4. Polynomial Functions and Their Graphs
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Chapter 1
4.
Polynomial Functions and Their Graphs
Polynomial functions and their graphs play a pivotal role in understanding the behavior of mathematical equations. The lesson delves deep into the nuances of graphing polynomial functions, emphasizing the significance of x-intercepts, end behavior, and turning points. By analyzing polynomial functions, one can determine the regions where the function increases or decreases, and where it reaches its maximum or minimum values. Furthermore, the symmetry of a graph can reveal if a function is even, odd, or neither. Through various examples and exercises, the lesson provides a holistic approach to mastering the art of graphing polynomial functions, ensuring a robust foundation in algebra.
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| | 16 Theory slides |
| | 12 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Polynomial Functions and Their Graphs
Slide of 16
Graphing functions is essential in different fields of research. This has led many people to develop programs and technologies that draw accurate graphs, such as graphing calculators. However, knowing how to draw a graph by hand is also essential. After all, that is how they did it not too many years ago!
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.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Investigating the End Behavior
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 |
Discussion
Polynomial Functions and End Behavior
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.
Concept
Polynomial Function
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
Concept
End Behavior
The end behavior of a function is the value to which f(x) tends as x extends to the left or the right infinitely. If f(x) keeps increasing without bound, it is said to tend to positive infinity. The end behavior of this case is stated as 
From the arrows on the graph, it can be seen that the left end of the graph extends downward, while the right end extends upward. The end behavior of g can then be expressed as follows. 
Note that when the degree of the polynomial is even, both ends have the same behavior, which depends on the sign of the leading coefficient. By contrast, when the degree is odd, both ends have opposite behaviors.
up.
f(x)→+∞
Conversely, if f(x) keeps decreasing without bound, it is said to tend to negative infinity. In this case, the end behavior is stated as down.
f(x)→-∞
For example, consider the graph of a function g(x). As x→ -∞,As x→+∞,g(x)→ -∞g(x)→+∞
To state the end behavior of a function in words, begin by stating the left-end behavior, then state the right-end behavior. A dash can also be used to separate the words. For instance, the end behavior of the graph of g(x) can be written as down and upor as
down-up.
End Behavior of Polynomial Functions
When the function is a polynomial function, the end behavior can be determined from the function rule.P(x)=anxn+an−1xn−1+⋯+a1x+a0
Particularly, the end behavior is given by the sign of the leading coefficient and the degree of the polynomial.
Pop Quiz
Determining the End Behavior
Determine the end behavior of the given polynomial function.
Example
Image Editor Running Time
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 |
a Emily used online software to plot both polynomials in an effort to clear things up before deciding which software to purchase.
{"type":"choice","form":{"alts":["Yes","No"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
b What is the end behavior of the graph of y=P(x)?
{"type":"choice","form":{"alts":["Up-Up","Up-Down","Down-Up","Down-Down"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":2}
c What is the end behavior of the graph of y=Q(x)?
{"type":"choice","form":{"alts":["Up-Up","Up-Down","Down-Up","Down-Down"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
Hint
b The end behavior of a polynomial graph can be determined by studying the sign of the leading coefficient and the degree of the polynomial rule. When the degree is odd, both ends of the graph have opposite behaviors.
c When the degree is even, both ends of the graph have the same behavior.
Solution
a Before analyzing Emily's graph, evaluate the polynomial functions at different values to compare algebraically the running times of both software.
| 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.
- For very small images, Software 1 is faster than Software 2.
- For images of about 240-by-240 pixels, both software have similar running times.
- For images with more realistic dimensions, 600-by-600 pixels images or larger, Software 2 is faster than Software 1.
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.
b Although the behavior of the right end of the graph of y=P(x) was studied in Part A, it can also be determined using the function rule.
P(x)=x3−520x2+10000x+20000000
Start by writing the end behavior of a polynomial graph based on the leading coefficient and degree. | 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.
c As in Part B, the end behavior of the graph of y=Q(x) can be obtained from its function rule.
Q(x)=100x2+1000000
Here, the leading term is 100x2, so the leading coefficient is 100 and the degree is 2. Since the leading coefficient is positive and the degree is even, the end behavior of the graph of y=Q(x) is Up-Up.
Again, the y-axis is measured in seconds, not in nanoseconds.
Discussion
Alternative Methods for Finding the x-intercepts
When graphing polynomial functions, the x-intercepts is a key feature. Such information can sometimes be derived from the function rule, but it is not always self-evident. Consider the following polynomial.
