1

Fórum 2 unit 4 (week 7)

John Alexander

Código

Algebra and Trigonometry

Group: MAT 111 - 1601 – S12

Director-Tutor

Doumaris Carolina Maldonado Lugo

National Open and Distance University - UNAD

School of Basic Sciences, Technology and Engineering

2024
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Introductión

This document was designed with the purpose of explaining quadratic functions to

improve understanding of the subject.

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page 297

23. 𝑓(𝑥) = 𝑥 2 + 4𝑥 + 2

Graph

𝑓(𝑥) = 𝑥 2 + 4𝑥 + 2

𝑦(𝑥) = 𝑥 2 + 4𝑥 + 2 we replace the f with y

𝑦(𝑥) = 𝑥 2 + 4𝑥 + 22 − 22 + 2 complete squared

𝑦(𝑥) = (𝑥 + 2)2 −22 + 2 use the binomial form

𝑦(𝑥) = (𝑥 + 2)2 − 2 simplify

(−2, −2) vertex

𝑦(𝑥) = (𝑥 + 2)2 − 2 canonical or standard form

Symmetry is = -2

And the minimum is = -2

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20. 𝑓(𝑥) = 2𝑥 2 + 4

Graph

𝑓(𝑥) = 2𝑥 2 + 4

𝑦(𝑥) = 2(𝑥 2 + 0 ∗ 𝑥 + 2) we solve for 2 of x squared

𝑦(𝑥) = 2(𝑥 2 + 0 ∗ 𝑥 + 02 − 02 + 2) we complete the square

𝑌(𝑥) = 2((𝑥 + 0)2 − 02 + 2) we use the binomial formula

𝑌(𝑥) = 2((𝑥 + 0)2 + 2) we simplify

𝑌(𝑥) = 2 ∗ (𝑥 + 0)2 + 4 we apply the distributive property

(0,4) vertex

Symmetry = 0 and minimum = 4

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The shift of the function is up 4 units

Understanding is expanded

39. 𝑓(𝑥) = 2𝑥 2 − 𝑥 + 2

Symmetry

39.) 𝑓(𝑥) = 2𝑥 2 − 𝑥 + 2

We identify the coefficients of a and b (a = 2, b = -1)

𝑏
We look for the axis of symmetry by substituting 𝑎 = 2 𝑦𝑏 = −1 𝑒𝑛 𝑥 − 2𝑎

−1
𝑥=−
2𝑥2

We solve the equation for

1
𝑥=
4

The alternative way is 𝑥 = 0,25, 𝑥 = 2−2

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Vertex

𝑓(𝑥) = 2𝑥 2 − 𝑥 + 2

We identify the coefficients of a y b (a = 2, b = -1)

We look for the axis of symmetry by substituting

𝑏
𝑎 = 2 𝑦𝑏 = −1 𝑒𝑛 𝑥 −
2𝑎

−1
𝑥=−
2𝑥2

We solve the equation for x

1
𝑥=
4
1
We evaluate the function for 𝑥 = 4 and we calculate the value

1
𝑓(𝑥) = 2𝑥 2 − 𝑥 + 2, 𝑥 =
4

1 15
𝑓( ) =
4 8

Decimal alternative form (0.25,1.875)

Intersection

𝑓(𝑥) = 2𝑥 2 − 𝑥 + 2

Para encontrar el corte con el eje y sustituyendo x por 0 (so we can say that any multiplication by

0 is equal to 0)

𝑓(0) = 2 ∗ 02 − 0 + 2

𝑓(0) = 0 + 2

Then we remove the 0 from the expression

𝑓(0) = 2
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So the intercept is (0,2)

Domain

𝑓(𝑥) = 2𝑥 2 − 𝑥 + 2

The domain of a quadratic function is the set of all real numbers.

(∞, ∞), {𝑥 =∊ 𝑅)

Range

−𝑏 −𝑏
To find the range we will apply the following formula 𝑣 (2∗𝑎 , 𝑓 2∗𝑎)

So we say that a = 2 b = -1, c = 2 then we substitute the values

−𝑏 −(−1) 1
𝑋= = = = 0.25
2∗𝑎 2.2 4

𝑥 = 0.25

Then we apply the function and replace the value of x with 0.25

𝑦 = 2(0.25)2 − (0.25) + 2

𝑦 = 0,125 − 0.25 + 2

15
= 1.875 =
8

So we say that the range is

15
[ , ∞)
8
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41.) 𝑓(𝑥) = −2𝑥 2 + 2𝑥 − 3

Domain and range

𝑓(𝑥) = −2𝑥 2 + 2𝑥 − 3

We apply the following formula to find the range

−𝑏 −𝑏
𝑣( ,𝑓 )
2∗𝑎 2∗𝑎

So we say that a = -2 b = 2, c = 3 then we substitute the values

−𝑏 −2 −2
𝑋= = = = 0.5
2 ∗ 𝑎 2. (−2) −4

𝑥 = 0.5

Then we apply the function and replace the value of x with 0.5

𝑦 = −2(0.5)2 + (0.5) − 3
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𝑦 = −0,5 + 1 − 3

5
= −2.5 = −
2

So we say that the range is

15
(∞, ]
8

The domain of a quadratic function is the set of all real numbers.

