Sign In
| 16 Theory slides |
| 12 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Unfortunately, graphing an arbitrary function usually requires knowing advanced algebraic techniques. However, when the function is a polynomial, some key features of the graph can be derived directly from the function rule, leading to an accurate graph. As such, this lesson aims to teach how to graph polynomials.
Here are a few recommended readings before getting started with this lesson.
In order to improve her graphing polynomial skills, Emily decided to try graphing the fourth-degree polynomial function f(x)=x4−4x3−39x2+46x+80. To start, she made a table of values.
x | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
f(x) | 560 | 0 | -216 | -220 | -120 | 0 | 80 | 84 | 0 | -160 | -360 | -540 | -616 |
Functions are usually named after the algebraic expression that defines them. For example, if the function rule is a real number, the function is called a constant function. Something similar happens when the function rule is a linear, quadratic, or exponential expression.
Function Rule | Name |
---|---|
f(x)=-7 | Constant Function |
g(x)=2x+1 | Linear Function |
h(x)=x2+3x−1 | Quadratic Function |
t(x)=2e3x | Exponential Function |
The same holds true to those functions whose function rule is a polynomial.
A polynomial function is a function whose rule is a polynomial. In general, a polynomial function has the following form.
f(x)=anxn+an−1xn−1+⋯+a1x+a0
up.
down.
down and upor as
down-up.
Determine the end behavior of the given polynomial function.
Emily loves to record every trip she takes on her blog, including amazing photos. However, since she does not have a professional camera, some of her photos need some retouching. To help with this, she plans to buy an image editing software. She is torn between two options whose running times to process an x-by-x pixels image are given by the following polynomials.
Running Time (nanoseconds) | |
---|---|
Software 1 | P(x)=x3−520x2+10000x+20000000 |
Software 2 | Q(x)=100x2+1000000 |
Image Size (pixels) | Running Times (nanoseconds) | Conclusion |
---|---|---|
100-by-100 | P(100)Q(100)=16800000=2000000 ✓
|
Software 2 is faster |
240-by-240 | P(240)Q(240)=6272000 ✓=6760000 ✓
|
Similar Running Times |
300-by-300 | P(300)Q(300)=3200000 ✓=10000000
|
Software 1 is faster |
600-by-600 | P(600)Q(600)=54800000=37000000 ✓
|
Software 2 is faster |
1200-by-1200 | P(1200)Q(1200)=1011200000=145000000 ✓
|
Software 2 is faster |
2000-by-2000 | P(2000)Q(2000)=5960000000=401000000 ✓
|
Software 2 is faster |
The table suggests the following conclusions.
Emily's graph also suggests the first two conclusions. However, that graph corresponds only to a small portion of the domain. Therefore, before making any decisions about end behavior, graph both polynomials in a bigger domain.
It can be seen that the right-end behavior of the graph of y=P(x) is up, and not down as Emily concluded. Additionally, for images with 600-pixel sides or bigger, the running time of Software 2 is considerably shorter than the running time of Software 1.
Conclusion
In general, Software 2 is faster.
Although the right-end behavior of both graphs is up, the graph of y=P(x) increases way much faster than the graph of y=Q(x). Therefore, Emily should buy Software 2.
Leading Coefficient | Degree | End Behavior |
---|---|---|
Positive | Even | Up-Up |
Positive | Odd | Down-Up |
Negative | Odd | Up-Down |
Negative | Even | Down-Down |
The leading term of P(x) is x3, which means that the leading coefficient is 1 and the degree is 3. Since the leading coefficient is positive and the degree is odd, the end behavior of the graph of y=P(x) is Down-Up.
Despite the y-axis being measured in seconds rather than nanoseconds, the conclusion about the end behavior or the fastest software remains the same.
Up-Up.
