Post on 02-Oct-2020
Warm Up Chapter 2
2
Graph 1 3 4y x x x
A rectangular dog pen is constructed using a barn wall
as one side and 60 m of fencing for the other three sides.
Find the dimensions of the pen that give the greatest area.
1.
2. Write an equation for the graph.
3.
4. Factor a. 2𝑥2 + 𝑥 − 28 b. 6𝑥2 − 24
Warm Up Chapter 2
2
Graph 1 3 4y x x x 1.
2. Write an equation for the graph.
3
Graph 1 3y x x
A rectangular dog pen is constructed using a barn wall
as one side and 60 m of fencing for the other three sides.
Find the dimensions of the pen that give the greatest area.
2
60 2 , 0 30
2 60
Maximum value will occur at 2
A x x x x
A x x x
bx
a
Problem
3.
2
6015
2 2
the maximum area occurs when the dimensions are
15 m by 30 m
15 450 m
x
A
4. Factor a. 2𝑥2 + 𝑥 − 28 b. 6𝑥2 − 24
𝟐𝒙 − 𝟕 𝒙 + 𝟒 𝟔(𝒙𝟐 − 𝟒)
𝟔 𝒙 − 𝟐 𝒙 + 𝟐
1. Solve and graph on a number line: −𝟑𝒙 − 𝟗 ≥ 𝟗
2. Solve on a number line: 𝟐𝒙 − 𝟓 𝟐(𝒙 + 𝟑)(𝒙 + 𝟐) > 𝟎
3. A factory produces short and long sleeved shirts. A short sleeved shirt requires 30 minutes of labor, a long sleeved shirt requires 45 minutes of labor. There are 240 hours of labor available each day. The maximum number of shirts that can be packaged is 400. If the profit on a short sleeved shirt is $11 and a long sleeved shirt is $16, find the maximum daily profit.
*Write the constraints, profit equation, and make a graph
1. Solve and graph on a number line: −𝟑𝒙 − 𝟗 ≥ 𝟗
−3𝑥 − 9 ≥ 9 −3𝑥 − 9 ≤ −9
−3𝑥 ≥ 18
𝑥 ≤ −6
−3𝑥 ≤ 0𝑥 ≥ 0
2. Solve on a number line: 𝟐𝒙 − 𝟓 𝟐(𝒙 + 𝟑)(𝒙 + 𝟐) > 𝟎
2. Solve on a number line: 𝟐𝒙 − 𝟓 𝟐(𝒙 + 𝟑)(𝒙 + 𝟐) > 𝟎
𝑥 < −3 𝑥 > −2 𝑥 ≠ 2.5
3. A factory produces short and long sleeved shirts. A short sleeved shirt requires 30 minutes of labor, a long sleeved shirt requires 45 minutes of labor. There are 240 hours of labor available each day. The maximum number of shirts that can be packaged is 400.
If the profit on a short sleeved shirt is $11 and a long sleeved
shirt is $16, find the maximum daily profit.
Let x = short sleeved shirts
Let y = long sleeved shirts
𝒙 ≥ 𝟎𝐲 ≥ 𝟎
𝟏
𝟐𝒙 +
𝟑
𝟒𝒚 ≤ 𝟐𝟒𝟎
𝑷 = 𝟏𝟏𝒙 + 𝟏𝟔𝒚
𝒙 + 𝒚 ≤ 𝟒𝟎𝟎
𝒙 ≥ 𝟎𝒚 ≥ 𝟎
𝒙 + 𝒚 ≤ 𝟒𝟎𝟎
𝒙
𝒚
50
50 150
150
250
250
350
350
450
450
100
100
200
200
300
300
400
400 𝟏
𝟐𝒙 +
𝟑
𝟒𝒚 ≤ 𝟐𝟒𝟎
(0,400)
(240,160)
(0,320)
𝑷 = 𝟏𝟏𝒙 + 𝟏𝟔𝒚
(0,320)
P = 11 0 + 16 320 = $5120
(240,160)
P = 11 240 + 16 160 = $5200
(400,0)
P = 11 400 + 16 0 = $4400
The maximum daily
profit is $5200
Warm Up Chapter 41. Find the inverse of the function
𝑓(𝑥) = 2 +3𝑥 − 5
2. Sketch the graph of g and g-1.Then find a rule for
g-1(x).
3. A manufacturer wants to design an open top box
with a square base and a surface area of 400 in2.
a. Express the volume V of the box as a function of
the base width w.
b. What dimensions will produce a box with
maximum volume?
