Quadratic Functions. Expanding to Standard Form A quadratic function is a function that can be written in the standard form below and where quadratic.

UNIT 6Quadratic Functions

QUADRATICS: DAY 1

Expanding to Standard Form

A quadratic function is a function that can

be written in the standard form below and where

2( )f x ax bx c

0a

quadraticterm

linearterm

constantterm

If a = 0, then the function has no quadratic term & it is not a quadratic function.

RECALL:

F.O.I.L.First,Outer,Inner,Last

EXAMPLE:

Simplify

a.) 7253 xx

7253 xx

First,Outer,Inner,Last26x x21 x10 35

26x x11 35

Combine Like

Terms!!

CONTINUED…b.) 5243 xx

5243 xx

First,Outer,Inner,Last26x x15 x8 20

26x x23 20

Combine Like

Terms!!

EXAMPLE: Expand to standard form & determine whether

each function is linear or quadratic.

a.) 2 3 4y x x b.) 2 2( ) 3 2 3 2f x x x x

CONTINUED…c.) 2 2( ) 5f x x x x

d.) ( ) 3f x x x

e.) ( ) 5 3 1f x x x

HOMEWORK

Worksheet: Expanding to Standard Form

QUADRATICS: DAY 2

Quadratic Graphs

QUADRATIC GRAPHS The graph of a quadratic function is a parabola.

The axis of symmetry is the line that divides a parabola into two parts that are mirror images.

axis of symmetry

CONTINUED… The vertex of a parabola is the point at which

the parabola intersects the axis of symmetry.

The y-value of the vertex represents the maximum or minimum value of the function.

vertex

THE X2 TERM If the x2 term is positive, the parabola is

concave up & the vertex is the minimum point

If the x2 term is negative, the parabola is concave down & the vertex is the maximum point

EXAMPLE: Identify the vertex, tell whether it’s a minimum or a

maximum, identify the axis of symmetry, & describe the graph’s concavity.

a.)

Vertex: )2,1( Minimum

b.)

Vertex: )3,4(Maximum

Axis of Symmetry: x=1

Concave Up

Axis of Symmetry: x=-4

Concave Down

x y=1/2x2 y

EXAMPLE: Make a table of values & graph the function

Draw in your axis of symmetry Reflect the points across the axis of symmetry & draw line

a.) 2

2

1xy

x y=1/2x2 y

0

1

2

x y=1/2x2 y

0 y=1/2(0)2

1

2

x y=1/2x2 y

0 y=1/2(0)2 0

1

2

x y=1/2x2 y

0 y=1/2(0)2 0

1 y=1/2(1)2

2

x y=1/2x2 y

0 y=1/2(0)2 0

1 y=1/2(1)2 1/2

2

x y=1/2x2 y

0 y=1/2(0)2 0

1 y=1/2(1)2 1/2

2 y=1/2(2)2

x y=1/2x2 y

0 y=1/2(0)2 0

1 y=1/2(1)2 1/2

2 y=1/2(2)2 2

x y=2x2 y

EXAMPLE:

b.) 22xy

x y=2x2 y

0

1

2

x y=2x2 y

0 y=2(0)2

1

2

x y=2x2 y

0 y=2(0)2 0

1

2

x y=2x2 y

0 y=2(0)2 0

1 y=2(1)2

2

x y=2x2 y

0 y=2(0)2 0

1 y=2(1)2 2

2

x y=2x2 y

0 y=2(0)2 0

1 y=2(1)2 2

2 y=2(2)2

x y=2x2 y

0 y=2(0)2 0

1 y=2(1)2 2

2 y=2(2)2 8

Make a table of values & graph the function

Draw in your axis of symmetry Reflect the points across the axis of symmetry & draw line

THE “C” TERM

caxy 2

+ up- down

narrow/wide

translates(shifts)

graph up/down

x y=x2-4 y

EXAMPLE: Graph the function

42 xyx y=x2-4 y

0

1

2

x y=x2-4 y

0 y=(0)2-4

1

2

x y=x2-4 y

0 y=(0)2-4 -4

1

2

x y=x2-4 y

0 y=(0)2-4 -4

1 y=(1)2-4

2

x y=x2-4 y

0 y=(0)2-4 -4

1 y=(1)2-4 -3

2

x y=x2-4 y

0 y=(0)2-4 -4

1 y=(1)2-4 -3

2 y=(2)2-4

x y=x2-4 y

0 y=(0)2-4 -4

1 y=(1)2-4 -3

2 y=(2)2-4 0

HOMEWORK:

Worksheet: Quadratic Graphs

QUADRATICS: DAY 3

Graph y = ax2+bx+c

Axis of Symmetry

This will also give you the x-coordinate of the vertex

2

bx

a

FORMULA


563 2 xxya.)

