Find the value of x 2 + 4x + 4 if x = –2.

A. –8

B. 0

C. 4

D. 16

Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7.

A. –9

B. 9

C. 19

D. 41

Factor 8xy 2 – 4xy.

A. 2x(4xy 2 – y)

B. 4xy(2y – 1)

C. 4xy(y 2 – 1)

D. 4y 2(2x – 1)

A.

B.

C.

D.

1.1 Functions

Objectives

•Use Set Notation

•Use Interval Notation

Set – a collection of objects

Example: Colors, Cars

Element – are the objects that belong to a set.

Example: red, orange, blue, ….

Nissan, Audi, Jeep, …

Infinite Set

A set that has an unending list of elements

Countable – a collection of objects

Uncountable – are the objects that belong to a set.

Use Set-Builder Notation

A. Describe {2, 3, 4, 5, 6, 7} using set-builder notation.

The set includes natural numbers greater than or equal to 2 and less than or equal to 7.

This is read as the set of all x such that 2 is less than or equal to x and x is less than or equal to 7 and x is an element of the set of natural numbers.

Use Set-Builder Notation

B. Describe x > –17 using set-builder notation.

The set includes all real numbers greater than –17.

Use Set-Builder Notation

C. Describe all multiples of seven using set-builder notation.

The set includes all integers that are multiples of 7.

Describe {6, 7, 8, 9, 10, …} using set-builder notation.

A.

B.

C.

D.

Interval Notation

Is a method of writing numbers in a set.

Recall the Number Line

Use Interval Notation

A. Write –2 ≤ x ≤ 12 using interval notation.

The set includes all real numbers greater than or equal to –2 and less than or equal to 12.

Answer: [–2, 12]

Use Interval Notation

B. Write x > –4 using interval notation.

The set includes all real numbers greater than –4.

Answer: (–4, )

Use Interval Notation

C. Write x < 3 or x ≥ 54 using interval notation.

The set includes all real numbers less than 3 and all real numbers greater than or equal to 54.

Answer:

Write x > 5 or x < –1 using interval notation.

A.

B.

C. (–1, 5)

D.

Review

Homework

1.1 Functions Continued

•What is a function?

•How do we use the Vertical Line Test?

x represents the domain

y represents the range

Turn your pencil vertically. Does you pencil pass through the graph

more than once?

Identify Relations that are Functions

B. Determine whether the table represents y as a function of x.

Answer: No; there is more than one y-value for an x-value.

Identify Relations that are Functions

C. Determine whether the graph represents y as a function of x.

Answer: Yes; there is exactly one y-value for each x-value. Any vertical line will intersect the graph at only one point. Therefore, the graph represents y as a function of x.

Practice

WKST

Review

Homework

And

Worksheet

Quiz on Section 1.1

Tuesday, September 16

Day 4Extra Help – Second half of Lunch

1.1 – Functions

Objectives

•Determine if the equation is a function

•Find function values

•Find the domain of the function

Identify Relations that are Functions

D. Determine whether x = 3y 2 represents y as a

function of x.

To determine whether this equation represents y as a function of x, solve the equation for y.

x = 3y 2 Original equation

Divide each side by 3.

Take the square root of each side.

Identify Relations that are Functions

Answer: No; there is more than one y-value for an x-value.

This equation does not represent y as a function of x because there will be two corresponding y-values, one positive and one negative, for any x-value greater than 0.

Let x = 12.

Determine whether 12x 2 + 4y = 8 represents y as a

function of x.

A. Yes; there is exactly one y-value for each x-value.

B. No; there is more than one y-value for an x-value.

Find Function Values

A. If f (x) = x 2 – 2x – 8, find f (3).

To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8.

f (x) = x 2 – 2x – 8 Original function

f (3) = 3 2 – 2(3) – 8 Substitute 3 for x.

= 9 – 6 – 8 Simplify.

= –5 Subtract.

Answer: –5

Find Function Values

B. If f (x) = x 2 – 2x – 8, find f (–3d).

To find f (–3d), replace x with –3d in f (x) = x 2 – 2x – 8.

f (x) = x 2 – 2x – 8 Original function

f (–3d)= (–3d)2 – 2(–3d) – 8 Substitute –3d for x.

= 9d 2 + 6d – 8 Simplify.

Answer: 9d 2 + 6d – 8

Find Function Values

C. If f (x) = x2 – 2x – 8, find f (2a – 1).

To find f (2a – 1), replace x with 2a – 1 in f (x) = x 2 – 2x – 8.

f (x) = x 2 – 2x – 8 Original function

f (2a – 1) = (2a – 1)2 – 2(2a – 1) – 8 Substitute 2a – 1 for x.

= 4a 2 – 4a + 1 – 4a + 2 – 8 Expand

(2a – 1)2 and 2(2a – 1).

= 4a 2 – 8a – 5 Simplify.

Answer: 4a 2 – 8a – 5