Note that the fact that f(k)=0 has two implications — that k is a zero of the polynomial function and that x=k is a solution to the equation f(x)=0.
Remember, in an equivalence, both the conditional statement and its converse are true.
f(x)=x4−3x3−15x2+19x+30
At first glance, it is not clear whether the graph of y=f(x) intercepts the x-axis, and if so, where. However, if such a point existed, then it would have the form (k,0), which means that f(k)=0. f(k)=0⇕k4−3k3−15k2+19k+30=0
Consequently, finding the x-intercepts of the graph of y=f(x) is equivalent to finding the real zeros of the polynomial f(x), which in turn is equivalent to finding the real solutions to the equation f(x)=0. The following diagram explains what this equivalence means.
Example
Graphing a Polynomial Function
Consider the following polynomial function.
Comparing the graph obtained with the given ones, it can be concluded that the graph of y=f(x) is Graph 4.
f(x)=x4−3x3−15x2+19x+30
a List all the values where the graph of y=f(x) intercepts the x-axis.
{"type":"text","form":{"type":"list","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false,"ordermatters":false,"numinput":4,"listEditable":true,"hideNoSolution":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["-3","-1","2","5"]}}
b List the missing numbers in the following table of values. Write them in the order they should be in the table from left to right.
| x | -4 | -2 | 0 | 1 | 3 | 4 | 6 |
|---|---|---|---|---|---|---|---|
| f(x) | -28 | -48 | 252 |
{"type":"text","form":{"type":"list","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false,"ordermatters":true,"numinput":4,"listEditable":false,"hideNoSolution":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["162","30","32","-70"]}}
c What is the end behavior of the graph of y=f(x)?
{"type":"choice","form":{"alts":["Up-Up","Down-Up","Up-Down","Down-Down"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
d Which of the following is the graph y=f(x)?
{"type":"choice","form":{"alts":["Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><\/span><\/span><\/span>","Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>","Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><\/span><\/span><\/span>","Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">4<\/span><\/span><\/span><\/span>","Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">5<\/span><\/span><\/span><\/span>","Graph <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">6<\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}
Hint
a Finding the x-intercepts is the same as finding the solutions to the equation f(x)=0. Factor the polynomial and use the Zero Product Property.
b Evaluate the polynomial at the missing values.
c Study the sign of the leading coefficient and the degree of the polynomial. When the degree is even, both ends have the same behavior.
d Start by plotting the x-intercepts obtained in Part A. Next, plot the points from the table of values. Connect all the points using a smooth curve and extend the graph according to the end behavior.
Solution
a Determining the places where the graph of y=f(x) intercepts the x-axis is equivalent to finding the solutions to the equation f(x)=0.
f(x)=0⇓x4−3x3−15x2+19x+30=0
To solve the polynomial equation, factor the polynomial. This can be done by grouping, but some rewrites are needed first.
-3x3-15x219x=-4x3+x3=-11x2−4x2=30x−11x
Substitute these expressions into the polynomial equation.
x4−3x3−15x2+19x+30=0
▼
Factor
SubstituteExpressions
Substitute expressions
x4−4x3+x3−11x2−4x2+30x−11x+30=0
CommutativePropAdd
Commutative Property of Addition
(x4−4x3−11x2+30x)+(x3−4x2−11x+30)=0
FactorOut
Factor out x
x(x3−4x2−11x+30)+(x3−4x2−11x+30)=0
FactorOut
Factor out (x3−4x2−11x+30)
(x3−4x2−11x+30)(x+1)=0
-4x2-11x=-2x2−2x2=-15x+4x
Now, substitute the expressions into the cubic polynomial.
(x3−4x2−11x+30)(x+1)=0
▼
Factor
SubstituteII
-4x2=-2x2−2x2, -11x=-15x+4x
(x3−2x2−2x2−15x+4x+30)(x+1)=0
CommutativePropAdd
Commutative Property of Addition
((x3−2x2−15x)+(-2x2+4x+30))(x+1)=0
FactorOut
Factor out x & -2
(x(x2−2x−15)−2(x2−2x−15))(x+1)=0
FactorOut
Factor out (x2−2x−15)
(x2−2x−15)(x−2)(x+1)=0
(x2−2x−15)(x−2)(x+1)=0
(x−5)(x+3)(x−2)(x+1)=0
(x−5)(x+3)(x−2)(x+1)=0⇓⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧x−5=0x+3=0x−2=0x+1=0⇒⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧x=5x=-3x=2x=-1
Consequently, the solutions to the polynomial equation, and therefore the x-intercepts, are x=-3, -1, 2, and 5.