Again, the y-axis is measured in seconds, not in nanoseconds.
x | -4 | -2 | 0 | 1 | 3 | 4 | 6 |
---|---|---|---|---|---|---|---|
f(x) | -28 | -48 | 252 |
Substitute expressions
Commutative Property of Addition
Factor out x
Factor out (x3−4x2−11x+30)
-4x2=-2x2−2x2, -11x=-15x+4x
Commutative Property of Addition
Factor out x & -2
Factor out (x2−2x−15)
x=-4
Calculate power and product
Multiply
Add and subtract terms
x | f(x)=x4−3x3−15x2+19x+30 | Result |
---|---|---|
0 | f(0)=04−3(0)3−15(0)2+19(0)+30 | 30 |
1 | f(1)=14−3(1)3−15(1)2+19(1)+30 | 32 |
4 | f(4)=44−3(4)3−15(4)2+19(4)+30 | -70 |
Consequently, the complete table of values looks as follows.
x | -4 | -2 | 0 | 1 | 3 | 4 | 6 |
---|---|---|---|---|---|---|---|
f(x) | 162 | -28 | 30 | 32 | -48 | -70 | 252 |
up-up.
up-upin Part C.
down-up.In contrast, the zeros are not easy to identify simply by looking at the function rule. Usually, the polynomial must be factored first. However, not all polynomials have a suitable factorization. Therefore, it is worth considering whether there is a simple way to estimate the zeros of a function. The following principle answers this.
Let f be a polynomial function and a and b be two real numbers. If f(a) and f(b) have opposite signs, then there is at least one zero between a and b.
Notice that if f(a) and f(b) have opposite signs, the principle guarantees one zero between a and b, but there could be more. Conversely, the fact that f(a) and f(b) have the same sign does not imply that there are no zeros between a and b.
Let a and b be two real numbers with a<b and let f be a polynomial function such that f(a)>0 and f(b)<0. A sketch of the graph of y=f(x) around x=a and around x=b could look as follows.
For f(x)=8x3−12x2 −26x +15, it can be found that f(0)=15>0 and f(1)=-15<0. Notice that f(0) and f(1) have opposite signs. Consequently, f has at least one zero between x=0 and x=1.
The interval where the zero is located can be reduced by considering closer testing values.
Use the Location Principle to indicate whether the given polynomial function is guaranteed to have a zero in the indicated interval.
Sometimes knowing that a graph passes through a point does not give enough information about how the curve is before and after the point. However, when the point is either a maximum or a minimum, there is a clear view of the behavior of the graph around it.
The value f(c) is a relative minimum, or local minimum, of a function if f(c) is the least output of f around x=c. Likewise, the value f(d) is a relative maximum, or local maximum, of a function if f(d) is the greatest output of f around x=d.
When a relative extrema is greater or lower than any other point in the graph, it has a particular name.
The absolute minimum, or global minimum, of a function is the least output in its whole domain.
The absolute maximum, or global maximum, of a function is defined in a similar way. It is the greatest output of the function in its whole domain.
The absolute maximum of a function is also a relative maximum, and the absolute minimum is also a relative minimum. If a function increases indefinitely, it does not have an absolute maximum. Likewise, if a function decreases indefinitely, it does not have an absolute minimum. The function might still have relative extrema.
Relative or absolute extrema are very useful when graphing functions, but the method for finding them is beyond the scope of this lesson. That said, when working with polynomial functions, there is a way to estimate the maximum number of relative extrema by using the turning points of the graph.
A turning point is a point where the graph of a function changes from increasing to decreasing or vice versa. In other words, the turning points correspond to the relative extrema of the graph of a function.
The graph corresponds to a fifth-degree polynomial with four turning points. When dealing with polynomial functions, there is a close connection between the degree, the zeros, and the turning points.
Consider the given polynomial function. Determine either the maximum or the exact number of turning points that its graph may have.
