What this means is that the degree is the largest exponent that can be found on the independent variable1 when the polynomial is simplified (all parentheses gone and such). So, for the polynomial $y=x^2+3x-2$, the degree is 2 since that is the largest exponential value. What is the degree of $y=2x^2-3x+x^4$? The degree of this particular polynomial is 4 since it is the largest exponent.

How about the degree of $y = (x+1)(x-2)$? This polynomial function is not simplified, so we first simplify it to $y=x^2-x-2$ so that the degree becomes obvious, 2.

The degree is a very important part of the polynomial function. It allows us to know in a way how the polynomial will generally look like when graphed. The most important thing it tells us is the number of possible roots, or x-intercepts. Whenever you first begin observing a polynomial in preparation to graph, always examine the degree first.

Defining Features

Polynomial Features
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The function is outlined in a dark gray. The x-intercepts are the red circles. The y-intercept is the purple circle. The extrema are the blue X's. The derivative is the green function. The end behaviors are shown with red arrows. The imaginary roots cannot be pictured on the coordinate plane, so they are not present. Actually, this particular function doesn't have any anyway.

There are a few defining features that all polynomial functions have in common that are extremely useful to note when graphing. These things since they are common for all polynomials are used as the base for graphing the polynomial in an easy and quick fashion. It also will ensure an accurate rendering of the graph.

These things are:

  • x-intercepts: where the function crosses or touches the x-axis on the coordinate plane
  • y-intercept: where the function crosses the y-axis on the coordinate plane
  • End Behavior: what the function does at the negative end and positive end graphically and mathematically
  • Extrema: the tops of the "hills" and bottoms of the "valleys" in the middle behavior
  • Derivative: a secondary function to describe the slope of the polynomial at various parts of the curve
  • Imaginary Roots: roots that do not exist on the real coordinate plane

The list seems long, but when handled one at a time systematically, it isn't too much to worry about.

The graph to the right shows an example of these features except for the imaginary roots.

Graphing the Polynomial

To actually graph a polynomial function, it is best to find all the defining features defined above. This gives a bunch of information about the shape and positioning of the polynomial which makes it possible to sketch a graph. Here, we will discuss how to find these defining features and tell how they help.

The Degree

The degree of the polynomial is absolutely key to graphing it. The degree tells a good amount of information about the graph. The first thing it tells us is the general shape of the polynomial, such as does it start from the bottom and continue to the top, or does it start at the top and curve its way back to the top? The second thing it tells us is the number of possible x-intercepts the graph might have.

If the degree of the polynomial is 1, like in the equation $y = 3x+2$, then the graph is a line. If it has a degree of 2, like in $y = x^2+3x-2$, then the shape is a parabola, which is like a U. All odd degree functions have unlimited range, which means they seem to go on forever in both up and down directions. Even degree functions have either an absolute minimum or maximum, so they don't go on forever in both directions, just one. This will be discussed further in the topic of End Behavior.

The number of possible x-intercepts is also given by the degree. Depending on the polynomial, its graph can cross the x-axis more than once. The degree is maximum number of times the graph can cross the x-axis. That doesn't mean it will, but it means that it can, and this tells a lot about the shape. For example, the equation $y = 2x^3+4x^2-5x+12$ is able to cross the x-axis three times. This will be discussed further in the topic of x-intercepts.

End Behavior

The end behavior of a polynomial function is a description of how the polynomial behaves as it approaches positive and negative infinity. In other words, what does the polynomial do at the two ends of the graph? Some go up to infinity on one side and down to negative infinity on the other. Some have each end go to positive infinity. The end behavior is totally dependent on the leading term of the polynomial function when simplified2.

There are four possible end behaviors that could occur assuming the degree of the function is not 0. The end behavior is determined by the sign of the coefficient of the leading term (is it positive or negative) and the degree. If the degree is even, then we know that the polynomial will be sort of shaped like a "U" or an "n". If the degree is odd, it will be shaped like a "/" or a "\". The sign further differentiates between the two remaining options. Observe below the four outcomes and how they are obtained.