1.) Find the axis of symmetry/vertex

a

b

2

)3(2

6

6

6

1

Plug in x-coordinate to get y-coordinate

5)1(6)1(3 2 y563 y

8y

Vertex = (1,8)

x

CONTINUED…2.) Find the y-intercept

Plug in 0 for x to find y-intercept

5)0(6)0(3 2 y

5y

y-intercept = (0,5)

563 2 xxy

CONTINUED…3.) Find a 3rd point that is on the same side of the axis of symmetry as the y-intercept

Could plug in for x: -1, -2, -3, etc

Let x = -1

5)1(6)1(3 2 y

563 2 xxy

563 y

4y

3rd Point:(-1,-4)

CONTINUED…4.) Reflect points across axis of symmetry & draw graph


962 xxyb.)

1.) Find the axis of symmetry/vertex

a

b

2

)1(2

6

2

63

Plug in x-coordinate to get y-coordinate

9)3(6)3( 2 y9189 y

0y

Vertex = (3,0)

x

CONTINUED…

2.) Find the y-intercept

Plug in 0 for x to find y-intercept

9)0(602 y

9y

y-intercept = (0,9)

962 xxy

CONTINUED…

3.) Find a 3rd point that is on the same side of the axis of symmetry as the y-intercept

Could plug in for x: 2, 1, -1, etc

Let x = 1

9)1(6)1( 2 y

962 xxy

961 y

4y

3rd Point:(1, 4)

CONTINUED…4.) Reflect points across axis of symmetry & draw graph

EXAMPLE: Graph

2 2 5y x x c.)

1. Axis of Sym.:

a

bx

2

12

2 2

21

y 5121 2 521 6

Vertex = (-1, -6)

2. y-intercept: 5020 2 y 5

y-intercept = (0, -5)

3. 3rd Point:Let x = 1

5121 2 y 521 2

3rd Point = (1, -2)

EXAMPLE: Graph

2 4 4f x x x d.)

1. Axis of Symm:

a

bx

2

12

4

2

4

y 4242 2 484 0

Vertex = (2, 0)

2. y-intercept: 4040 2 y 4

y-intercept = (0, -4)

3. 3rd Point:Let x = 1

4141 2 xf 441 1

3rd Point = (1, -1)

2

QUADRATICS: DAY 4

Solving Quadratics Equations

SOLVING QUADRATIC EQUATIONS Recall: Standard Form of Quadratic Equation

To solve a quadratic equation, we want to know when y = 0; therefore, we set the equation equal to 0

Imagine the graph of a quadratic equation.

What happens when y = 0?

cbxaxy 2

cbxax 20 OR 02 cbxax

x-intercepts!

The solutions to a quadratic equation are the x-intercepts!!

x y

EXAMPLE: Solve by graphing

042 xa.)

Treat it as if it’s y=x2-4

x Y

0

1

2

x y

0 -4

1

2

x y

0 -4

1 -3

2

x y

0 -4

1 -3

2 0

y = 0 at the x-intercepts, which are:

-2 & 2

Therefore, the solutions are:

-2 & 2

x y


092 xb.)x Y

0

1

2

x y

0 -9

1

2

x y

0 -9

1 -8

2

x y

0 -9

1 -8

2 -5

The graph doesn’t cross the x-axis.But, will it eventually?

The solutions are: -3 & 3

Add more #’s to your table!

x y

0 -9

1 -8

2 -5

3

x y

0 -9

1 -8

2 -5

3 0

x y


042 xc.)

x Y

0

1

2

x y

0 4

1

2

x y

0 4

1 5

2

x y

0 4

1 5

2 8

Therefore, the solution is:

No solution

The graph doesn’t cross the x-axis.

But, will it eventually?

