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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 fourthdegree polynomial function $f(x)=x_{4}−4x_{3}−39x_{2}+46x+80.$ 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 
$h(x)=x_{2}+3x−1$  Quadratic Function 
$t(x)=2e_{3x}$  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.
$f(x)=a_{n}x_{n}+a_{n−1}x_{n−1}+⋯+a_{1}x+a_{0}$
up.
down.
down and upor as
downup.
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 $xbyx$ pixels image are given by the following polynomials.
Running Time (nanoseconds)  

Software $1$  $P(x)=x_{3}−520x_{2}+10000x+20000000$ 
Software $2$  $Q(x)=100x_{2}+1000000$ 
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 rightend 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 rightend 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  UpUp 
Positive  Odd  DownUp 
Negative  Odd  UpDown 
Negative  Even  DownDown 
The leading term of $P(x)$ is $x_{3},$ 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 DownUp.
Despite the $y$axis being measured in seconds rather than nanoseconds, the conclusion about the end behavior or the fastest software remains the same.
UpUp.
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 $(x_{3}−4x_{2}−11x+30)$
$4x_{2}=2x_{2}−2x_{2}$, $11x=15x+4x$
Commutative Property of Addition
Factor out $x&2$
Factor out $(x_{2}−2x−15)$
$x=4$
Calculate power and product
Multiply
Add and subtract terms
$x$  $f(x)=x_{4}−3x_{3}−15x_{2}+19x+30$  Result 

$0$  $f(0)=0_{4}−3(0)_{3}−15(0)_{2}+19(0)+30$  $30$ 
$1$  $f(1)=1_{4}−3(1)_{3}−15(1)_{2}+19(1)+30$  $32$ 
$4$  $f(4)=4_{4}−3(4)_{3}−15(4)_{2}+19(4)+30$  $70$ 
Consequently, the complete table of values looks as follows.
$x$  $4$  $2$  $0$  $1$  $3$  $4$  $6$ 

$f(x)$  $162$  $28$  $30$  $32$  $48$  $70$  $252$ 
upup.
upupin Part C.
downup.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.
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 $b.$For $f(x)=8x_{3}−12x_{2}$ $−26x$ $+15,$ 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.
All real zeros of a polynomial can be estimated in this fashion.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 $x=d.$
If the function is continuous, the function switches from increasing to decreasing at a relative maximum or from decreasing to increasing at a relative minimum.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 fifthdegree 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 $f(x)=0.5x_{5}−3.5x_{4}$ $+91x_{3}+374x_{2}$ $−3556x−7840.$ 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
$f(x)=0.5x_{5}−3.5x_{4}+91x_{3}+374x_{2}−3556x−7840$  

$x=9$  $x=8$ 
$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  UpUp 
Positive  Odd  DownUp 
Negative  Odd  UpDown 
Negative  Even  DownDown 
For the given polynomial, the leading term is $0.5x_{5}.$ The leading coefficient is negative and the degree is odd. Consequently, the end behavior of the graph of $f$ is UpDown.
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.