Even Degree

Odd Degree


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This shape results if the leading term is positive and the degree of the polynomial is even. An example would be y = 2x2+4x-3. The degree of the equation is 2 (even), and its leading coefficient, 2x2, is positive.

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This shape results if the leading term is positive and the degree of the polynomial is odd. An example would be y = 5x5+3x2-4x-2. The degree of the equation is 5 (odd), and its leading coefficient, 5x5, is positive.


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This shape results if the leading term is negative and the degree of the polynomial is even. An example would be y = -2x4+3x3+x-7. The degree of the equation is 4 (even), and its leading coefficient, -2x4, is negative.

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This shape results if the leading term is negative and the degree of the polynomial is odd. An example would be y = -x3+3x2-8x+1. The degree of the equation is 3 (odd), and its leading coefficient, -x3, is negative.

So, let us take the following polynomial function and use it as an example to determine its end behavior.

\begin{equation} f(x) = (x+3)(x^2-3x+1)(2x-1)^2 \end{equation}

The first thing to note is that this polynomial is not in standard form. It is in factored form. In order to expose that leading coefficient, we need this polynomial in standard form. To do that, distribute the terms to one another. When this is expanded, you get the following:

\begin{equation} f(x) = 4x^5-4x^4-7x^3+44x^2-20x+3 \end{equation}

The leading term is 4x5. Now we determine the degree. The degree is 5, so it is odd. The coefficient is a positive 4, so the end behavior of this polynomial is positive-odd, shaped like a "/".


The x-intercepts of a polynomial are where the polynomial intersects the x-axis on the real coordinate plane. Mathematically speaking, these x-intercepts only occur when y is equal to 0. Polynomials can have multiple x-intercepts because of the way they curve. The number of x-intercepts a certain polynomial can have is the degree of the polynomial. A first degree polynomial can only have one x-intercept. A fourth degree can have up to four, but it doesn't have to have four. For even degree polynomials, it is possible that there are no x-intercepts. Odd degree polynomials must have at least one x-intercept.

The x-intercepts are key to graphing a polynomial. They are points that you can connect that lie on the x-axis. These x-intercepts are also known as solutions to the polynomial. So, how do we find these x-intercepts? Simply, these points are where y = 0. So, you simply solve the polynomial for x when y or f(x) is 0.

For example:

\begin{equation} y = x^2-5x+6 \end{equation}

To find the x-intercepts, we set y = 0 and then solve for x. So, we solve:

\begin{equation} 0 = x^2-5x+6 \end{equation}

The only problem is solving the polynomial. There is no easy way to do this except by factoring, and in most cases the polynomial won't factor properly3. You can learn more about solving polynomial equations here. This particular polynomial is quadratic, so it is easy to solve. It factors into $0 = (x-2)(x-3)$. Now we can set each factor to zero.

\begin{align} 0 = x-2, \hspace{5pt} 0 = x-3 \\ x = 2, \hspace{5pt} x = 3 \\ (2,0), \hspace{5pt} (3,0) \\ \end{align}

Notice that the degree was 2, and we ended up with two x-intercepts.

Sometimes, the polynomial is already factored for you, and so it is easy to identify the x-intercepts. For example, observe the polynomial:

\begin{equation} y = (x-5)(x^2+2x+3) \end{equation}

The degree is 3, so we are expecting to get three x-intercepts. First, we set y equal to 0.

\begin{equation} 0 = (x-5)(x^2+2x+3) \end{equation}

Now we set each factor equal to the 0.

\begin{align} 0 = x-5, \hspace{5pt} 0 = x^2+2x+3 \end{align}

Now solve for x. The first one is easy; we get the point (5,0). However, the second one is quadratic and doesn't factor. And when we use the quadratic formula, we get $-1\pm i\sqrt{2}$. This is important. Whenever you end up with an imaginary part (notice the $i$), then no x-intercepts result. Therefore, this polynomial has one x-intercept, (5,0).


Usually, x-intercepts cross the x-axis straight through. However, there is more than one way that the polynomial can intercept the x-axis. There are actually three total ways that the graph intercepts the x-axis. In the first, it passes straight through no problem. In the second, it goes down and touches the x-axis and then rebounds off it. In the third, the graph sort of lingers around the interception point before crossing.