(∞, ∞), {𝑥 =∊ 𝑅)

Symmetry

We identify the coefficients a and b of the quadratic function and locate the axis of symmetry by
𝑏
substituting in 𝑋 = − 2∗𝑎

𝑓(𝑥) = −2𝑥 2 + 2𝑥 − 3

𝑎 = −2, 𝑏 = 2

2
𝑥=−
2 ∗ (−2)

1
𝑥=
2

Decimal value 0.5

Vertex

We identify the coefficients a and b of the quadratic function and locate the axis of symmetry by
𝑏
substituting in 𝑋 = − 2∗𝑎

𝑓(𝑥) = −2𝑥 2 + 2𝑥 − 3

𝑎 = −2, 𝑏 = 2

2
𝑥=−
2 ∗ (−2)
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1
𝑥=
2
1
We will find the coordinate of y of the vertex, substituting in 𝑥 = 2

1 1 2 1
𝑓 ( ) = −2 ( ) + 2 ( ) − 3
2 2 2

Then we cancel the greatest common factor and perform the power operation

1 1 2 1
𝑓 ( ) = −2 ( ) + 2 ( ) − 3
2 2 2

1 1 2
𝑓 ( ) = −2 ( ) + 1 − 3
2 2

1 1
𝑓 ( ) = −2 ∗ + 1 − 3
2 4

1 1
𝑓( ) = − +1−3
2 2

We perform the difference operation

1 1
𝑓( ) = − +1−3
2 2

1 1
𝑓( ) = − −2
2 2

1 5
𝑓( ) = −
2 2

Page 298

65.) 𝑓(𝑥) = −𝑥 2 + 4; 𝑔(𝑥) = −2𝑥 + 1

Solution; To solve the problem we remember the formula (𝑓 ∗ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) and we

substitute g(x) by -2x + 1 in the equation and replace x with -2x + 1 in 𝑓(𝑥) = −𝑥 2 + 4

(𝑓 ∗ 𝑔)(𝑥) = 𝑓(−2𝑥 + 1)

𝑓(−2𝑥 + 1) = −(−2𝑥 + 1)2 + 4

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We use the commutative property to rearrange the terms and use (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 to

expand the expression

𝑓(−2𝑥 + 1) = −(1 − 2𝑥)2 + 4

𝑓(−2𝑥 + 1) = −(1 − 4𝑥 + 4𝑥 2 ) + 4

When we have a – in front of an expression we change the sign of each term to eliminate the

parentheses

𝑓(−2𝑥 + 1) = −(1 − 4𝑥 + 4𝑥 2 ) + 4

𝑓(−2𝑥 + 1) = −1 + 4𝑥 − 4𝑥 2 + 4

We calculate the operation

𝑓(−2𝑥 + 1) = 3 + 4𝑥 − 4𝑥 2

We rearrange the equation

𝑓(−2𝑥 + 1) = −4𝑥 2 + 4𝑥 + 3

66.) 𝑓(𝑥) = −𝑥 2 + 9; 𝑔(𝑥) = 2𝑥 + 1

Solution; To solve the problem we remember the formula (𝑓 ∗ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) and we

substitute g(x) by 2x + 1 in the equation and we substitute x for 2x + 1 in 𝑓(𝑥) = −𝑥 2 + 9

𝑓(2𝑥 + 1) = −(2𝑥 + 1)2 + 9

We use the commutative property to rearrange the terms and use (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 to

expand the expression

𝑓(2𝑥 + 1) = −(4𝑥 2 + 4𝑥 + 1) + 9

When we have a – in front of an expression we change the sign of each term to eliminate the

parenthesis

𝑓(2𝑥 + 1) = −4𝑥 2 − 4𝑥 − 1 + 9
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We calculate the sum

𝑓(2𝑥 + 1) = −4𝑥 2 − 4𝑥 ∓ 8

Page 299

75. Maximizing Revenue Suppose that the manufacturer of a gas clothes dryer has found that,

when the unit price is p dollars, the revenue R (in dollars) is

𝑅(𝑝) = −4𝑝2 + 4000𝑝

What unit price should be established for the dryer to maximize revenue? What is the maximum

revenue?

