Module 4 Lecture Notes - People

Module 4 Lecture Notes

MAC1105

Fall 2019

4 Quadratic Functions

4.1 Factor Trinomials

Rules for Positive Exponents

For all positive integers m and n and all real numbers a and b:

Product Rule

anam =

Power Rules

(an)m =

(ab)nm =

⇣ab

⌘n= b 6= 0

Zero Exponent

a0 =

Definition

An expression of the form akxk where k � 0 is an integer, ak is a constant, and x is a variable,

is called a . The constant ak is called the and k is

the of the monomial if k 6= 0.

ntma

n ma

grimbumna

b

7

MONOMIAL COEFFICIENTDEGREE

Note 1. The sum of monomials with di↵erent degrees forms a . The

monomials in the polynomial are called the . A polynomial with exactly

two terms is called a and a polynomial with exactly 3 terms is called a

.

Polynomial in One Variable in Standard Form:

anxn + an�1x

n�1 + ...+ a1x+ a0

where a0, ..., an�1, an are real numbers and n � 0 is an integer.

Definition

A is a polynomial function of degree 2.

Note 2. x2 + 5x + 2 is a quadratic function, but x + 2 is not a quadratic function because

.

Operations on Polynomials

Adding and Subtracting Polynomials

Polynomials are added and subtracted by combining like terms.

Multiplying Polynomials

Two polynomials are multiplied by using the properties of real numbers and the rules for exponents.

2

POLYNOMIAL

TERMS

BINOMIAL

TRINOMIAL

Ex 5 5 7 4t 17 31 12 2 t x t 2921 17 lx't 12X

QUADRATIC FUNCTION

THE DEGREE OF Xt2 IS 1 NOT 2

Example 1. Perform the operation:

(2x4 � 3x2 + 1)(4x� 1)

Note 3. When multiplying two binomials, use .

Definition

The greatest common factor (GCF) of a polynomial is the

that divides evenly into the polynomials.

3

or

man

U2 4114 3 2 4 71 3 2 C 1 1 1 4x 1 t

8 4 2 4 12 2 3 2 4 1

8 5 2 4 123 3 2 4 1

FOIL OR THEBOXMETHODFIRST OUTER INNER LAST

EXAMPLE Xt 2 X Z

X Xt 2 X 2 X 2

X 2 2 4 1 244X Z

X F 2xX 2 1 2 1 L 4

Z 2x 4 XZ 4

LARGESTPOLYNOMIAL

EX THE GLF OF 16X AND 20 2 IS 4xTHE GCE OF 2 2 4 22 IS 2

How to Factor out the Greatest Common Factor

1. Identify the GCF of the .

2. Identify the GCF of the .

3. Combine 1 and 2 to find the GCF of the expression.

4. Determine what the GCF needs to be multiplied by to obtain each term in the polynomial.

5. Write the factored polynomial as the product of the GCF and the sum of the terms we need

to multiply by.

Example 2. Factor 6x3y3 + 45x2y2 + 21xy by factoring out the greatest common factor.

Factor a Trinomaial with Leading Coe�cient 1

A trinomial of the form x2 + bx + c can be factored as (x + p)(x + q), where pq = and

p+ q = .

Note 4. Not every polynomial can be factored. Some polynomials cannot be factored, in which

case we say the polynomial is prime.

4

COEFFICIENTS

VARIABLES

1 GCFOF THE COEFFICIENTS 6 45,21 Is 3

2 GCF OF THE VARIABLES X3y3 X yZ XY IS Xy

3 THE GCF OF THE EXPRESSION IS 3 Xy 3Xy

4 FACTOROUTTHE GCF 3xy 2 242 t 15Xy t 7

C

b

How to Factor a Trinomial of the Form x2 + bx+ c

1. Determine all possible factors of c.

2. Using the list found in 1, find two factors p and q, in which pq = and p+ q = .

3. Write the factored expression as .

Note 5. The order in which you write the factored polynomial does not matter. This is because

multiplication is .

Example 3. Factor the trinomial:

x2 + 24x+ 140

5

C b

x p Xt g

COMMUTATIVE

1ALL POSSIBLEFACTORSOF 140 140 I70 235 428 520 714 10

2 FINDTWOFACTORS p AND 4 FOUND IN 1 WHERE ptg 24

141 10 24

3 Xt 14 Xt10

CANCHECKOURANSWERBYFOILXt14 Xt10 XZ t 10 14 1140

XZ t 24 1 140 V

Factor a Trinomial by Grouping

To factor a trinomial in the form ax2 + bx+ c by grouping, we find two numbers with a product of

and a sum of .