Why are there three types of intercepts? This is governed by a mathematical thing called multiplicity. Multiplicity is the number of times a particular x-intercept or solution appears. What if you ended up with the same x-intercept twice? That means that that particular x-intercept has a multiplicity of 2. It occurs twice, and it therefore has a multiplicity of 2. An x-intercept that occurs 3 times has a multiplicity of 3.

Let us look at the following example.

\begin{equation} y = (x-1)(x+2)(x+2) \end{equation}

We could commence the normal procedure for finding x-intercepts by setting the y equal to zero and solving. However, we end up with $0 = x+2$ twice, which means we get the intercept, (-2,0), twice. The x-intercept, (-2,0) has a multiplicity of 2.

So what does multiplicity do the shape of the graph? Depending on the multiplicity of an x-intercept, you obtain one of the three types of x-intercepts from the first paragraph in this section. How do you know what type?

Multiplicity 1
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The graph passes straight through when the x-intercept has a multiplicity of one.

Normally, an x-intercept has a multiplicity of one, or it only occurs once. When this happens, the graph simply passes straight through the x-axis. It occurs once, so it passes through and continues along with the normal path that it takes.

Even Multiplicity
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The graph touches and rebounds the x-axis when the particular x-intercept has an even multiplicity, like 2, 4, 6, etc.

If the intercept has an even multiplicity, meaning it occurs twice, four times, eight times, etc., then the graph appears to touch the x-axis and then bounces off in the same direction it came from. The graph never passes through the x-axis, it simply touches it and goes back. As multiplicity increases, the valley will become flatter and flatter.

Odd Multiplicity
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The graph sort of lingers around the point of interception before actually crossing. This happens when the multiplicity of the x-intercept is odd, like 3, 5, 7, etc. It doesn't happen if the multiplicity is 1.

If the x-intercept has an odd multiplicity, meaning it occurs three times, five times, etc., then the graph kind of lingers around the interception point before passing through. The graph does actually pass through, but it is sort of delayed before actually passing through, like in the image. As multiplicity increases, the deflection becomes closer and closer to the x-axis.


The y-intercept of a polynomial is where the graph crosses the y-axis on the real coordinate plane. Unlike with x-intercepts, there can only be one y-intercept, and there always is one y-intercept. No matter what polynomial you are given, there will be a y-intercept.

Mathematically, the y-intercept occurs when the independent variable, x, is equal to 0. This is a contrast from x-intercepts where y was equal to 0; this time the other variable is 0. The y-intercept is a point like with x-intercept, except it is in the form (0,y). To find the y-intercept, simply plug in 0 for x.

So, let us say we want to identify the y-intercept of the following polynomial.

\begin{equation} y = 2x^2+4x-7 \end{equation}

Set every x equal to 0, then simplify. Whatever you end up with, that is the y-intercept.

\begin{align} y = 2(0)^2+4(0)-7 \\ y = -7 \\ (0,-7) \\ \end{align}

There really isn't any special case of a y-intercept. Simply set x equal to 0 and solve. If the y-intercept happens to be (0,0), then it is an x-intercept as well.


The derivative is more of a calculus topic, but nonetheless it is extraordinarily helpful when extracting information from a polynomial in order to graph it. The derivative is a secondary function that exposes the slope of the original polynomial at given points. What the heck does that mean?

Tangent Line
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The green line is tangent to the red point on the graph of y=x2.

Think about a first degree polynomial, or a simple line. For example, we will use $y = 3x+1$. This line is said to have a slope, a sort of definition to how steep it is compared to the x-axis. The slope is 3 everywhere since the value of y increases by three for every unit increase of x.

Consequentially, all other polynomials have multiple slopes. Take the quadratic polynomial, $y = x^2$, for instance. We cannot say it has a slope since the slope constantly changes. It continues to curve. However, every single point on the curve has a definitive slope. Take a look at the image to the right. A line touches a single point on the curve, but it doesn't go through it. It simply touches the point. And it keeps going.

The slope of the line represents the slope of the curve at that particular point. This special line is called a tangent line. Every point has a slope. The job of the derivative is to tell us specifically the slope of every point on the curve4. Therefore, there is a very large graphical connection between the function and its derivative.

