Graphing Polynomial Functions Graphing Parabolas End-Behavior Definitions and Theorems Vertical and Horizontal Asymptotes Des Cartes’ Rule of Signs copyright

copyright © 2013 by Lynda Aguirre 1

Graphing Polynomial Functions

Graphing ParabolasEnd-Behavior

Definitions and TheoremsVertical and Horizontal Asymptotes

Des Cartes’ Rule of Signs


𝑓 (𝑥 )=𝑥2 𝑓 (𝑥 )=−𝑥2 𝑓 (𝑥 )=𝑥2+1

𝑓 (𝑥 )=−𝑥2+1

Graphing ParabolasUsing a graphing calculator, enter the following functions and look at the graphs and tables.

Determine what part of each equation is creating each change.

𝑓 (𝑥 )=𝑥2−2 𝑓 (𝑥 )=−𝑥2+3

𝑓 (𝑥 )=(𝑥−1)2 𝑓 (𝑥 )=−(𝑥+2)2 𝑓 (𝑥 )=−(𝑥+2)2

𝑓 (𝑥 )=(𝑥−1)2−3 𝑓 (𝑥 )=−(𝑥−1)2+4

𝑓 (𝑥 )=2𝑥2𝑓 (𝑥 )=−𝑥2 𝑓 (𝑥 )=3 𝑥2

𝑓 (𝑥 )=12𝑥2

𝑓 (𝑥 )=−2 𝑥2−1 𝑓 (𝑥 )=−2(𝑥+4)2−1


End Behavior

p p

p1

p

p1

p1

p

p1

2) Describe what happens on the far left and far right hand side of each graph 3) How does changing the sign of the first term affect each graph?

1) Use a graphing calculator to draw each function on graph paper

4

End Behavior

Describes whether the left and right ends of the polynomial are going up or down. If the first

term in the polynomial has a power of: -Even degree: Both ends go up -Odd degree: Left end goes down, right end goes up

copyright © 2013 by Lynda Aguirre

*Note: If the first term in the polynomial is negative, these directions reverse.

(up becomes down and vice versa)


DefinitionsFactor Theorem:

If you plug in the value of x or use synthetic division and get a remainder of zero, then “c” is a zero (x-intercept) and the factor

is (x – c) and vice versa..

𝑝 (𝑥 )=𝑥3+𝑥2−16 𝑥+20

𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 h h𝑤 𝑒𝑡 𝑒𝑟 𝑝 (3 )𝑖𝑠 𝑎 𝑧𝑒𝑟𝑜 𝑓𝑜𝑟

𝑝 (3 )=(3)3+(3)2−16 (3)+20

𝑝 (3 )=27+9−48+20𝑝 (3 )=8

Since , then 3 is not a zero


DefinitionsRemainder Theorem: p(c) = remainder

𝑝 (𝑥 )=𝑥3+𝑥2−16 𝑥+20

0

The remainder = zero, and 2 is a factor of p(x)

Use the remainder theorem to find p(2)

2211−1620

1 36−10−200

See notes on how to

do synthetic division

or use synthetic division to find the remainder which will be equal to p(c).

You can find p(c) either by plugging in “c” for all the x’s in the polynomial (as we did in the previous slide)


DefinitionsFundamental Theorem of Algebra (existence thm):

The number of zeros (and factors) of a polynomial is equal to the degree of the polynomial

Degree of a polynomial = number of zeros

𝑝 (𝑥 )=2𝑥4−3 𝑥3− 𝑥2+5

The highest power is the degree

This polynomial has a degree of 4, so it has 4 zeros

In this context, “Zeros” are the x-intercepts


DefinitionsIntermediate Value theorem: Plug in two x-values, if the answers have different signs, then the function (graph) crossed the x-axis

somewhere in between those x’s.

𝑝 (𝑥 )=𝑥3+𝑥2−16 𝑥+20We think there is a zero at x= -5, so we are going to test it by plugging in

values on either side of it to see if they have opposite signs.

𝑝 (−6 )=−64

−611−1620

1−530

14−84−64

−6

𝑝 (−4 )=+36

−411−1620

1 3

12

−416

+36

−4

At x=-6, the value is negative(under the x-axis) and at x= -4, it is positive (above the x-axis)

So this means that the function crosses the x-axis somewhere in between the two numbers (maybe at x=-5, like we thought)


Definitions

Rational Zeros (or Location Theorem): Find b/c (all possible factors of p(x)), test them with synthetic division to find those

that have a remainder of zero-these are the zeros.