After purchasing the image editing software, Emily was curious about the online software she used to plot the polynomials and wanted to try graphing a polynomial by hand. To challenge herself, she chose the polynomial function f(x)=-0.5x5−3.5x4 +91x3+374x2 −3556x−7840. So far, she has created the following table of values.
x | -15 | -10 | -7 | -3 | 0 | 6 | 8 | 11 |
---|---|---|---|---|---|---|---|---|
f(x) | 25025 | -10880 | 4165 | 3575 | -7840 | -4480 | 3520 | -12350 |
She next plans to find the zeros of the polynomial and determine the end behavior of the graph. She also knows that (-11.83,-16791.37), (-5.16,7312.41), (2.67,-13181.92), and (8.71,4482.69) are turning points. Emily thinks it is enough information to make an accurate sketch of the graph. Follow Emily's plan and graph y=f(x).
x | -15 | -10 | -7 | -3 | 0 | 6 | 8 | 11 |
---|---|---|---|---|---|---|---|---|
f(x) | 25025 | -10880 | 4165 | 3575 | -7840 | -4480 | 3520 | -12350 |
x=-14
Calculate power and product
Multiply
Add and subtract terms
f(x)=-0.5x5−3.5x4+91x3+374x2−3556x−7840 | |
---|---|
x=-9 | x=-8 |
f(-9)=-0.5(-9)5−3.5(-9)4+91(-9)3+374(-9)2−3556(-9)−7840⇓f(-9)=-5320
|
f(-8)=-0.5(-8)5−3.5(-8)4+91(-8)3+374(-8)2−3556(-8)−7840⇓f(-8)=0
|
The polynomial has a second zero at x=-8. The remaining zeros can be found by applying a similar reasoning.
After testing some integers that belong to each of the mentioned intervals, the remaining zeros can be found at x=-2, x=7, and x=10.
Zeros
-14,-8,-2,7,10
Since the polynomial has degree five, it has no more zeros.
Leading Coefficient | Degree | End Behavior |
---|---|---|
Positive | Even | Up-Up |
Positive | Odd | Down-Up |
Negative | Odd | Up-Down |
Negative | Even | Down-Down |
For the given polynomial, the leading term is -0.5x5. The leading coefficient is negative and the degree is odd. Consequently, the end behavior of the graph of f is Up-Down.
Now, graph the points from the table.
Before connecting the points with a smooth curve, plot the turning points (-11.83,-16791.37), (-5.16,7312.41), (2.67,-13181.92), and (8.71,4482.69). At these points, the graph changes from increasing to decreasing or vice versa.
According to the distribution of the points, from left to right, the first turning point is a relative minimum, the second one a relative maximum, the third another relative minimum, and the fourth and final one another relative maximum. There are no absolute extrema. With all this information, the graph can finally be drawn.
Some functions have the property that their graphs look the same on both sides of the y-axis, as if the y-axis were a mirror. Such a property can be easily identified once the graph is drawn but, can it be deduced algebraically? The next concepts can answer this question.
An even function is a function for which the value of f(-x) is equal to the value of f(x) for all the values in its domain. That is, opposite inputs have the same output.
f(-x)=f(x)
x=-x
(-a)4=a4
(-a)2=a2
3x4−2x2+1=f(x)
Some other functions seem to be reflected across the origin. Functions with this property are called odd functions.
An odd function is a function for which the value of f(-x) is equal to the value of -f(x) for all the values in its domain. It is like if the function allows moving the negative sign from the input to the output.
f(-x)=-f(x)
x=-x
(-a)2=a2
Put minus sign in front of fraction
x2−1x=f(x)
Be aware that these concepts are exclusive — in other words, if a function is even, it cannot be odd, and vice versa. However, unlike integers, a function may be neither even nor odd.
At the beginning of the lesson, Emily graphed the polynomial function f(x)=x4−4x3−39x2+46x+80 using a table of values, but she was not so sure about the end behavior of the graph. She concluded that the end behavior was "Up-Down."
up-up.Consequently, Emily's graph was not correct. She had to extend the right part upward, and not downward.
Describe the end behavior of each of the following graphs.