When the slope of the function is positive, the derivative is above the x-axis. When the slope is negative, the derivative is under the x-axis. When the slope is zero, this means that the point is either the top of a hill or bottom of a valley, the derivative is crossing the x-axis. This helps a lot when trying to find the extrema.

How do you find the derivative of a function? There is a complex calculus formula that defines the derivative of a function, but our concern is simply graphing a polynomial function. So, we will take the shortcut way. First, the polynomial should be in standard form. Then, we apply this formula to every term and add the terms together.

\begin{align} ax^n \rightarrow anx^{n-1} \end{align}

Let us say we have a polynomial $f(x) = 4x^3+2x^2-7x+1$. To find its derivative, we apply the formula to every single term and add them together Ignore the constant term. We get this:

\begin{align} f'(x) = 4(3)x^{3-1}+2(2)x^{2-1}-7(1)x^{1-1} \\ f'(x) = 12x^2+4x-7 \\ \end{align}

The derivative will always have a degree one less than its parent function. The next section will explain the importance of the derivative and how it can be used to graph the original polynomial.


The extrema are the minimum points and maximum points within the middle behavior of the polynomial. That is, they are the tops of the hills and bottoms of the valleys. For example, the vertex of a quadratic polynomial is considered an extrema. A polynomial can have up to one less than their degree extrema. So a fourth degree polynomial can have up to three extrema. An odd degree polynomial can have no extrema, but an even degree function must have at least one.

The extrema are very important when graphing a polynomial. They are like limits telling you when to turn around and draw the other direction. The extrema are unfortunately not usually located on either of the axes; they are usually some point in one of the four quadrants. So how do we find these extrema?

In the previous section, the derivative was discussed. Remember that the derivative relates to the slope of the original polynomial. As a matter of definition, an extrema mathematically occurs when the slope of the polynomial at a given x location is 0. To find the extrema, we must solve the derivative by setting it equal to 0.

If you don't understand, here is what is happening. An extrema occurs if slope is zero. The derivative tells the slope of the polynomial. If the slope is zero, the derivative must be equal to zero. So to find the extrema, we set the derivative equal to zero.

Let us say we have the polynomial $f(x) = \frac{1}{3}x^3-2x^2+3x+1$. We want to find the extrema. First, the derivative must be identified. We will use the power rule from before to achieve this.

\begin{align} ax^n \rightarrow anx^{n-1} \\ f'(x) = \frac{1}{3}(3)x^{3-1}-2(2)^{2-1}+3(1)x^{1-1} \\ f'(x) = x^2-4x+3 \\ \end{align}

Now, we solve the derivative for x if f'(x) is 0.

\begin{align} 0 = x^2-4x+3 \\ 0 = (x-1)(x-3) \\ x = 1, \hspace{5pt} x = 3 \\ \end{align}

Therefore, the extrema happen when x is 1 and x is 3. But, that is not all.

Solving the derivative for x is not enough. The extrema are points, or x-y pairs. Therefore, we need to find the appropriate y part. Fortunately, this is easy. Once you find the x part, simply substitute the values into the original polynomial, and the respective y parts will emerge. In this case, substituting 1 for x gives 7/3, and substituting 3 for x yields 1. Therefore, the extrema occur at (1,7/3) and (3,1).

Remember, extrema will not always turn up with nice numbers like this one did. And also, higher degree polynomials are much harder to find extrema for since it requires the solving of higher degree derivatives.

What happens if we don't get any extrema? Consider the case $y = x^3+7x+1$. The power algorithm yields a derivative of $f'(x) =3x^2+7$, and when we solve for 0, then we find there are no real solutions for the derivative. In this case the polynomial has no extrema, and you therefore cannot plot any.

So going over the steps:

  1. Find the derivative of the polynomial
  2. Solve the derivative for x when set equal to 0
  3. Substitute solutions into original polynomial
  4. Write as coordinates

Note: An x-intercept of even multiplicity is considered an extremal point.