SOLVE USING SQUARE ROOTS

Some equations can be solved by simply solving for the variable

EXAMPLE: Solve using square roots

a.) 0982 2 x Solve for x!

98 9822x 98

2 22x 49 To un-do squaring of x, square

root!492 x

x 7

(must get x2 alone first)

Be sure to include both solutions!

EXAMPLE:

b.) 12123 2 n12 12

23n 03 3

2n 0

02 n

n 0

c.) 22 32 0g 32 32

22g 322 2

2g 162 16g

4g

FACTORED FORM Expand to standard form

2 5y x x

2 5 2 10y x x x 2 3 10y x x

An equation in this form expands to a quadratic equation.

Therefore, this is a quadratic equation in factored form

ZERO-PRODUCT PROPERTY

For every real number, a & b,

a = 0 or b = 0if ab = 0, then

Ex:If (x+3)(x+2) = 0, thenx+3 = 0 or x+2 = 0

EXAMPLE: Solve

a.) 0625 xx

05 x OR 062 x5 5

5x6 6

62 x2 2

3x5x OR

b.) 032 xx

02 x OR 03 x2 2

0x

3 3

3xOR

HOMEWORK

Worksheet: Solving Quadratic Equations

QUADRATICS: DAY 5

Factoring x2+bx+c

FACTORING X2+BX+C

Since 3 x 5 = 15, 3 & 5 are factors of 15

Since ,

& are factors of

15853 2 xxxx

3x 5x 1582 xx

*What do you notice about 3 & 5?

EXAMPLE:

Factora.) 1272 xx

Find factors of +12 that add up to +7

12

1 & 122 & 63 & 4

1+12 = 132+6 = 83+4 = 7

1272 xx (x+ )(x+ )

3 4

CONTINUED…

b.) 30132 aaFind factors of +30 that add up to +13

30

1 & 30 1+30 = 31

2 & 15 2+15 = 17

3 & 10 3+10 = 13

30132 aa (a+ )(a+ )

3 10

CONTINUED…

c.) 42172 ddFind factors of +42 that add up to -17

42

-2 & -21 -2+(-21) = -23

-3 & -14 -3+(-14 )= -17

42172 dd

(d- )(d- )

3 14

Both factors will have to be negative!

CONTINUED…

d.) 18112 xxFind factors of +18 that add up to -11

18

-1 & -18 -1+(-18) = -19

-2 & -9 -2+(-9) = -11

18112 xx

(x- )(x- )2 9

Both factors will have to be negative!

CONTINUED…

e.) 2762 mm

2762 mm(m+ )(m- )

39

-27-1 & 27 -1+27 = 26

-3 & 9 -3+9 = 6

Find factors of -27 that add up to +6

One will have to be positive and one has to be negative. The larger factor will need to be

positive!

CONTINUED…

f.) 1832 pp

-181 & -18 1+(-18) = -172 & -9

Find factors of -18 that add up to -3

2+(-9) = -73 & -6 3+(-6) = -3

1832 pp

(p+ )(p- )

3 6

One will have to be positive and one has to be negative. The larger factor will need to be

negative!

CONTINUED…

g.) 2082 mmFind factors of -20 that subtract to get +8

-20-1 & 20 -1+20 = 19-2 & 10 -2+10 = 8

2082 mm(m+ )(m- )

10 2

One has to be positive and one has to be

negative. The larger factor will need to be

positive!

CONTINUED…

h.) 562 yyFind factors of -56 that add up to -1

-561 & -56 1+(-56) = -552 & -28 2+(-28) = -264 & -14 4+(-14) = -107 & -8 7+(-8) = -1

562 yy

(y+ )(y- )

87

One has to be positive and one has to be

negative. The larger factor will need to be

negative!