b To find the values missing from the table, evaluate f(x) at the corresponding x value. For example, begin by substituting x=-4.
f(x)=x4−3x3−15x2+19x+30
▼
Substitute -4 for x and simplify
Substitute
x=-4
f(-4)=(-4)4−3(-4)3−15(-4)2+19(-4)+30
CalcPowProd
Calculate power and product
f(-4)=256−3(-64)−15(16)−76+30
Multiply
Multiply
f(-4)=256+192−240−76+30
AddSubTerms
Add and subtract terms
f(-4)=162
| 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 |
c To determine the end behavior of the graph of a polynomial, start by identifying the leading term.
f(x)=x41−3x3−15x2+19x+30
Since the degree is 4, which is an even number, both ends of the graph have the same behavior. Additionally, since the leading coefficient is 1, which is a positive number, the end behavior of the graph of y=f(x) is up-up.
d To determine which of the graphs is correct, plot the graph of y=f(x) by using the information obtained in the previous parts.
- Begin by plotting the x-intercepts from Part A.
- Next, graph the points from the table of values in Part B.
- Connect the points using a smooth curve.
- Extend both ends of the graph upwards, since the end behavior of the graph of y=f(x) was found to be
up-up
in Part C.
Discussion
Locating the x-Intercepts of a Graph
Given a polynomial function, the end behavior of the graph can be determined directly from the function rule without doing anything to it. For example, consider the following function.
The end behavior of the graph of y=f(x) is 
Since graphs of polynomial functions are continuous — that is, they have no gaps — the points (a,f(a)) and (b,f(b)) must be connected by a smooth curve that will cross the x-axis at least once, implying that there is at least one zero between a and b. 
All real zeros of a polynomial can be estimated in this fashion.
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.
Rule
The Location Principle
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.
Proof
Informal JustificationLet 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.
Pop Quiz
Locating Zeros
Use the Location Principle to indicate whether the given polynomial function is guaranteed to have a zero in the indicated interval.
Discussion
Maximum and Minimum Points of a Graph
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.
Concept
Relative Minimum and Maximum
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.
Concept
Absolute Minimum and Maximum
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.
Discussion
Turning Points of Polynomial Functions
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.
Concept
Turning Point
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.
Pop Quiz
Determining the Maximum Number of Turning Points
Consider the given polynomial function. Determine either the maximum or the exact number of turning points that its graph may have.
Example
Graphing a Fifth-Degree Polynomial Function
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).
a Find all the zeros of the polynomial function.
b What is the end behavior of the graph of f?
c Sketch of the graph of f.
Answer
a Zeros: -14,-8,-2,7,10
b End behavior: Up-Down
c Graph:
Hint
a Use the Location Principle.
b Take a look at the degree and the sign of the leading coefficient of the polynomial.
c Use the fact that the zeros are the x-intercepts of the graph. Plot the turning points. Connect all the points with a smooth curve and extend it according to the end behavior.
Solution
a One way of finding the zeros of the function is by factoring the given polynomial. However, it does not seem easy to factor. Alternatively, the Location Principle can help locate where the zeros could be. In the given table, highlight the positive values with one color and the negative values with a different color.
| x | -15 | -10 | -7 | -3 | 0 | 6 | 8 | 11 |
|---|---|---|---|---|---|---|---|---|
| f(x) | 25025 | -10880 | 4165 | 3575 | -7840 | -4480 | 3520 | -12350 |
f(x)=-0.5x5−3.5x4+91x3+374x2−3556x−7840
▼
Substitute -14 for x and evaluate
Substitute
x=-14
f(-14)=-0.5(-14)5−3.5(-14)4+91(-14)3+374(-14)2−3556(-14)−7840
CalcPowProd
Calculate power and product
f(-14)=-0.5(-537824)−3.5(38416)+91(-2744)+374(196)+49784−7840
Multiply
Multiply
f(-14)=268912−134456−249704+73304+49784−7840
AddSubTerms
Add and subtract terms
f(-14)=0
| 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.
- The third sign change occurs between -3 and 0.
- The fourth sign change occurs between 6 and 8.
- The fifth sign change occurs between 8 and 11.
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.
b The end behavior of the graph of a polynomial function can be determined by the sign of the leading coefficient and the degree of the polynomial.
| 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.
c Since the zeros of a polynomial are the x-intercepts, start by plotting these points on the coordinate plane. It also may help to make a sketch reflecting the end behavior of the graph.
