# 3.4 Graphs of polynomial functions  (Page 5/13)

 Page 5 / 13

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.

## Sketching the graph of a polynomial function

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=-2{\left(x+3\right)}^{2}\left(x-5\right).$

This graph has two $\text{\hspace{0.17em}}x\text{-}$ intercepts. At $\text{\hspace{0.17em}}x=-3,\text{\hspace{0.17em}}$ the factor is squared, indicating a multiplicity of 2. The graph will bounce at this $\text{\hspace{0.17em}}x\text{-}$ intercept. At $\text{\hspace{0.17em}}x=5,\text{\hspace{0.17em}}$ the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.

The y -intercept is found by evaluating $\text{\hspace{0.17em}}f\left(0\right).$

The $\text{\hspace{0.17em}}y\text{-}$ intercept is $\text{\hspace{0.17em}}\left(0,90\right).$

Additionally, we can see the leading term, if this polynomial were multiplied out, would be $\text{\hspace{0.17em}}-2{x}^{3},\text{\hspace{0.17em}}$ so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See [link] .

To sketch this, we consider that:

• As $\text{\hspace{0.17em}}x\to -\infty \text{\hspace{0.17em}}$ the function $\text{\hspace{0.17em}}f\left(x\right)\to \infty ,\text{\hspace{0.17em}}$ so we know the graph starts in the second quadrant and is decreasing toward the $\text{\hspace{0.17em}}x\text{-}$ axis.
• Since $\text{\hspace{0.17em}}f\left(-x\right)=-2{\left(-x+3\right)}^{2}\left(-x–5\right)\text{\hspace{0.17em}}$ is not equal to $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ the graph does not display symmetry.
• At $\text{\hspace{0.17em}}\left(-3,0\right),\text{\hspace{0.17em}}$ the graph bounces off of the $\text{\hspace{0.17em}}x\text{-}$ axis, so the function must start increasing.

At $\text{\hspace{0.17em}}\left(0,90\right),\text{\hspace{0.17em}}$ the graph crosses the $\text{\hspace{0.17em}}y\text{-}$ axis at the $\text{\hspace{0.17em}}y\text{-}$ intercept. See [link] .

Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at $\text{\hspace{0.17em}}\left(5,0\right).\text{\hspace{0.17em}}$ See [link] .

As $\text{\hspace{0.17em}}x\to \infty \text{\hspace{0.17em}}$ the function $\text{\hspace{0.17em}}f\left(x\right)\to \mathrm{-\infty },\text{\hspace{0.17em}}$ so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.

Using technology, we can create the graph for the polynomial function, shown in [link] , and verify that the resulting graph looks like our sketch in [link] . The complete graph of the polynomial function   f ( x ) = − 2 ( x + 3 ) 2 ( x − 5 )

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{4}x{\left(x-1\right)}^{4}{\left(x+3\right)}^{3}.$ ## Using the intermediate value theorem

In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x -axis, we can confirm that there is a zero between them. Consider a polynomial function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ in the domain of $\text{\hspace{0.17em}}f,$ if $\text{\hspace{0.17em}}a and $f\left(a\right)\ne f\left(b\right),$ then the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ takes on every value between $\text{\hspace{0.17em}}f\left(a\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(b\right).\text{\hspace{0.17em}}$ We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ lies above the $\text{\hspace{0.17em}}x\text{-}$ axis and another point at $\text{\hspace{0.17em}}x=b\text{\hspace{0.17em}}$ lies below the $\text{\hspace{0.17em}}x\text{-}$ axis, there must exist a third point between $\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=b\text{\hspace{0.17em}}$ where the graph crosses the $\text{\hspace{0.17em}}x\text{-}$ axis. Call this point This means that we are assured there is a solution $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ where $f\left(c\right)=0.$

find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what By By By By Edward Biton  By   By  By By