Imaginary Roots

So, the degree of a polynomial tells the most number of possible x-intercepts the function can have. However, sometimes the function has less x-intercepts than that maximum amount. What happened to the other ones? These non-existent x-intercepts are called imaginary roots. These imaginary roots are solutions to the original polynomial, but they contain an imaginary part, or an $i$/$\sqrt{-1}$. Therefore, they cannot be graphed on the real x-axis.

Here is an example of when imaginary roots occur. Consider the polynomial function $y = x^2+2x+3$. We must solve for x when y is equal to zero. However, the polynomial is not factorable, so we must use the quadratic formula.

\begin{align} {-b\pm\sqrt{b^2-4ac} \over 2a} \\ -2\pm\sqrt{2^2-4(1)(3)} \over 2(1) \\ -2\pm\sqrt{-8} \over 2 \\ \end{align}

It is impossible to get a real number result when taking the square root of a negative number. Thus, this represents two imaginary roots, $-1\pm i\sqrt{2}$. The $i$ is a symbolic representation of $\sqrt{-1}$. An imaginary root cannot be graphed as an x-intercept, and they do not provide much visual clue when the polynomial is graphed. Nonetheless, they are still an important part of the function, and it is best to identify the imaginary roots whenever possible.

Stringing Together the Information

Once you have acquired all of the information from above, then you can "connect the dots", literally. The above information is all that is needed to really sketch an accurate drawing of the function. All the information gathering before this was the hard part; the actual graphing is easy.

So, there are a few major steps in this final part. The first is to plot the y-intercept and any extrema that is known. These are simply points. If you do not know any extrema, then just plot the y-intercept.

Next, plot the x-intercepts. Do not forget to note multiplicity. If a certain x-intercept has a multiplicity other than one, write out the multiplicity beside the point.

Now, note the end behavior. Write out little arrows indicating where the graph will enter from and where it will leave.

Finally, connect the dots! Start where the end behavior on the left side is and go to each point. Every time you reach an extremal point, turn around and go the other direction. Whenever you hit a point of even multiplicity, turn around. When you hit a point of odd multiplicity, put the delay. You are done once you reach the other end behavior mark. After that, you are completely finished with the graph! Simply label it with the function.

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This illustration shows the basic steps of drawing the graph. Plot the points, note multiplicities, note end behaviors, and simply connect the points together.

It is possible, however, to end up trying to graph a polynomial function that has one x-intercept and no extrema, or no x-intercepts and one extrema. These graphs can be a bit difficult to draw because of the lack of information. To graph something described as such, you simply have to imagine. Use what you do know to graph the polynomial.

Example Walk Through

This section will walk you through the steps in graphing a polynomial function.

Graph the following:

\begin{equation} f(x) = 2(x-2)(x+3)^2(x^2+1) \end{equation}

Step 1

First, observe the problem. Find out what kind of problem it is so you can better prepare your strategy for graphing it. The first thing to notice is that the polynomial is in factored form. How do we know it is in factored form? There are only factors being multiplied by each other. Notice that there are no addition or subtraction operators outside of the parentheses. Therefore, it is factored. It is fully factored since there are no other factors that can be further simplified or factored.

There are five distinct factors: 2, x-2, x+3, x+35, and x2+1. Each factor is prime, meaning it is entirely factored, so no additional work is needed.

Since it is in factored form and not in standard form, we cannot identify things like y-intercept or end behavior easily. This will help with our strategy. When a problem is given in any form other than standard, it is always best to turn it into that form. So we will start expanding the polynomial using distribution. It is a little more advanced than quadratic FOIL, but it is very much the same concept; every term in a factor gets multiplied by every term in another.

We will turn this polynomial into standard form:

\begin{align} f(x) = 2(x-2)(x+3)^2(x^2+1) \\ f(x) = 2(x-2)(x^2+1)(x+3)^2 \hspace{5pt} \operatorname{rearrange} \\ f(x) = 2(x^3-2x^2+x-2)(x^2+6x+9) \\ f(x) = 2(x^5+4x^4-2x^3-14x^2-3x-18) \\ f(x) = 2x^5+8x^4-4x^3-28x^2-6x-36 \\ \end{align}

Keep both forms handy. The factored form will be very useful later.