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.
Discussion
Even and Odd Functions
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.
Concept
Even Function
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)
Extra
Determining If a Function is EvenTo determine algebraically whether a function is even, substitute -x into the function rule and simplify. If the resulting expression is equal to f(x), then the function is even; otherwise, it is not. For example, consider the following function.
Since f(-x)=f(x), the given function is even.
f(x)=3x4−2x2+1
Substitute -x for x and simplify.
f(x)=3x4−2x2+1
Substitute
x=-x
f(-x)=3(-x)4−2(-x)2+1
NegBaseToPosPow
(-a)4=a4
f(-x)=3x4−2(-x)2+1
NegBaseToPosBase
(-a)2=a2
f(-x)=3x4−2x2+1
Substitute
3x4−2x2+1=f(x)
f(-x)=f(x)
Some other functions seem to be reflected across the origin. Functions with this property are called odd functions.
Concept
Odd Function
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)
Extra
Determining If a Function is OddTo determine algebraically whether a function is odd, substitute -x into the function rule and simplify. If the resulting expression is equal to -f(x), then the function is odd; otherwise, it is not. For example, consider the following function.
Since f(-x)=-f(x), the given function is odd.
f(x)=x2−1x
Substitute -x for x and simplify.
f(x)=x2−1x
Substitute
x=-x
f(-x)=(-x)2−1-x
NegBaseToPosBase
(-a)2=a2
f(-x)=x2−1-x
MoveNegNumToFrac
Put minus sign in front of fraction
f(-x)=-x2−1x
Substitute
x2−1x=f(x)
f(-x)=-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.
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Classifying Functions
Closure
Verifying End Behavior
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."
f(x)=x41−4x3−39x2+46x+80⇓Leading Coefficient:1Degree:4
The leading coefficient is positive and the degree is even. Therefore, the end behavior is 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 polynomial functions.
f(x)=4x3−3x2−2x−100
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g(x)=-8x7+32x5+2x2−6
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h(x)=3x3+51−2x2+4x4
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Let's begin by writing the end behavior of a polynomial graph based on the leading coefficient and degree of the function rule.
| Leading Coefficient | Degree | End Behavior |
|---|---|---|
| Positive | Even | Up-Up |
| Positive | Odd | Down-Up |
| Negative | Odd | Up-Down |
| Negative | Even | Down-Down |
Now, let's take a look at the polynomial function and identify its leading coefficient and degree. f(x) = 4x^3-3x^2-2x-100 The leading coefficient is 4, which is a positive number. The degree of the polynomial is 3, which is an odd number. With this information, we can say that the end behavior of the graph of f(x) is down-up.
To determine the end behavior of the graph of g(x), we will use the table we wrote in Part A. Let's start by identifying the leading coefficient and degree of the function rule.
g(x) = -8x^7 + 32x^5 + 2x^2 - 6
The leading coefficient is -8, which is a negative number. The leading coefficient is 7, which is an odd number. Therefore, the end behavior of the graph of g(x) is up-down.
We will determine the end behavior of the graph of h(x) the same way we did in the previous parts. However, notice that the given polynomial function is not written in standard form. Let's begin by rearranging the terms.
h(x) = 3x^3 + 51 - 2x^2 + 4x^4
⇕
h(x) = 4x^4 + 3x^3 - 2x^2 + 51
With the function rule written in standard form, we can see that the leading coefficient is positive and the degree is even — both are equal to 4. Consequently, the end behavior of the graph of g(x) is up-up.
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Consider the following polynomial functions.
Pair the function rules with the their graphs.
f(x)g(x)h(x)m(x)=0.02x4+0.04x3−0.62x2−0.64x+1.2=0.02x4+0.06x3−0.56x2−1.2x=0.04x3−0.04x2−0.48x=-0.02x4−0.06x3+0.5x2+1.5x
Additionally, consider the following four graphs.