𝑔 𝑥 = 𝑥2 + 2, 𝑥 ≥ 0
Warm Up Chapter 4
1. Find the inverse of the function 𝑓(𝑥) = 2 +3𝑥 − 5,
𝑦 = 2 +3𝑥 − 5
𝑥 = 2 +3𝑦 − 5
𝑥 − 2 =3𝑦 − 5
𝑥 − 2 3 = 𝑦 − 5
𝑥 − 2 3 + 5 = 𝑦
𝑓−1(𝑥) = 𝑥 − 2 3 + 5
Warm Up Chapter 4
2. Sketch the graph of g and g-1.
Then find a rule for g-1(x).
𝑔 𝑥 = 𝑥2 + 2,𝑥 ≥ 0
𝑥 = 𝑦2 + 2
𝑥 − 2 = 𝑦2
𝑥 − 2 = 𝑦
𝑔−1 𝑥 = 𝑥 − 2,
𝑥 ≥ 2
g g
g-1
Warm Up Chapter 4A manufacturer wants to design an open top box
with a square base and a surface area of 400 in2.
a. Express the volume V of the box as a function of
the base width w.
3.
w w
h2( , )V h hw w Write in terms of .h w
𝑆𝐴: 𝑤2 + 4𝑤ℎ = 400
4𝑤ℎ = 400 −𝑤2
ℎ =400 −𝑤2
4𝑤
𝑉 𝑤 = 𝑤2400 −𝑤2
4𝑤
𝑉 𝑤 =400𝑤 −𝑤3
4
Warm Up Chapter 4A manufacturer wants to design an open top box
with a square base and a surface area of 400 in2.
b. What dimensions will produce a box with
maximum volume?
3.
ww
h
𝑉 𝑤 =400𝑤 −𝑤3
4
Calculate max value
on the calculator.
(11.55,769.80)
w = 11.55 𝑖𝑛 ℎ =400 −𝑤2
4𝑤=400 − 11.552
4(11.55)= 5.77 𝑖𝑛
1. Simplify.15𝑥−2 𝑥4𝑦
5
9𝑥5 𝑥𝑦02. Solve. 275𝑥 =
1
9
𝑥−3
3. Solve. a. 𝑙𝑜𝑔5𝑥 = 2 b. 𝑙𝑜𝑔𝑥121 = 2
4. Expand. 𝑙𝑜𝑔3𝑥4𝑦
𝑧5.
6. Which plan yields the most interest? Invest $100
Plan A: A 7.5% annual rate compounded monthly for 4 years
Plan B: A 7.2% annual rate compounded daily for 4 years
Plan C: A 7% annual rate compounded continuously for 4 years
Warm Up Chapter 5
2 2Solve log log ( 2) 3.x x
WARM UP CHAPTER 5
1. Simplify. 15𝑥−2 𝑥4𝑦
5
9𝑥5 𝑥𝑦0=15𝑥−2 𝑥4𝑦 5
9𝑥5 𝑥=15 𝑥4𝑦 5
9𝑥5 𝑥 𝑥2
=15𝑥20𝑦5
9𝑥5 𝑥 𝑥2 =15𝑥20𝑦5
9𝑥8=15𝑥12𝑦5
9
=5𝑥12𝑦5
3
2. Solve. 275𝑥 =1
9
𝑥−3WARM UP CHAPTER 5
33 5𝑥 = 3−2 𝑥−3
315𝑥 = 3−2 𝑥−3
315𝑥 = 3−2𝑥+6
15𝑥 = −2𝑥 + 6
17𝑥 = 6
𝑥 =6
17
3. Solve. a. 𝑙𝑜𝑔5𝑥 = 2 b. 𝑙𝑜𝑔𝑥121 = 2
WARM UP CHAPTER 5
52 = 𝑥
𝑥 = 25 𝑥 = 11
𝑥2 = 121
4. Expand. 𝑙𝑜𝑔3𝑥4𝑦
𝑧
WARM UP CHAPTER 5
𝑙𝑜𝑔3(𝑥4𝑦) − 𝑙𝑜𝑔3𝑧
𝑙𝑜𝑔3𝑥4 + 𝑙𝑜𝑔3𝑦 − 𝑙𝑜𝑔3𝑧
4𝑙𝑜𝑔3𝑥 + 𝑙𝑜𝑔3𝑦 − 𝑙𝑜𝑔3𝑧
2 2Solve log log ( 2) 3.x x
2log ( 2) 3x x
3 22 2x x 20 2 8x x
0 4 2x x
2 or 4x
4x
WARM UP CHAPTER 5
5.
