# 5.3 Graphs of polynomial functions  (Page 4/13)

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Use the graph of the function of degree 5 in [link] to identify the zeros of the function and their multiplicities.

The graph has a zero of –5 with multiplicity 1, a zero of –1 with multiplicity 2, and a zero of 3 with even multiplicity.

## Determining end behavior

As we have already learned, the behavior of a graph of a polynomial function    of the form

$f\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+...+{a}_{1}x+{a}_{0}$

will either ultimately rise or fall as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ increases without bound and will either rise or fall as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say –100 or –1,000.

Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, $\text{\hspace{0.17em}}{a}_{n}{x}^{n},\text{\hspace{0.17em}}$ is an even power function, as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ increases or decreases without bound, $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ increases without bound. When the leading term is an odd power function, as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ decreases without bound, $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ also decreases without bound; as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ increases without bound, $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ also increases without bound. If the leading term is negative, it will change the direction of the end behavior. [link] summarizes all four cases.

## Understanding the relationship between degree and turning points

In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function $\text{\hspace{0.17em}}f\left(x\right)={x}^{4}-{x}^{3}-4{x}^{2}+4x\text{\hspace{0.17em}}$ in [link] . The graph has three turning points.

This function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is a 4 th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function.

## Interpreting turning points

A turning point    is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).

A polynomial of degree $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ will have at most $\text{\hspace{0.17em}}n-1\text{\hspace{0.17em}}$ turning points.

## Finding the maximum number of turning points using the degree of a polynomial function

Find the maximum number of turning points of each polynomial function.

1. $\text{\hspace{0.17em}}f\left(x\right)=-{x}^{3}+4{x}^{5}-3{x}^{2}+1$
2. $\text{\hspace{0.17em}}f\left(x\right)=-{\left(x-1\right)}^{2}\left(1+2{x}^{2}\right)$
1. First, rewrite the polynomial function in descending order: $\text{\hspace{0.17em}}f\left(x\right)=4{x}^{5}-{x}^{3}-3{x}^{2}+1$

Identify the degree of the polynomial function. This polynomial function is of degree 5.

The maximum number of turning points is $\text{\hspace{0.17em}}5-1=4.$

2. First, identify the leading term of the polynomial function if the function were expanded.

Then, identify the degree of the polynomial function. This polynomial function is of degree 4.

The maximum number of turning points is $\text{\hspace{0.17em}}4-1=3.$

## Graphing polynomial functions

We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.

Given a polynomial function, sketch the graph.

1. Find the intercepts.
2. Check for symmetry. If the function is an even function, its graph is symmetrical about the $\text{\hspace{0.17em}}y\text{-}$ axis, that is, $\text{\hspace{0.17em}}f\left(-x\right)=f\left(x\right).\text{\hspace{0.17em}}$ If a function is an odd function, its graph is symmetrical about the origin, that is, $\text{\hspace{0.17em}}f\left(-x\right)=-f\left(x\right).$
3. Use the multiplicities of the zeros to determine the behavior of the polynomial at the $\text{\hspace{0.17em}}x\text{-}$ intercepts.
4. Determine the end behavior by examining the leading term.
5. Use the end behavior and the behavior at the intercepts to sketch a graph.
6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
7. Optionally, use technology to check the graph.

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
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Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
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Abhi
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