§ 1-2 Functions

§ 1-2 Functions

The student will learn about:

functions,

quadratic functions,

domain, range,

linear functions,

functions used in a business setting,and some new calculator techniques.

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Definition of a Function

Example – For the sale of x items there corresponds a revenue.

Def. – a function f is a rule that assigns to each number x in a set a number f (x). That is, for each element in the first set there corresponds one and only one element in the second set.

The first set is called the domain, and the set of corresponding elements in the second set is called the range.

That’s one – not two, not three, but one, 1, uno, un, eine, no more no less, but exactly one.

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A function f may be thought of as a numerical procedure or “machine” that takes an “input” number x and produces an “output” number f(x),

as shown below. The permissible input numbers form the domain, and the resulting output numbers form the range.

Functions as a Machine

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Functions Defined by Equations

2x – y = 1

x2 + y2 = 25 (Not a function – graph it.)

Note if x = 3 then y = both 4 and – 4.

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USING THE VERTICLE LINE TEST

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• The domain and range can be illustrated graphically.

Functions

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Functions Defined by Equations

Finding the domain:1. Eliminate square roots of negatives

2. Eliminate division by zero

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f (x) x 3

This implies that x – 3 ≥ 0 or x ≥ 3. 1

f (x)x 5

This implies that x – 5 ≠ 0 or x ≠ 5.

Introduction to Linear Functions

A straight line has an equation of the form Ax + By = C.

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Linear Functions And Equations.

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Cost Function

Finding a company’s cost function

Cost Function C (x) = mx + b

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C is the total cost and

m (slope) is the unit cost.

b (the y-intercept) is the fixed cost.

mx is the variable cost.

Example

A company manufactures bicycles at a cost of $55 each. If the fixed cost are $900, express the company’s cost as a linear function.

Cost Function C (x) = mx + b

Where m (slope) is the unit cost and b (the y-intercept) is the fixed cost.

The unit cost is $55 and the fixed cost is $900 hence

C (x) = 55 x + 900

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Example

The graph of C (x) = 55 x + 900 is a line with slope 55 and y-intercept 900, as shown below.

12units

Cos

t

C (x) = 55 x + 900

slope 55

y-intercept 900

The graph of C (x) = 55 x + 900 is a line with slope 55 and y-intercept 900, as shown below. The slope is the unit cost which is the same as the rate of change of the cost, which is also the company’s marginal cost.

The graph of C (x) = 55 x + 900 is a line with slope 55 and y-intercept 900, as shown below. The slope is the unit cost which is the same as the rate of change of the cost, which is also the company’s marginal cost. The y-intercept is the fixed cost.

Introduction to Quadratic Functions

A quadratic is an equation of the form

f (x) = a x2 + b x + c , and graphs as a parabola.

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Definition

NOTE:

a is the coefficient of the x 2 term,

b is the coefficient of the x term, and

c is the constant term.

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Functions

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Fact of Interest

A quadratic function graph can either open up or open down. If a > 0 the graph opens up and if a < 0 the graph opens down.

f (x) = - 2x 2 + 4x + 6 a = - 2

f (x) = x2 + x – 2. a = 1

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Functions

The vertex of a parabola is its “central” point, the lowest point on the parabola if it opens up and the highest point if it opens down.

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Definition

The vertex (lowest or highest point) is of importance and the x value can be found as follows:

BUT - there is an easier way!

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2. Enter the function Y1 = x 2 + x – 2 and then press ZOOM and 6 for a standard window.

x-intercepts using CALC and zero giving 1 and -2

Graphing a Quadratic Function: Calculator

1. Turn the calculator on and press the y = button. If something is there press clear.

3. Find the x and y intercepts using the calculator.

y-intercept using CALC and value giving -2OR use table! 19

5. Press CALC and 3 for a minimum.Giving x = - 0.5 and y = - 2.25

Graphing (Continued)

4. The vertex of the parabola can be found on the calculator using the minimum or maximum menu options.

NOTE: There is no need to use the previous formula for the x-value.

y = x 2 + x – 2

2

1

2

1

a2

bx

I love my calculator! 20

Solving a Quadratic Equation

f (x) = x2 + x – 2.