How to Factor a Trinomial of the Form ax2 + bx+ c by Grouping

1. Determine all possible factors of .

2. Using the list found in 1, find two factors p and q, in which pq = and p+ q = .

3. Rewrite the original polynomial as .

4. Pull out the GCF of .

5. Pull out the GCF of .

6. Factor out the GCF of the expression.

6

ac b

4C

ac b

ax7pxtqxt C

ax't px

Ext C

Example 4. Factor the polynomial:

35x2 + 48x+ 16


21x2 + 40x+ 16


36x2 + 19x� 6

7

FINDTHE FACTORSOF 35 16 560 WHOSE SUM IS 48560 I 20 28 48280 2140 4 REWRITEAS 35 2 t 20 1 28 16

o fiTIIYITiEiiEiiG70 8JG 10 l I40 14 OFEXPRESSION3516 7 1 4 5 42028J

FINDTHEFACTORSOF 21 16 336 WHOSE sun IS 40336 I n28t12 40168 2 REWRITE AS 21 728 12 16112 3

8,4 4gtkfjtc.ee P8EfGcF

487 7 3 4 t 413 4428 GCF EXPRESSION28012

FINDFACTORSOF 36316 216WHOSE SUN IS19216 I 27 8216 I zy

27 8 19REWRITEAS 365 27 8 6

108 2108 Z IPU d pullOUT OUTGLE72 3 GCF

72 3 9 4 3 214 354 4 GCFOFEXPRESSION54 436 636 6

Factor a Perfect Square Trinomial

A perfect square trinomial can be written as the square of a binomial:

=

or

=

How to Factor a Perfect Square Trinomial

1. Confirm that the first and last term are perfect squares.

2. Confirm that the middle term is the product of .

3. Write the factored form as .


100x2 + 60x+ 9

8

a Zab tb2 atb

92 Zab b ca b2

TWICE ab

at b 2

FIRSTANDLASTTERM ARE PERFECTSQUARES 100 2 110 32 AND 9 32

Q IS THE MIDDLETERM GX TWICETHE PRODUCTOF 10 313

2110 1 33 20 13GOXYES

PERFECTSQUARETRINOMIAL100 2 60 9 10 5 240 113 132 1110 1372 1

Factor a Di↵erence of Squares

A di↵erence of squares is a perfect square subtracted from a perfect square. We can factor a

di↵erence of squares by: = .

How to Factor a Di↵erence of Squares

1. Confirm that the first and last term are perfect squares.

2. Write the factored form as .


49x2 � 16

9

al b atb G b

at b a b

w wI I

17 5 42

44 2 16 7 12 42

1FFERENCEOFSQUARES

Ra2tb2 atb7Ca

yGOESNOT EQUAL

4.2 Graphing Quadratic Functions

Definitions:

The graph of a quadratic function is a U-shaped curve called a . The

extreme point of a parabola is called the . If the parabola opens up, the vertex

represents the lowest point on the graph, called the of

the quadratic function. If the parabola opens down, the vertex represents the highest point on the

graph, called the . The graph of a quadratic function

is symmetric, with a vertical line drawn through the vertex called the

. The - are the points where the parabola

crosses the x-axis.

10

PARABOLA

VERTEX

MINIMUM VALUE

MAXIMUM VALUE

AXIS OF

SYMMETRY X INTERCEPTS

1 PARABOLA1 OPENSUP

I

l

l

l

l

1 AXISOFSYMMETRYX 3

gI NOX INTERCEPTSHERE

I ERTEX1 AT 3,1

Parabolas and Quadratic Functions

General Form of a Quadratic Function: The general form of a quadratic function is

, where a, b, c are real numbers and a 6= 0. If , the parabola

opens up. If , the parabola opens down.

Standard Form of a Quadratic Function: The standard form of a quadratic function is

, where a 6= 0. If , the parabola opens up. If ,

the parabola opens down.

Axis of Symmetry: The equation to find the axis of symmetry is given by

x = .

Vertex: The vertex is located at (h, k), where

h = , and k = f(h) = f⇣ ⌘

How to Find the Equation of a Quadratic Function Given its Graph

1. Identify the coordinates of the vertex, (h, k).

2. Substitute the values of h and k (found in 1) into the equation f(x) = a(x� h)2 + k.

3. Substitute the values of any other point on the parabola (other than the vertex) for x and

f(x).