HOMEWORK

Worksheet: Factoring Day 1 x2+bx+c

QUADRATICS: DAY 6

Factoring ax2+bx+c

EXAMPLE: Factor

a.) 7236 2 nn1.) Multiply the 1st & last

terms, find factors of that # that

combine to get middle term

76 42 Factors of 42:

1 & 42 1 + 42 = 43

2 & 21 2 + 21 = 23

2.) Split the middle term

72126 2 nnn3.) Factor out the GCF from the 1st two terms, then the last two terms

n2 ( n3 )1 7 ( n3 )1

72 n 13 n

CONTINUED…b.) 8267 2 xx 87 56 Factors of -56:

1 & -56 1 + (-56) = -552 & -28 2 + (-28) = -26

27x x2 x28 8

x ( x7 )2 4 ( x7 )2

4x 27 x

1.) Multiply the 1st & last terms,

find factors of that # that combine to get middle term


3.) Factor out the GCF from the 1st two terms, then the last two terms

CONTINUED…c.) 252 2 yy 22 4 Factors of 4:

-2 & -2 -2 + (-2) = -4-1 & -4 -1+ (-4) = -5

22y y y4 2

y ( y2 )1 2 ( y2 )1

2y 12 y





CONTINUED…d.) 32 2 nn 32 6 Factors of -6:

-1 & 6 -1+ 6 = 5-2 & 3 -2+ 3 = 1

22n n2 n3 3

n2 ( n )1 3 (n )1

32 n 1n





CONTINUED…e.) 358020 2 xx



74 28 Factors of 28:

1 & 28 1+ 28 = 29

2 & 14 2+ 14 = 163.) Split the middle term

24x x2 x14 74.) Factor out the GCF from the 1st two terms, then the last two terms

x2 ( x2 )1 7 ( x2 )1 72 x 12 x

1.) First, factor out the GCF

5 [ 24x x16 ]7

Now, just factor what’s left in

brackets & just carry down the 5

5

CONTINUED…f.) 10122 2 vv



51 5 Factors of 5:

-1& -5 -1+ -5 = -6


2v v1 v5 54.) Factor out the GCF from the 1st two terms, then the last two terms v (v )1 5 (v )1

5v 1v

1.) First, factor out the GCF

2 [ 2v v6 ]5

Now, just factor what’s left in

brackets & just carry down the 2

2

HOMEWORK

Worksheet: Factoring Day 2 ax2+bx+c

QUADRATICS: DAY 7

Factoring a Difference of Two Squares

DIFFERENCE OF TWO SQUARES

22 ba baba

Ex: 812x 99 xx

4916 2x 7474 xx

EXAMPLE:

Factor

a.) 642x 88 xx

b.) 1002m 1010 mm

c.) 1214 2x 112112 xx

d.) 6425 2x 8585 xx

e.) 4010 2x 410 2x 2210 xx

f.) 753 2c 253 2c 553 cc

HOMEWORK

Worksheet: Factoring Day 3 All types

QUADRATICS: DAY 8

Solving Quadratics by Factoring

EXAMPLE: Solve using factoringa.) 04882 xx

(x )(x ) = 0 12 4012 x OR 04 x

12 1212x

4 44xOR

b.) 8852 2 xx8888

08852 2 xx22x x16 x11 88 0

(2x x )8 11 x( 0)8 112 x 08 x

0112 x 08 x11 11

112 x2 2

2

11x OR

8 8

8x

CONTINUED…

a.) 2 24 20 10 3 4x x x Must get into standard form first!23x23x

2 20 10 4x x 10x10x

2 10 20 4x x 44

2 10 24 0x x

6 4 0x x

6 4 0x x

6 0x OR 4 0x 6 6

6x OR

4 44x

HOMEWORK

Worksheet: Solving Quadratics by Factoring

QUADRATICS: DAY 9

Simplifying Square Roots

RECALL: FACTOR TREES Square Roots

48 48

2 24

2 12

2 6

2 3

2 2 3 342

We’re taking thesquare root, therefore,the index is 2. We justdon’t write it.

PROPERTY:

Multiplication Property of Square Roots

baab

Ex: 43 43

EXAMPLE: Simplifya.) 54

Step 1: List factors of 54, find pair with largest perfect square

54

2 & 273 & 18

6 & 9

69 Step 2: Break up the square root into itstwo factors (write perfect square first)

Step 3: Simplify

63 63

b.) 50

50

2 & 255 & 10

225

25

CONTINUED…

d.) 184 18418

2 & 93 & 6

4

93

12

250

c.) 5005 5005500

2 & 2505 & 10025 & 20

5 100 5

5 10 5

5

4

2

2

HOMEWORK:

Worksheet: Simplifying Radicals

QUADRATICS: DAY 10

Using the Quadratic Formula

QUADRATIC FORMULA

If ax2+bx+c = 0, and a 0, then

a

acbbx

2

42

*Equation always has to be in standard form first

EXAMPLE: Solvea.) 0253 2 xx Already in standard form

a = -3 b = 5 c = -2

a

acbbx

2

42

32

23455 2

6

24255

6

15

6

15

6

15

OR6

15

6

4

OR6

6

3

2OR 1

EXAMPLE: Solve

b.) xx 562 Put into standard form

x5x50652 xx a = 1 b = -5 c = 6

a

acbbx

2

42

12

61455 2 2

24255

2

152

15 2

15OR

2

15 2

6OR

2

4 3 OR 2

EXAMPLE: Solvec.) 22 6 1 0x x Already in standard form

a = 2 b = 6 c = 1

a

acbbx

2

42

26 6 4 2 1

2 2

6 36 8

4

6 28

4

6 4 7

4

6 2 7

4

3 7

2

EXAMPLE: Solved.) 742 2 xx Put into standard form

770742 2 xx a = 2 b = 4 c = -7

a

acbbx

2

42

22

72444 2 4

56164

4

724 4 36 2

4

4 6 2

4

2 3 2

2

HOMEWORK

Worksheet: Using the Quadratic Formula

QUADRATICS: DAY 11

Completing the Square

You can solve equations in which one side is a perfect square trinomial by taking the square root of each side.

EXAMPLE: Solve

2 10 25 36x x The left side is a perfect square trinomial,so factor!

5 5 36x x

Square root each side 25 36x

25 36x

5 6x

5 6x OR 5 6x 5 5 55

1x OR 11x

EXAMPLE: Solve

2 14 49 81x x The left side is a perfect square trinomial,so factor!

7 7 81x x

Square root each side 27 81x

27 81x

7 9x

7 9x OR 7 9x 7 7 77

16x OR 2x

COMPLETING THE SQUARE

If one side of an equation is not a perfect square trinomial, you can convert it into one by rewriting the

constant term.

This is called completing the square.

Use the following relationship to find the term

that will complete the square.2 2

2

2 2

b bx bx x

EXAMPLE: Find the missing value to complete the squarea.) 2 8 ?x x

2

2

b

28

2

24 16

CONTINUED…

b.) 2 7 ?x x

2

2

b

27

2

49

4

EXAMPLE: Solve by completing the squarea.) 2 12 5 0x x 1.) Find

2

2

b

2

2

b

212

2

26 36

2.) Re-write so all terms containing x are on one

side

2 12 5 0x x 5 5

2 12 5x x 3.) Complete the square by adding 36 to each side 36 36

2 12 36 31x x 4.) Factor

6 6 31x x 2

6 31x 5.) Square root each side

2( 6) 31x 6 31x 6.) Solve for x

6 66 31x

EXAMPLE:

b.) 2 4 4 0x x 1.) Find 2

2

b

2

2

b

24

2

22 4

2.) Re-write so all terms containing x are on one

side

2 4 4 0x x 4 4

2 4 4x x 3.) Complete the square by adding 4 to each side 4 4

2 4 4 8x x 4.) Factor

2 2 8x x 2

2 8x 5.) Square root each side

2( 2) 8x 2 8x 6.) Solve for x

2 22 8x 2 2 2x

EXAMPLE:c.) 22 4 3 0x x

Since there is a value for a that is greater than 1, the order of steps changes!

1.) Re-write so all terms containing x are on one side

22 4 3 0x x 3 3

22 4 3x x 2.) Divide both sides by the

coefficient of x2 term

2 22 2 1.5x x

3.) Find 2

2

b

22

2

1

4.) Complete the square by adding 1 to

each side

1 12 2 1 2.5x x

1 1 2.5x x

5.) Factor 2

1 2.5x

21 2.5x 1 2.5x

7.) Solve for x

1 2.5x

1 1

6.) Square root both sides

HOMEWORK:

Worksheet: Completing the Square

QUADRATICS: DAY 12

Translating Parabolas: Part I

Just as we graphed absolute value functions as translations of their parent function , we can graph a quadratic function as a translation

of the parent function .

y x

2y ax

VERTEX FORMTo translate the graph of a quadratic function,

we’ll use the vertex form of a quadratic function.