Which plan yields the most interest? Invest $100
Plan A: A 7.5% annual rate compounded monthly for 4 years
Plan B: A 7.2% annual rate compounded daily for 4 years
Plan C: A 7% annual rate compounded continuously for 4 years
𝟏𝟎𝟎(𝟏 +. 𝟎𝟕𝟓
𝟏𝟐)𝟏𝟐∙𝟒= 𝟏𝟑𝟒. 𝟖𝟔
𝟏𝟎𝟎(𝟏 +. 𝟎𝟕𝟐
𝟑𝟔𝟓)𝟑𝟔𝟓∙𝟒= 𝟏𝟑𝟑. 𝟑𝟕
𝟏𝟎𝟎𝒆.𝟎𝟕∙𝟒 = 𝟏𝟑𝟐. 𝟑𝟏
WARM UP CHAPTER 5
WARM UP CHAPTER 61. Find the center and radius of the circle.
2. Find the coordinates of the vertices and the foci, and
sketch the ellipse with equation 9𝒙𝟐 + 𝟕𝒚𝟐 = 𝟔𝟑
3. Find an equation for the hyperbola that has center at (0,0), a vertex at (0,-12), and a focus at (0,-13).
4. Find the coordinates of the vertex and the focus, and the
equation of the directrix, of the parabola 𝐲 = −𝟑
𝟓𝒙𝟐
5. Solve the system algebraically to find the intersection points.
𝒙𝟐 + 𝟖𝒙 + 𝒚𝟐 − 𝟐𝟐𝒚 + 𝟔 = 𝟎
𝟒𝒙𝟐 + 𝒚𝟐 = 𝟏𝟔
𝒙𝟐 − 𝒚𝟐 = −𝟒
1. Find the center and radius of the circle.
𝑥2 + 8𝑥 + 𝑦2 − 22𝑦 + 6 = 0
𝑥2 + 8𝑥 + 16 + 𝑦2 − 22𝑦 + 121 = −6 + 16 + 121
𝑥 + 4 2 + 𝑦 − 11 2 = 131
Center = (−4,11) Radius = 131
2. Find the vertices and foci of 9𝑥2 + 7𝑦2 = 63
9𝑥2 + 7𝑦2 = 63
63 63 63
𝑥2
7+𝑦2
9= 1
Vertices = (0,3) & (0,-3)
𝑐2 = 𝑎2 − 𝑏2
𝑐2 = 9 − 7
𝑐2 = 2
Foci = (0, 2) & (0,- 2)
2 2
2 21.
y x
a b
3. Find an equation for the hyperbola that has center
at (0,0), a vertex at (0,-12), and a focus at (0,-13).
𝑦2
144−𝑥2
25= 1
𝑐2 = 𝑎2 + 𝑏2
𝑏2 = 𝑐2 − 𝑎2
𝑏2 = 169 − 144
𝑏2 = 25𝑦2
144−𝑥2
𝑏2= 1
4. Find the coordinates of the vertex and
the focus, and the equation of the
directrix, of the parabola y = −3
5𝑥2
𝒚 = −𝟏
𝟒𝒑𝒙𝟐
𝑫: 𝒚 = 𝒑
𝑭 = (𝟎,−𝒑)
𝒑 =𝟏
𝟒𝒂=
𝟏
𝟒 −𝟑𝟓
=𝟏
−𝟏𝟐𝟓
= −𝟓
𝟏𝟐=
𝟓
𝟏𝟐
F = (0,−5
12) D: y =
5
12
Vertex = (0,0)
5. Solve the system algebraically to find the
intersection points.
𝟒𝒙𝟐 + 𝒚𝟐 = 𝟏𝟔𝒙𝟐 − 𝒚𝟐 = −𝟒
𝟓𝒙𝟐 = 𝟏𝟐
𝒙𝟐 =𝟏𝟐
𝟓
𝒙 = ±𝟏𝟐
𝟓∗= ±
12
5= ±
4 ∗ 3
5= ±
2 3
5
5
5= ±
2 15
5
𝒚𝟐 = 𝒙𝟐 + 𝟒
𝒚𝟐 =2 15
5
𝟐
+ 𝟒
𝒚𝟐 =𝟔𝟎
𝟐𝟓+ 𝟒 =
𝟏𝟔𝟎
𝟐𝟓𝒚 = ±
𝟏𝟔𝟎
𝟐𝟓= ±
𝟒 𝟏𝟎
𝟓