Solution

The income R in dollars is:

𝑅(𝑝) = −4𝑝2 + 4000𝑝

To find out the unit price for maximum revenue, we find the vertex. In the quadratic equation

given a = 4 and b = 4000

The formula to find the x coordinate of the vertex is

−𝑏
𝑥=
2𝑎
−4000
𝑥=
2(−4)

𝑥 = 500

So p = $500 is the unit price to obtain maximum income

Now we find the maximum income, we substitute 500 for p in the given equation𝑅(𝑝) = −4𝑝2 +

4000𝑝

𝑅(500) = −4(500)2 + 4000(500)

𝑅(500) = 1000000
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Therefore, the maximum income is = $1,000,000

77. Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of

producing one additional unit of output. For example, if the marginal cost of producing the 50th

product is $6.20, it cost $6.20 to increase production from 49 to 50 units of output. Suppose the

marginal cost

𝐶(𝑥) = 𝑥 2 − 140𝑥 + 7400

C (in dollars) to produce x thousand mp3 players is given by

the function

(a) How many players should be produced to minimize the

marginal cost?

(b) What is the minimum marginal cost?

Answer

a.) 80.000 MP3 players

b.) $1.400

Solution

To solve the following equation we must C(x)=x^2-140x+7400

Step 1

As it is a quadratic function (parabola), the minimum value for x=(-b)/2a, where b = -1600; a= a =

−(−160)
𝑥= = 80
2∗1

So we should produce 80,000 MP3 players

Step 2
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Then we substitute x = 80

𝐶(80) = 802 − 140(80) + 7400 = 1.400

So the minimum marginal cost to produce 1000 MP3 players is $1,400.

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Conclusions

Thanks to the development of this activity, it was possible to understand and deepen the

topic of quadratic functions, which allows them to be used in everyday life.

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Bibliográphic references

Text 2: Algebra & Trigonometry. (Sullivan, Michael 2012).URL

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Assignment 1 – unit 4

John Alexander Ríos Hernández

Código 1075653285

Algebra and Trigonometry

Group: MAT 111 - 1604 - S3

Director-Tutor

Pablo Lopez

National Open and Distance University - UNAD

School of Basic Sciences, Technology and Engineering

2023
 18

Introductión

This document seeks to provide support to the reader in the process of analyzing linear

quadratic equations, in order to develop skills in providing solutions to the problems expressed in

this document.
 19

In Problems 13–20, a linear function is given.

13. f(x) = 2x + 3

a) Determine the slope and y-intercept of each function.

As a first step we rewrite the equation of the function substituting f (x) for y

𝑦 = 2𝑥 + 3

then we identify the slope, as the coefficient of the variable x

𝑦 = 2𝑥 + 3, 𝑚=2

Then we identify the cut with the y axis of the line as the constant term and in this way we

determine the slope and the cut with the y axis

𝑦 = 2𝑥 + 3, 𝑚 = 2 , 𝑏 = 3

𝑚 = 2, 𝑏 = 3

then we find the intercept by equaling (y) equal to 0

𝑦 = 2𝑥 + 3

0 = 2𝑥 + 3

We subtract 3 and solve for x

−3 = 2𝑥

3
− =𝑥
2

𝑥 = 1.5

That's why we say that the intersection is

(𝑥, 𝑦) → (1.5,0)

b) Use the slope and y-intercept to graph the linear function.

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c) Determine the average rate of change of each function.

∆𝑑 2
𝑒𝑎𝑟𝑟𝑖𝑛𝑔 = =
∆𝑡 3

we deduce that if a x = 0 then

𝑦 = 2(0) + 3

𝑦=3

(0,3)

𝑥=1

𝑦 = 2(1) + 3

𝑦=1

(1,1)

reason to change

3−1 2
= = −2 𝑎𝑛𝑑 2
0 − 1 −1
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The coefficient of x is 2, therefore the rate of change is 2. This means that for every

increase of 1 in x, y increases by 2 units.

d) Determine whether the linear function is increasing, decreasing, or constant.

We can say that our function is an increasing function since its slope is positive, therefore,

as the value of x increases, the value of y increases.

𝑥1 < 𝑥2 → 𝑓(𝑥1 ) < 𝑓(𝑥2 )

As a first step we rewrite the equation of the function substituting f (x) for y

𝑦 = 2𝑥 + 3

then we identify the slope, as the coefficient of the variable x

𝑦 = 2𝑥 + 3, 𝑚=2

Then we identify the cut with the y axis of the line as the constant term and in this way we

determine the slope and the cut with the y axis

𝑦 = 2𝑥 + 3, 𝑚 = 2 , 𝑏 = 3

𝑚 = 2, 𝑏 = 3
1
17 𝑓(𝑥) = 4 𝑥 − 3

a) Determine the slope and y-intercept of each function.