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To determine which function rule corresponds to which graph, let's begin by determining the end behavior of the polynomial functions. We will first write the end behavior of a polynomial graph based on the leading coefficient and the degree of the function rule.
| Leading Coefficient | Degree | End Behavior |
|---|---|---|
| Positive | Even | Up-Up |
| Positive | Odd | Down-Up |
| Negative | Odd | Up-Down |
| Negative | Even | Down-Down |
Now let's take a look at all the given functions and identify their leading coefficients and degrees.
| Function | Leading Coefficient | Degree | End Behavior |
|---|---|---|---|
| f(x) = 0.02x^4 + 0.04x^3 - 0.62x^2 - 0.64x + 1.2 | 0.02→ Positive | 4→ Even | Up-Up |
| g(x) = 0.02x^4 + 0.06x^3 - 0.56x^2 - 1.2x | 0.02→ Positive | 4→ Even | Up-Up |
| h(x) = 0.04x^3 - 0.04x^2 - 0.48x | 0.04→ Positive | 3→ Odd | Down-Up |
| m(x) = -0.02x^4 - 0.06x^3 + 0.5x^2 + 1.5x | -0.02→ Negative | 4→ Even | Down-Down |
If we take a look at the four given graphs and pay close attention to their end behaviors, we can get the following information. A &→ Down-Down B &→ Up-Up C &→ Up-Up D &→ Down-Up With this, we can make a new table where we write the possible graphs for each function rule.
| Function | End Behavior | Possible Graphs |
|---|---|---|
| f(x) = 0.02x^4 + 0.04x^3 - 0.62x^2 - 0.64x + 1.2 | Up-Up | B and C |
| g(x) = 0.02x^4 + 0.06x^3 - 0.56x^2 - 1.2x | Up-Up | B and C |
| h(x) = 0.04x^3 - 0.04x^2 - 0.48x | Down-Up | D |
| m(x) = -0.02x^4 - 0.06x^3 + 0.5x^2 + 1.5x | Down-Down | A |
From the table, we can pair two function rules to their graphs: graph D corresponds to h(x) and graph A corresponds to m(x). GraphA &→ m(x) GraphD &→ h(x) However, functions f(x) and g(x) can be paired either to graph B or graph C. Therefore, we need to go beyond the end behavior. Let's identify their x-intercepts.
As we can see, both graphs have almost the same x-intercepts.
| Graph | x-intercepts |
|---|---|
| B | -6, -2, 0, 5 |
| C | -6, -2, 1, 5 |
Let's use the non-matching x-intercepts to determine which function rule corresponds to which graph. Let's substitute x=0 into both function rules. The polynomial whose output is 0 for x=0 is the function that corresponds to graph B.
| Function | Substitute x=0 | Evaluate |
|---|---|---|
| f(x) = 0.02x^4 + 0.04x^3 - 0.62x^2 - 0.64x + 1.2 | f( 0) = 0.02( 0)^4 + 0.04( 0)^3 - 0.62( 0)^2 - 0.64( 0) + 1.2 | f(0)=1.2 |
| g(x) = 0.02x^4 + 0.06x^3 - 0.56x^2 - 1.2x | g( 0) = 0.02( 0)^4 + 0.06( 0)^3 - 0.56( 0)^2 - 1.2( 0) | g(0)=0 |
Since f(0)≠ 0, then x=0 is not an x-intercept of the graph of f(x). Conversely, g(0) is equal to 0. Therefore, graph B corresponds to g(x) and graph C corresponds to f(x). A &→ m(x) B &→ g(x) C &→ f(x) D &→ h(x)
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Determine whether the functions are even, odd, or neither.
f(x)=x5−34x3+225x
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g(x)=x4−20x2+64
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To determine if a function is even or odd, let's substitute -x for x into the function rule and simplify the right-hand side. If the resulting expression is equal to f(x), the function is even. Conversely, if the resulting expression is equal to -f(x), the function is odd. Otherwise, the function is neither even nor odd.
As we can see, the resulting expression is not equal to f(x). Therefore, the function is not even. However, we can continue simplifying the expression on the right-hand side by factoring out -1.
We found that f(-x) is equal to -f(x), which implies that the function is odd.
As we did in Part A, let's substitute -x for x into the function rule and simplify the right-hand side.
As we can see, g(-x) is equal to g(x). Therefore, the function is even. Because a function cannot be even and odd at the same time, we do not need to check if the function is odd.
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f(x)=x4+4x3−x2−4x
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g(x)=(x+9)(x−5)(x+1)3
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{"type":"choice","form":{"alts":["Yes, it is possible.","No, it is not possible."],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
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Let's begin by recalling that a turning point is a point where the graph of a function changes from increasing to decreasing or vice versa. However, this definition is not so helpful if we do not have the graph of the function. Nevertheless, the following fact can also help us.
Polynomial Functions and Turning Points |- The graph of a polynomial function of degree n can have at most n-1 turning points.
Since the polynomial function f(x) has degree 4, we can say that the graph of f(x) has at most 3 turning points.
