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