Step 2

Identify the degree and end behavior. How do we find the degree? Remember that the degree of a polynomial is the largest exponent that a variable has. This is best done looking at the polynomial in standard form. The degree is five since that is the largest exponent present. This means that there can be at most five x-intercepts. That doesn't mean there will be five x-intercepts; it just means there can be at most five x-intercepts, so expect that.

Now, we must identify the end behavior. We do this by looking at the leading term when in standard form. The leading term is the one with the largest exponent, or in this case, 2x5. The degree is odd, and the coefficient is positive. Therefore, we will have a positive-odd end behavior, or one that looks like "/". Write it out somewhere so you don't forget, like this:

eb: / /

Step 3

Identify the x-intercepts and note multiplicity. To find the x-intercepts, you must solve the polynomial when f(x) is 0. This is where the factored form is perfectly handy. The factored form allows us to set every factor to zero therefore making the entire thing possible to solve.

Solve for x:

\begin{align} 0 = 2(x-2)(x+3)^2(x^2+1) \\ 0 \ne 2, \hspace{5pt} 0 = x-2, \hspace{5pt} 0 = x+3, \hspace{5pt} 0 = x+3, \hspace{5pt} 0 = x^2+1 \\ x = \{2,-3,-3\} \\ \end{align}

There are three real solutions. 0 = x2+1 cannot be solved (yet…). The solutions are 2, -3, and -3 again. Remember, when a solution appears more than once, it has multiplicity. So the solution of -3 has a multiplicity of 2. There are therefore 2 x-intercepts: (2,0) and (-3,0). Mark these down for later, and do not forget to indicate that the intercept (-3,0) has a multiplicity!

eb: / /
x-int: (2,0); (-3,0)m2

Step 4

Next, identify the y-intercept. This is probably the easiest step in the whole process. Simply make every x 0. When looking at standard form, the y-intercept is simply the constant term. Observe:

\begin{align} f(x) = 2(0)^5+8(0)^4-4(0)^3-28(0)^2-6(0)-36 \\ f(x) = -36 \\ \end{align}

The y-intercept is (0,-36). Write it down for later.

eb: / /
x-int: (2,0); (-3,0)m2
y-int: (0,-36)

Step 5

Now the extrema must be found. Before we can find the extrema, we need to identify the derivative first. The derivative tells information about the slope of the polynomial, so it will help a ton. To find the derivative, we use the power rule, $ax^n \rightarrow anx^{n-1}$ on each term when the polynomial is in standard form.

\begin{align} f'(x) = 2(5)x^{5-1}+8(4)x^{4-1}-4(3)x^{3-1}-28(2)x^{2-1}-6(1)x^{1-1} \\ f'(x) = 10x^4+32x^3-12x^2-56x-6 \\ \end{align}

When the slope of the polynomial is zero, an extrema occurs, so we will solve the derivative when it is set equal to 0. Begin by factoring.

\begin{equation} 0 = 2(5x^4+16x^3-6x^2-28x-3) \end{equation}

The factor in the parentheses looks a bit daunting to factor, and there is of course the possibility that it will not factor. But recall the special property of x-intercepts with even multiplicity. An x-intercept of even multiplicity is itself an extremal point. Yes! We have an extrema! But there might be more. The degree is five, so there might be four extrema. Since we have one, we know that it is a solution to the derivative. So, we can factor it out.

\begin{equation} 0 = 2(x+3)(5x^3+x^2-9x-1) \end{equation}

Polynomial Division was used to obtain each factor. Now we need to factor the cubic term, but I will already tell you it isn't factorable. So what do we do? How can we solve this monster?

Unfortunately, there is no easy way to solve it by hand. The task is best handled by a calculator. Using a calculator, find the three solutions to $y = 5x^3+x^2-9x-1$. Look here to find out more on how this can be done. Using the calculator, we obtain three solutions. They are -1.39, -0.11, and 1.3, rounded of course.