Let's take a look at the ends of the given graph. We can see that the left-end arrow points downward, while the right-end arrow points upward.
Therefore, the left-end behavior of the graph is down
and the right-end behavior is up.
Consequently, the end behavior of the given graph is down-up.
As we did in Part A, let's consider the ends of the graph. This time both the left-end arrow and the right-end arrow point upward.
Therefore, both the left-end behavior and the right-end behavior of the graph are up.
Consequently, the end behavior of the given graph is up-up.
Let's check the ends of the final graph like we did in the previous parts. This time we can see that the left-end arrow points upward while the right-end arrow points downward.
Therefore, the left-end behavior of the graph is up
and the right-end behavior is down.
Consequently, the end behavior of the given graph is up-down.
We can use the Location Principle to determine whether f is guaranteed to have a zero in the interval [ 2, 3]. This principle states that if the polynomial function has different signs at the endpoints of an interval, then the function has at least one zero on that interval. Let's start by evaluating f(x) at x= 2.
Similarly, let's evaluate f(x) at x= 3.
After evaluating the function f at the endpoints of the interval [ 2, 3] we got two negative outputs. f( 2) < 0 and f( 3) < 0 Since f( 2) and f( 3) have the same sign, we cannot use the Location Principle. Therefore, we cannot guaranteed that f has a zero in the interval [ 2, 3].
As in Part A, we can use the Location Principle to determine if f could have a zero in the given intervals. Let's start by listing the intervals.
I_1 &= [-13,-12] & I_4 &= [0,1]
I_2 &= [-12,-10] & I_5 &= [1,2]
I_3 &= [-3,-2] & I_6 &= [3,5]
Next, let's evaluate the function at each of the endpoints of the intervals. We can being by finding f(-13).
We can evaluate f(x) at the remaining endpoints by following the same procedures for each value. The following table summarizes the results.
Interval | Endpoint | Evaluate | Sign |
---|---|---|---|
I_1 = [-13,-12] | -13 | f(-13)=1334 | Positive |
-12 | f(-12)=-928 | Negative | |
I_2 = [-12,-10] | -12 | f(-12)=-928 | Negative |
-10 | f(-10)=-2866 | Negative | |
I_3 = [-3,-2] | -3 | f(-3)=-136 | Negative |
-2 | f(-2)= 102 | Positive | |
I_4 = [0,1] | 0 | f(0)=164 | Positive |
1 | f(1)= 60 | Positive | |
I_5 = [1,2] | 1 | f(1)=60 | Positive |
2 | f(2)= -46 | Negative | |
I_6 = [3,5] | 3 | f(3)=-58 | Negative |
5 | f(5)= 704 | Positive |
According to the Location Principle, f has at least one zero in [a,b] if f(a) and f(b) have different signs. Therefore, let's look for the intervals in which the outputs of the endpoints have opposite signs. From the table, we can identify four intervals meeting this condition. I_1 &= [-13,-12] & ✓ & & I_4 &= [0,1] * I_2 &= [-12,-10] & * & & I_5 &= [1,2] ✓ I_3 &= [-3,-2] & ✓ & & I_6 &= [3,5] ✓ From these results, we can guarantee that f has at least one zero in the intervals [-13,-12], [-3,-2], [1,2], and [3,5].
Consider the following graph of a polynomial function.
Let's start by recalling that the phrase relative extrema involves both relative maximums and relative minimums. Therefore, we want to count the number of relative maximums and minimums in the graph of f(x). At relative extrema, the graph changes from increasing to decreasing or vice versa.
From the graph, we get the following information.
Comparing the information we gathered to the given statements, we can say that only I and III are true.
Item | Statement | True/False |
---|---|---|
I | It has five relative extrema. | True |
II | It has both absolute minimum and absolute maximum. | False |
III | It has an absolute minimum but not an absolute maximum. | True |
IV | It has an absolute maximum but not an absolute minimum. | False |
At a local maximum, the graph of a function changes from increasing to decreasing. In other words, the local maximums are the y-coordinates of the tops of the hills in the graph of a function.