2y a x h k

- Right+ Left

+ Up- Down

The vertex is (h, k) & the axis of symmetry is the line x=h

EXAMPLE: Graph 21

2 32

y x

Step 1: Graph the vertex 2,3Step 2: Draw the axis of symmetry 2x

Step 3: Find & graph the y-intercept

210 2 3

2y 21

2 32

14 3

2 2 3 1

0,1

Step 4: Find 3rd point Let x = -2

212 2 3

2y 21

4 32

116 3

2 8 3 5

2, 5

Step 5: Reflect points & draw graph

EXAMPLE: Graph 2

2 1 4y x

Step 1: Graph the vertex 1, 4 Step 2: Draw the axis of symmetry 1x Step 3: Find & graph the y-intercept

22 0 1 4y 2

2 1 4

2 1 4 2 4 2 0, 2

Step 4: Find 3rd pointLet x = 1

22 1 1 4y 2

2 2 4 2 4 4 8 4 4 1,4

Step 5: Reflect points & draw graph

EXAMPLE: Write the equation of the parabola

Vertex: (3, 4) Other point: (5, -4)h, k x, y

Use vertex form 2y a x h k

Substitute for h, k, x, & y

24 5 3 4a

Solve for a 24 2 4a

4 4 4a 44

8 4a 442a

Write equation in vertex form,using values for a, h, & k

22 3 4y x

EXAMPLE: Write the equation of the parabola

Vertex: (-1, 0) Other point: (-2, 2)h, k x, y

Use vertex form 2y a x h k

Substitute for h, k, x, & y

22 2 1 0a

Solve for a 22 1a

2 1a2a

Write equation in vertex form,using values for a, h, & k

22 1y x

HOMEWORK

Textbook p. 251 #2-20 even(printout)

Use Graph paper!!

QUADRATICS: DAY 13

Translating Parabolas: Part II

EXAMPLE: Write in vertex forma.) 23 12 5y x x

1.) Find the x-coordinateof the vertex 2

bx

a

12

2( 3)

12

6

2

2.) Plug that value in to getthe y-coordinate of the vertex

23(2) 12(2) 5y 3(4) 24 5y 12 24 5y

17y Vertex is (2, 17)

3.) Plug values for a (found in original equation), h, & k into vertex form

2y a x h k

23 2 17y x

h, k

CONTINUED…b.)

2 4 3y x x 1.) Find the x-coordinateof the vertex 2

bx

a

4

2(1)

4

2

2

2.) Plug that value in to getthe y-coordinate of the vertex

2( 2) 4( 2) 3y 4 8 3y

7y Vertex is (-2, -7)

3.) Plug values for a (found in original equation), h, & k into vertex form

2y a x h k

22 7y x

h, k

21 ( 2) ( 7)y x

EXAMPLE: Identify the vertex & the y-intercept of the

graph of the functiona.) 2

2 1 1y x 2

2 1 ( 1)y x

h k

Vertex is (1, -1)

y-intercept: plug in 0 for x

22 0 1 1y

22 1 1y

2 1 1y 2 1y

1y y-intercept is (0, 1)

CONTINUED…b.) 2

3 2 4y x

23 ( 2) 4y x

h k

Vertex is (-2, 4)

y-intercept: plug in 0 for x

23 0 2 4y

23 2 4y

3 4 4y 12 4y

8y y-intercept is (0, -8)

HOMEWORK

Textbook p. 251-252 (printout) #1, 3, 17, 19, 21-24, 27-30

(Use Graph paper for #1 & #3)

QUADRATICS: DAY 14

Real-World Application

The number of items a company sells frequently is a function of the item’s price.

The revenue from sales of the item is the product of the price and the number sold.

EXAMPLE: The number of unicycles a company sells can be

modeled by the function, , where p = price

What price will maximize the company’s revenue from unicycles?

What is the maximum revenue?

2.5 500p

Revenue = Price x Number sold

R = p x (-2.5p+500)

( 2.5 500)R p p 22.5 500R p p

CONTINUED… How can we find the maximum value of the

function? Since a<0, the graph opens down, and the vertex

represents a maximum value. Instead of (x, y), we’re using (p, r), so we need to

find p at the vertex.

Now, find the value of r at the vertex

2

bp

a

500

2( 2.5)

500

5

100

22.5 500R p p 22.5(100) 500(100) 25,000

The price of $100 will maximize revenue at $25,000.