4. Solve for the stretch factor, |a|.

5. Determine if a is positive or negative.

11

4 2 1 Xtc a O

GLO

fCx alX h K a O GLO

b OR x hza

b b24 29

6. Expand and simplify to write in general form.

Example 9. Write the equation of the graph below in the form ax2 + bx + c, assuming a = 1 or

a = �1:

Example 10. Write the equation of the graph below in the form ax2 + bx+ c, assuming a = 1 or

a = �1:

12

PARABOLAOPENS f xj aLX h 2tKUP SO a 701417 7 fix 7 X C lift to

f X Xt 2 5

f X Xt1 XH 5fiiii.IEis

F E s t12 5

MfT fks aLx h5tk

PARABOLA OPENS f X l X 4 12 3DOWN

a 20 f G 1 132 3I f 47 1 131 131 3

f 4 X2tXtXt1 1 3f X 421 2 111 3f x XZ 2 11 3ff X2_2Xt

Example 11. Graph the equation f(x) = (x� 3)2 � 19

13

CAN USEDESMOS

THIS IS THE FORM f X a X h t Kh 3 AND K 14 SO VERTEX IS Ch K 3 140 1 so a 0 ANDTHUS THE PARABOLAOPENSUP

143 149VERTEX

AXIS OFSYMMETRY IS X h so 3

4.3 Solving Quadratics by Factoring

How to Find the x-Intercept and y-Intercept of a Quadratic Function:

1. To find the y-intercept, evaluate the function at .

2. To find the x-intercepts, solve the quadratic equation .

Note 6. Solving the quadratic equation f(x) = 0 can be done by factoring, or by using the

quadratic formula. First, we will solve quadratic equations by factoring. To solve f(x) = 0, we will

factor f(x) and set each factor equal to 0.

Example 12. Solve the quadratic equation by factoring:

15x2 + 9x� 6 = 0

14

X O

f X O

1 FACTOR 15 2 9 61511 6 90

90 I 7115 6 90AND 15 2 9 6 0

go I 15 6 4 15 2 15 6 6 045 2

15 1 1 61 13 0

30 3 Xt1 15 6 0

18 5 SET EACHFACTOREQUAL TO 018 5

1 15 6 0

1 0 15 6 015 G1511 6

EETTHESE ARE THESOLUTIONSTOf X 0 ie THEY ARE THE

ZEROSANDY INTERCEPTS


4x2 + 12x+ 9 = 0


350x2 + 30x� 8 = 0

15

1 FACTOR4 2 12 94 2 12 72 AND 4 32 SOTHIS MIGHT BE A PERFECTSQUARETRINOMIA2 2X 33 14 113 12X V

4 2 12 9 2 32 212 313 132 2 33 0

So 4 412 9 0 2x 3J12 332 0 x2 3 2 3

1 FACTOR350 2 30 8350 L 8 2800

FINDTWOFACTORSOF 2800WHOSESUM IS 30 70 AND 40WORK70 C403 2800 AND 7 Otc 403 30

350 2 30 8 0 TEACH FACTOR EQUAL

350 770 40 8 0 TO 0

JSE5 1 0 70 8 070 5 1 8 5 1 0 5 I 7011 8

5 1 70 8 0 X I X 870

xElTHESE ARETHE SOLUTIONSIZEROSI

X INTERCEPTS

4.4 Solving Quadratics using the Quadratic Formula

The Quadratic Formula:

To solve the quadratic function f(x) = 0, we can use the quadratic formula which is given by:

x =

Note 7. Recall that to find the x-intercepts of a quadratic function, we solve the quadratic equation

f(x) = 0. So, to find the x-intercepts, we can solve by factoring, or we can solve using the quadratic

formula. The quadratic formula will always work, but sometimes it is much more tedious to use.

Example 15. Solve the quadratic equation using the quadratic formula:

4x2 � 8x� 8 = 0

Example 16. Solve the quadratic equation using the quadratic formula:

2x2 � 8x+ 7 = 0

16

f x ax't b Xt C

b I THE24

x ltFs x t

4 b ooh 8 f

F

of 818

11531x btf C8 I Tt84hl7 8 I 76456242 2 4

Too 8 I 54.24 4

8 I 22T4

21,2 121127

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Module 4 Lecture Notes - People