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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
EXAMPLE: Graph the function
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
EXAMPLE: Graph the function
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
EXAMPLE: Solve by graphing
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
EXAMPLE: Solve by graphing
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
2.) Split the 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
1.) Multiply the 1st & last terms,
find factors of that # that combine to get middle term
2.) Split the middle term
3.) Factor out the GCF from the 1st two terms, then the last two terms
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
1.) Multiply the 1st & last terms,
find factors of that # that combine to get middle term
2.) Split the middle term
3.) Factor out the GCF from the 1st two terms, then the last two terms
CONTINUED…e.) 358020 2 xx
2.) Multiply the 1st & last terms,
find factors of that # that combine to get middle term
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
2.) Multiply the 1st & last terms,
find factors of that # that combine to get middle term
51 5 Factors of 5:
-1& -5 -1+ -5 = -6
3.) Split the middle term
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
Worksheet: Quadratics Real World Applications Day 2
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
Worksheet: Quadratics Real World Applications Day 3