EXAMPLE: The number of widgets the Woodget Company sells

can be modeled by , where p = price. What price will maximize revenue? What is the maximum revenue?

5 100p

Revenue = Price x Number soldR = p x (-5p+100)

( 5 100)R p p 25 100R p p

2

bp

a

100

2( 5)

100

10

10 25(10) 100(10)R 500

The price of $10 will maximize revenue at $500.

HOMEWORK

Worksheet: Quadratics Real World Applications Day 1

QUADRATICS: DAY 15

More Real-World Application

EXAMPLE: Smoke jumpers are in free fall from the time they

jump out of a plane until they open their parachutes. The function models a jumper’s

height in y feet at t seconds for a jump from 1600 ft. How long is a jumper in free fall if the parachute

opens at 1,000 ft?

216 1600y t

216 1600y t Plug in 1,000 for y & solve for t21000 16 1600t

160016002600 16t

1616237.5 t

6.1t The jumper is in free fall for about 6.1 seconds

EXAMPLE:•The figure shows a pattern for an open-top box. •The total area of the sheet of material used to manufacture the box is 288 in2. •The length is 2 in longer than the width.•The height of the box is 3 in. Therefore, 3in x 3in squares are cut from each corner. •Find the dimensions of the box.

Length x Width = Area3

3

Define:x = width of a side of the boxx

Width of Material:3 + x + 3 =

x+2

x + 6Length of Material:3 + x + 2 + 3 = x + 8

8x 6x 288

CONTINUED… 8x 6x 288

2x

Must equal 0 in order to factor

x6 x8 48 2882x x14 48 288

2882882x x14 240 0

(x )(x ) = 0 24 10024 x OR 010 x

24 24 10 1024x OR 10x

Not reasonableb/c negative

Dimensions of Box:

Width = x = 10 in.

Length = x + 2= 12 in.

Height = 3 in

EXAMPLE: The area of a square is Find the length of a side.

2 29 12 4g g cm

Is a perfect square trinomial29 12 4g g

3 2 3 2g g Factor!

The side of the square has a length of (3g + 2) cm

HOMEWORK


QUADRATICS: DAY 16

More Real-World Application

EXAMPLE: The volume (lwh) of a rectangular prism is Factor to find possible expressions for the length, width, &

height of the prism.

3 280 224 60x x x

3 280 224 60x x x Factor out GCF

24 20 56 15x x x Factor the quadratic trinomial

220 50 6 15x x x 15 x 20 = 300

Factors of 300 that combine to get 56?

10 2 5 3 2 5x x x

10 3 2 5x x 4x

The possible dimensions of the prism are 4x, (10x+3) & (2x+5) in

EXAMPLE: Write in vertex form 2 6 2y x x

2 6 2y x x factor out -1 from 1st two terms(in order to make x2 positive)

2( 6 ) 2y x x complete the square2

2

b

26

2

9

add & subtract 9 on the right side2( 6 9) 2 9y x x

factor the perfect square trinomial2( 3) 2 9y x

Simplify2( 3) 7y x

EXAMPLE: Write in vertex form 2 10 2y x x

2 10 2y x x complete the square2

2

b

210

2

25

add & subtract 25 on the right side2 10 25 2 25y x x

factor the perfect square trinomial2( 5) 2 25y x

Simplify2( 5) 27y x

EXAMPLE: The profit P from handmade sweaters depends on the price s at

which each sweater is sold. The function models the monthly profit

from sweaters for one custom tailor. Write the function in vertex form. Use the vertex form to find the

price that yields the maximum monthly profit and the amount of the maximum profit.

2 120 2000P s s

2 120 2000P s s Factor -1 from the 1st two terms2( 120 ) 2000P s s Complete the square

2

2

b

2120

2

3600

Add & subtract 3600on the right side

2[ 120 3600] 2000 3600P s s

2( 60) 2000 3600P s Factor the perfect Square trinomial

Simplify in vertex form2( 60) 1600P s

The vertex is (60, 1600), which means a price of $60 per sweater

gives a maximum monthly profit of $1600.

HOMEWORK


Quadratic Functions. Expanding to Standard Form A quadratic function is a function that can be written in the standard form below and where quadratic.

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