Since we know that the equation is written in the form. 𝑦 = 𝑚𝑥 + 𝑏 So we say that the

slope is:

1
𝑚=
4

To find the intercession with the y axis, we replace the x with 0. we multiply and remove the 0
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1
𝑓(0) = ∗0−3
4

𝑓(0) = 0 − 3

𝑓(0) = −3

So we say that:

(𝑥, 𝑦) = (0, −3)

Then we will find the intercept for the x axis, substituting f(x) for 0

1
0= 𝑥−3
4

Combine the multiplied terms into a single fraction. and Multiply by 1

1𝑥
0= −3
4
𝑥
0= −3
4

We look for the common denominator and combine the fraction with the common denominator

𝑥 4(−3)
0= +
4 4

𝑥 + 4(−3)
0=
4

We multiply the numbers and all the terms by the same value to eliminate the denominators of the

fractions

𝑥 + 4(−3)
0=
4

𝑥 − 12
0=
4
𝑥 − 12
0= 4∗
4
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0 = 𝑥 − 12

We add 12 to both sides and then simplify the expression

0 = 𝑥 − 12

0 + 12 = 𝑥 − 12 + 12

12 = 𝑥 − 12 + 12

12 = 𝑥

Finally we move the variable to the left

𝑥 = 12

b) Use the slope and y-intercept to graph the linear function.

c.) Determine the average rate of change of each function.

∆𝑑 1
𝑒𝑎𝑟𝑟𝑖𝑛𝑔 = = = 0.25
∆𝑡 4
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The coefficient of x is 0.25 so the rate of change is 0.25 since for every increase in 1 in (x)

there is an increase of 0.25 in (y)

d.) Determine whether the linear function is increasing, decreasing, or constant.

We can say that our function is an increasing function since its slope is positive, therefore,

as the value of x increases, the value of y increases.

𝑥1 < 𝑥2 → 𝑓(𝑥1 ) < 𝑓(𝑥2 )

37. CarRentals The costC, in dollars,of renting amoving truck for a day is modeled by the

function C(x) = 0.25x + 35, where x is the number ofmiles driven.

a.) What is the cost if you drive x = 40 miles?

If x is = 40 miles

𝑐(𝑥) = 0,25𝑥 + 35

𝑐(𝑥) = 0.25(40) + 35

= 10 + 35

= 45

Therefore, the cost of driving 40 miles is = $45

b.) If the cost of renting the moving truck is $80, how many miles did you drive?

80 = 0.25𝑥 + 35

45
80 − 35 = 0,25𝑥 = = 180
0,25

So we can say that the miles traveled are 180 miles.

c.) Suppose that you want the cost to be no more than $100. What is the maximum number of

miles that you can drive?

100 = 0.25 + 35
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100
100 − 35 = 0,25 = = 400
0,25

Therefore, it can be said that the maximum that can be driven is 400 miles with a cost that

does not exceed $100

d.) What is the implied domain of C?

The domain is the set of all values that x can take. Therefore, it does not make sense to

calculate the cost of negative miles traveled, so the implicit domain of the function is (0,∞)

46. Straight-line Depreciation Suppose that a company has just purchased a new machine

for its manufacturing facility for $120,000. The company chooses to depreciate the machine using

the straight-line method over 10 years.

a.) Write a linear model that expresses the book value V of the machine as a function of its

age x.

To express the book value of the machine based on age, it is necessary to determine the

amount of annual depreciation. In this case, the machine was purchased for $120,000 and

will be depreciated over 10 years using the straight line method.

The amount of annual depreciation can be calculated by dividing the initial cost of the

machine by the number of years in which it will be depreciated. In this case, the amount

that will be depreciated is 120,000/10 = 12,000.

Now we can write the linear function

Value in pounds = initial cost – (annual depreciation * age)

𝑉(𝑥) = 120000 − 12000𝑥

b.) What is the implied domain of the function found in part (a)?
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This formula is only relevant during the 10 years that the machine is depreciating, not

before it is purchased, so the reasonable domain of the function v(x) 12000x+120000 that

we have constructed is 0 ≤ x ≤ 10

c.) Graph the linear function.

d.) What is the book value of the machine after 4 years?

𝑉(4) = 120000 − 12000 ∗ (4)

= 72000

The value after 4 years is $ 72000

e.) When will the machine have a book value of $72,000?

when V (x) = 72000, which is in one year, when x = 1.

𝑉(𝑥) = 0 → 120000 − 12000𝑥 = 72000

→ −12000𝑥 = 72000 − 120000

→ −12000𝑥 = −48000
 27

−48000
→ =4
−12000

→= 4

So after 4 years the machine will have a book value of $72,000.

 28

Conclusions

Thanks to the development of this activity, it was possible to develop the skills of linear

quadratic operations