Substitute those values for x in the original polynomial. Then, simplify. This will leave you with three y values which you can pair up with their parent x value to obtain the extrema.

\begin{align} f(x) = 2(-1.39)^5+8(-1.39)^4-4(-1.39)^3-28(-1.39)^2-6(-1.39)-36 \\ f(x) = -10.38+29.86+10.74-54.1+8.34-36 \\ f(x) = -51.53 \\ f(x) = 2(-0.11)^5+8(-0.11)^4-4(-0.11)^3-28(-0.11)^2-6(-0.11)-36 \\ f(x) = 0+0+0.01-0.34+0.66-36 \\ f(x) = -35.67 \\ f(x) = 2(1.3)^5+8(1.3)^4-4(1.3)^3-28(1.3)^2-6(1.3)-36 \\ f(x) = 7.43+22.85-8.79-47.32-7.8-36 \\ f(x) = -69.63 \\ \end{align}

The four extrema are (-3,0), (-1.39,-51.53), (-0.11,-35.67), and (1.3,-69.63). Extremal points will normally be irrational. These points can now be logged with the other information.

eb: / /
x-int: (2,0); (-3,0)m2
y-int: (0,-36)
extrema: (-3,0); (-1.39,-51.53); (-0.11,-35.67); (1.3,-69.63)

Step 6

Now identify the complex solutions. Those are the ones with the $i$ in them. Remember from before how we couldn't solve $0 = x^2+1$? Now, we will explore the problem. This is a simple quadratic that can be solved with simple arithmetic. Subtract one from both sides, then take the square root. You are left with $\pm\sqrt{-1}$. This results in the imaginary solutions $x = \pm i$. You can log this down.

eb: / /
x-int: (2,0); (-3,0)m2
y-int: (0,-36)
extrema: (-3,0); (-1.39,-51.53); (-0.11,-35.67); (1.3,-69.63)
img: ±i

Step 7

Now all the information has been acquired. It is now time to actually graph the polynomial. Plot every known point, including the x-intercepts, y-intercept, and extrema. Write out the end behavior as a form of a couple of arrows.

There might be a problem, though. If you look at the y-intercept and extrema, there are very large y-values, such as the -36 and -69.63. These obviously will not fit on the standard 10 x 10 grid. We therefore must establish a different viewing window, or in other words, different dimensions. The graph needs to be scaled so that we can graph every available point. Therefore, the y-axis must at least extend down to -70. It is always best to scale the plane in order to get the best view of the polynomial.

The lowest value of y is -69.63, so the y-axis will go down to -100. The highest value is 0, so we will go up to positive 30. Notice that I give 30 units of leeway. The lowest x-value is -3, and the highest is 2. So the x-axis will extend from -5 to 4, 2 units of leeway. I give leeway so that end behaviors can be seen. It is up to your judgment as to how much slack you give.

So we will now plot the points on the newly scaled plane. The y-axis has a unit scale of 10, meaning that every line on the y-axis is 10 units. Here is our plot so far:

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Now the actual graph must be drawn. Start at the lower left corner and draw up. Since the first x-intercept has a multiplicity of two, we turn around there. Make a turn at every extrema, but cross through the y-intercept and other x-intercept. You should end at the top right corner because that is the end behavior of the polynomial. The final graph looks like this:

Image Unavailable

All done! The graph is complete. If you are graphing multiple equations on the same graph, it is best to color code the graphs or label them. Once the graph is drawn, there is no more needed to be complete. You can label the irrational points for clarification, but that is about it.

Practice Problems

The only way to get better at graphing polynomials is by actually graphing a bunch of them. Here are some practice problems to try out on your own so you can use the information in this tutorial firsthand. Good luck!

1) $y = x^2-3x+2$

2) $y = (x+1)(x-2)(x+3)$

3) $y = (x-3)^3$

4) $f(x) = -(x-2)(x^2+2x+5)$

5) $f(x) = 2x^3+11x^2-12x-36$

6) $y = (x-1)(x^2-7x+6)$

7) $f(x) = 2(x+1)(2x-5)(-x-3)(4x-7)$

8) $y = 2x^5-7x^4-14x^3+24x^2-148x+15$

9) $f(x) = x^2+(x-2)(x+1)^2-6$

10) $f(x) = x^3+3x+2$

Works Cited

1) Algebra II class

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