2. Solutions may also be found algebraically. Let f (x) = 0 and solve for x :

1. The solutions of a quadratic (the x-intercepts) are also called roots or zeros. We just found the x-intercepts using the calculator.

a. Factor : 0 = (x + 2)(x – 1) and x = 1 and – 2.

b. Complete the square – No thank you!

c. Use the quadratic formula – Next slide please!21

Solving a Quadratic Equation

f (x) = x2 + x – 2.

c. Use the quadratic formula : a = 1, b = 1, c = -2.

I love my calculator!

2,12

31

2

811x

OR use your calculator under calc and zeros.

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Quadratic Reviewy = ax 2 + bx + c

x intercepts CALC & ZERO (x, 0)

opens up if a > 0

y intercept CALC & VALUE (0, y)

opens down if a < 0

vertex at CALC & MAX or MIN

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Basic Business Applications.

Cost Function C = Unit cost · Quantity + Fixed cost Where m (slope) is the unit cost and b (the y-intercept) is the fixed cost.

Revenue FunctionR = (price) · (quantity sold) (Note: quantity is number of items sold at price of $p.)

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Profit Function

P = R – C

Basic Business Applications.

Let C represent cost, R represent revenue and P profit. Then one of three things can occur:

R > C P > 0 a profit



R = C P = 0 a break-even point, or



R = C P = 0 a break-even point, or

R < C P < 0 a loss.

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Application ExampleBreak-even Analysis. Use the revenue and cost functions, R (x) = x (2,000 – 60x) & C (x) = 4,000 + 500 x, where x is thousands of dollars. Both functions have domain 0 ≤ x ≤ 25.

1. Sketch the graph of both functions on the same coordinate system. (Algebraically)

2. Find the break-even point (R (x) = C (x)) (Algebraically).

Do you have any idea how long this would take or how difficult it is to do? Well do you?

Do you have any idea how long this would take or how difficult it is to do?

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Application Example by Calculator

Break-even Analysis. Use the revenue and cost functions, R (x) = x (2,000 – 60x) & C (x) = 4,000 + 500 x, where x is thousands of dollars. Both functions have domain 0 ≤ x ≤ 25.

2. Use Calc & intersect to find the break-even point.

1. Graph both functions on your calculator.

0 ≤ x ≤ 25 0 ≤ y ≤ 17,000

(3.03, 5518) (21.96, 14982)

3. When does profit occur?


Solving Quadratic EquationsThe relationship among the cost, revenue, and profit functions can be seen graphically as follows.

Notice that the break-even points correspond to a profit of zero, and that the maximum profit occurs halfway between the two break-even points.

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Break-even Analysis. This is the problem we just did! Use the revenue and cost functions,

R (x) = x (2,000 – 60x) & C (x) = 4,000 + 500 x, where x is thousands of dollars. Both functions have domain 0 ≤ x ≤ 25.

Application Example by Calculator

Embrace your calculator!

For homework, graph the profit equation with these two graphs – revenue and cost.

Spreadsheet Exploration

Number of units Revenue Cost Profit

x R(x) C(x) P(x)

0 0 4000 -4000

1 1940 4500 -2560

2 3760 5000 -1240

3 5460 5500 -40

4 7040 6000 1040

5 8500 6500 2000

6 9840 7000 2840

7 11060 7500 3560

8 12160 8000 4160

9 13140 8500 4640

10 14000 9000 5000

11 14740 9500 5240

12 15360 10000 5360

13 15860 10500 5360

14 16240 11000 5240

15 16500 11500 5000

16 16640 12000 4640

17 16660 12500 4160

18 16560 13000 3560

19 16340 13500 2840

20 16000 14000 2000

21 15540 14500 1040

22 14960 15000 -40

23 14260 15500 -1240

24 13440 16000 -2560

25 12500 16500 -4000

-5000

0

5000

10000

15000

20000

0 5 10 15 20 25

R(x)

C(x)

P(x)

A spreadsheet may be substituted for a graphing calculator.

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Profi t

Summary.

• We learned about the basic business functions.

• We learned about the linear functions.

• We learned about functions and the basic terms involved with functions.

• We learned about the quadratic functions.

• We learned how to use a graphing calculator to make our work much easier.

• We saw how spreadsheets can replace the work of a graphing calculator.


ASSIGNMENT

§1.2 on my website

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§ 1-2 Functions