We can use the same fact we used in Part A to find the maximum possible number of turning points that the graph of g(x) can have. However, this time the function rule is given in factored form.
g(x) = (x+9)(x-5)(x+1)^3
The good news is that we do not need to expand the cube and perform the multiplications to determine the degree of g(x). We can answer the question by finding the degree of each factor.
| Factor | Degree |
|---|---|
| x+9 | 1 |
| x-5 | 1 |
| (x+1)^3 | 3 |
The degree of g(x) is the sum of the degrees of each factor. As such, the degree of g(x) is 5. This means that the graph of g(x) has at most 5-1=4 turning points.
In Part A, we found that the graph of f(x) has at most 3 turning points. This means that k=3. Knowing this, let's read the question again.
Is it possible for the graph off(x) to have2 turning points?
Before answering this question, let's remember another aspect of the turning points of a polynomial graph.
Polynomial Functions and Turning Points |- If a polynomial function has n distinct real zeros, then its graph has exactly n-1 turning points.
With this fact in mind, let's find zeros of f(x). We can start by factoring out x.
Each of the factors of f(x) gives us a zero of the function. ccc Factor & & Zero x & → & 0 x+4 & → & -4 x+1 & → & -1 x-1 & → & 1 As we can see, f(x) has four distinct real zeros. This means that the graph of f(x) has exactly 3 turning points. For this reason, the answer to the initial question is no. It is not possible for the graph off(x) to have2 turning points.
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f(x)=(x+4)(x+2)(1−x)(x−3)(x−6)
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{"type":"choice","form":{"alts":["Up-Up","Up-Down","Down-Up","Down-Down"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
Which of the following is the graph of f(x)?
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Since the x-intercepts of a graph are the zeros of the function, our task is reduced to find the zeros of the given function. To do so, let's equate the function to zero.
To solve this polynomial equation, we can apply the Zero Product Property and split the equation into five linear equations. Then we can solve each one separately. (x + 4)(x + 2)(1 - x)(x - 3)(x - 6) = 0 ⇓ x+4 = 0 x+2=0 1-x=0 x-3=0 x-6=0 ⇔ x = -4 x=-2 x=1 x=3 x=6 We found that the x-intercepts of the graph of f(x) are x=-4, -2 1, 3, 6.
Let's begin by writing the end behavior of a polynomial graph based on the leading coefficient and degree of the function rule.
| Leading Coefficient | Degree | End Behavior |
|---|---|---|
| Positive | Even | Up-Up |
| Positive | Odd | Down-Up |
| Negative | Odd | Up-Down |
| Negative | Even | Down-Down |
In our case, we have a polynomial function in factored form, so we cannot see the leading coefficient and degree directly. However, we do not have to perform all the multiplications to find them. Instead, we can use the following fact.
When two or more polynomials are multiplied, the leading coefficient of the resulting polynomial equals the product of the leading coefficients, and the degree equals the sum of the degrees of the multiplied polynomials.
Let's identify the leading coefficient and the degree of each factor of f(x).
| f(x) = (x+4)(x+2)(1-x)(x-3)(x-6) | ||
|---|---|---|
| Factor | Leading Coefficient | Degree |
| x+4 | 1 | 1 |
| x+2 | 1 | 1 |
| 1-x | -1 | 1 |
| x-3 | 1 | 1 |
| x-6 | 1 | 1 |
From the table, we can conclude that the product of the leading coefficients is equal to -1 and the sum of the degrees is 5. Therefore, the leading coefficient is negative and the degree is odd. This tells us that the end behavior of the graph of f(x) is up-down.
To determine which of the graphs is the graph of f(x), we will sketch the graph of f(x) and compare it to the given choices. Let's plot the x-intercepts we found in Part A and the end behavior we found in Part B. Remember, the x-intercepts occur at -4, -2, 1, 3, and 6, and the end behavior is up-down.
Comparing this rough sketch with the four given graphs, we can disregard graph A because it has different x-intercepts, and graph B because it has a different end behavior.
As we can see, graphs C and D have the same x-intercepts and end behavior. To determine which one corresponds to the graph of f(x), let's estimate their y-intercepts.
| Graph | y-intercept |
|---|---|
| C | Less than 200 |
| D | About 200 |
Now, let's find the y-intercept of the graph of f(x) by evaluating the function rule at x=0.
Since f(0)<200, we conclude that the graph of f(x) is graph C.
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Polynomial Functions and Their Graphs
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