As we can see, the graph has two hills and, therefore, it has two local maximums. Let's identify the x-coordinate of each point.
The x-values at which f reaches a local maximum are -3 and 2.
While we looked for hills
in Part B, this time we are going to look for the valleys
in the graph. This is because at a local minimum, the graph of a function changes from decreasing to increasing. The local minimums are the y-coordinates of the bottoms of the valleys in the graph of a function.
As we can see, the graph has three valleys, so it has three local minimums. Let's identify the x-coordinate of each point.
The x-values at which f reaches a local minimum are -5, 0, and 4.
From Part C, we know that the graph of f(x) reaches a local minimum at x=-5, 0, and 4. This time, we were asked to state the local minimums. This means that we will look at the y-coordinates of the points labeled in Part C.
As we can see, local minimums of the graph of f are -2.5, 1, and 2.
Consider the following graph of a polynomial function.
A turning point is a point where the graph of a function changes from increasing to decreasing or vice versa. Let's begin by identifying where the function increases and where it decreases.
As we can see there are four points where the graph changes its behavior. Therefore, the graph has 4 turning points.
Let's label each of the given points.
ll
A(-6.5,118) & D(1,47)
B(-6,87) & E(4,-307)
C(-3,-200) & F(4.5,-337)
From the graph, we can tell that the x-coordinates of the turning points are x_1=-6.5, x_2=-3, x_3=1, and x_4=4.5. With this information, we can discard those points whose x-coordinates are not included in this list.
llll
A(-6.5,118) & & D(1,47) &
B(-6,87) & * & E(4,-307) & *
C(-3,-200) & & F(4.5,-337) &
Of the remaining points, we need to verify that their y-coordinates correspond to those of the turning points. Let's plot these points along with the graph.
As we can see, point C is not on the graph, meaning we can discard it. Notice that points A, D, and F are on the graph and correspond to turning points. This means that of the six given points, only these three are turning points. llll A(-6.5,118) & ✓ & D(1,47) & ✓ B(-6,87) & * & E(4,-307) & * C(-3,-200) & * & F(4.5,-337) & ✓ The fourth turning point (-3,-267) is not listed among the choices.
Determine whether the given graphs are even, odd, or neither.
Let's begin by checking whether the graph is even. If a function is even, then the y-axis should act like a mirror. However, we can see that the point (0.5,3) is on the graph but the point (-0.5,3) — its reflection on the y-axis — is not on the graph.
Therefore, we know that the graph is not even. Now let's check if the graph is odd. A function is odd if its graph is symmetric about the origin. This implies that if the point (x,y) is on the graph, then (-x,-y) must also be on the graph. However, we can see that (1.5,2) is on the graph but (-1.5,-2) is not.
This means that the graph is not odd. With this information, we can conclude that the given graph is neither even nor odd.
As we did in Part A, let's begin by checking if the given graph is even. We can see that the y-axis does not act like a mirror because the point (0.5,1.5) is on the graph but its reflection on the y-axis (- 0.5,1.5) is not.
This is how we know that the graph is not even. To check if the graph is odd, let's pick some arbitrary points on the graph and reflect them about the origin. We have to verify that the reflections are on the graph.
We can see that the reflections about the origin are also in the graph. Furthermore, this happens to every point on the graph we pick. As such, the given graph is odd. We can also confirm this claim by rotating the graph 180^(∘) about the origin.
At a glance, we can see that the given graph seems to be symmetric about the y-axis. If this is the case, the graph is even. Let's verify this assumption by picking some points on the graph and reflecting them in the y-axis.
As we can see, all the reflections are also in the graph. This happens to every point on the graph we pick, so we can say that the given graph is even. Since a graph cannot be even and odd at the same time, we do not need to verify if it is odd. In fact, it is not odd!