Upper and Lower Bounds Theorem: All zeros lie in-between these two extremes (x-axis)- Part of “Location Theorem” Notes

Bisection Theorem : Second part of location theorem, used to find factors of p(x) that are in between integers

(i.e. usually irrational numbers)

See notes on this on these procedures on the greenebox website


Definitions

Linear Factorization Thm: P(x) of degree n has n factors (x – c) Irreducible quadratic factors

(i.e. when factors won’t reduce without producing imaginary numbers)

𝑓 (𝑥 )=𝑥2+2

This equation can’t be factored into real number factors, it will produce imaginary zeros which we are not covering in this course


Definitions

Complex Conjugates: Irreducible quadratic factors produce imaginary numbers in pairs

(because in the last step where you take the square root of both sides and put a sign in front of it producing two answers)

Imaginary zeros are not covered in this course


Examples of polynomial functions


1) f(x) = x (x+2)2 or f(x) = x3 + 4x2 + 4x

Where are the x and y-intercepts?

What is the end-behavior?

The figure cuts through the x-axis at one zero and bounces off the x-axis for other one. Why?


2) f(x) = x2 - 2

The zeros are located at x2 – 2 = 0 or


3) f(x) = (x + 4)2(x2+ 1)

need to readjust window settings

Notice the bounce off of the zero at x= -4 (caused by the 2nd power) and the bounce in midair caused by the imaginary zeros from

the (x2+1) term


Attributes: MultiplicityIf the factor has a 1st power, (x – c), the function passes

straight through the x-axis at that zero (c).

If the factor has a 2nd power, , the function behaves like a parabola at that zero.

If the factor has a 3rd power, , the function behaves like a cubic function at that zero.


Polynomial Graphs-Rational Functions (Asymptotes)

𝑓 (𝑥 )= 𝑎𝑥−h

+𝑘

Vertical Asymptotes exist where the denominator of a fraction is equal to zeroProcess: 1) Set “bottom” equal to zero 2) Solve for x 3) Draw a vertical line at that x-value.


Vertical Asymptote

𝑓 (𝑥 )= 1𝑥−2

3) Draw a vertical line

at that x-value

1) Set “bottom” equal to zero

2) Solve for x

𝑥=2

+2+2𝑥−2=0

VerticalAsymptote

12345

Vertical Asymptotes exist where the denominator of a fraction is equal to zero


+𝑘


+𝑘


Horizontal Asymptotes

𝑎 ¿ 𝑓 (𝑥 )= 𝑥2

𝑥5

Process: Look at the powers of the first terms on the top and bottom

b) If n = m c) If n > m a) If n < m

𝑓 (𝑥 )=3 𝑥3

𝑥3 𝑓 (𝑥 )=𝑥5

𝑥3

horizontal asymptote: horizontal asymptote:

no horizontal asymptotes𝑦=

31

y=3

𝑦=0 (i.e. the x-axis)

12345 5


+𝑘


Polynomial Graphs-Rational Functions (Asymptotes)


+𝑘

Other information not included in this course:Oblique Asymptotes: Do polynomial long division and

graph the quotient (the answer). It will be a line


DesCartes’ Rule of Signs

Use: To determine how many real zeros (x-intercepts) exist in a function and how many are on each side

(positive or negative) of the origin.

Note: If some are imaginary zeros (they occur in pairs), this takes away 2 zeros at a time.


DesCartes’ Rule of SignsSteps: Find the positive zeros (to the right of the y-axis)1) Plug in (x) to the function p(x); 2) Count sign changes--# of positive zeros (going down by 2)

Find the negative zeros (to the left of the y-axis)3) Plug in (-x) to the function p(x); 4) count sign changes-# of negative zeros (going down by 2)



•Plugging in (x) gives us the original equation and count sign changes from left to right

Number of positive real zeros (to the right of y-axis): 3 or 1

Red arrows indicate sign changes

(+ to – and vice versa)

Find the positive zeros1) Plug in (x) to the function p(x)

2) Count sign changes--# of positive zeros (going down by 2)



Number of negative real zeros (to the left of y-axis): 2 or 0

Red arrows indicate sign changes

(+ to – and vice versa)

Find the negative zerosPlug in (-x) into the original equation and count sign changes from left to right

2) Count sign changes--# of negative zeros (going down by 2)

𝑷 (− 𝒙 )=𝟒(−𝒙)𝟓−𝟐 (−𝒙 )𝟒− (−𝒙 )𝟑+𝟗 (− 𝒙 )𝟐−𝟓

𝑷 (− 𝒙 )=−𝟒 𝒙𝟓−𝟐𝒙𝟒+𝒙𝟑+𝟗𝒙𝟐−𝟓

Figure out the new signs: positive power: negative cancels

negative power: negative remains

Documents

Graphing Polynomial Functions Graphing Parabolas End-Behavior Definitions and Theorems Vertical and Horizontal Asymptotes Des Cartes’ Rule of Signs copyright