**Example 1**

Find the **domain and range** of these relations according IB mathematics curriculum:

a) {(1,4),(2,7),(3,10),(4,13)}

b) {(-2,4),(-1,1),(0,0),(1,1),(2,4)}

**Answers**

a) The domain is {1,2,3,4}

The range is {4,7,10,13}

b) The domain is {-2,-1,0,1,2}

The range is {0,1,4}

So the domain is the **x value** of a pair and the **y value** represents the range

*Do not repeat value, eg even though in b) there are two 4s and two 1s in IB mathematics*

**Example 2**

Which of these sets of ordered pairs are functions?

a) {(1,4),(2,6),(3,8),(3,9),(4,10)}

b) {(1,3),(2,5),(3,7),(4,9),(5,11)}

c) {(-2,1),(-1,1),(0,2),(1,4),(2,6)}

**Answers**

a) Not a function because the number 3 appears** twice** in the domain

b) A function because all of the elements are **different**

c) A function because all of the elements are** different**

- It does not matter is there are
**y values**which are the same

**Example 3**

What are the **graphical characteristics **that a relation must have in order to be classified as a function?

**Answer**

Using the vertical line test, there must be no point at which it crosses the **curve more than once**. Therefore, there must be no point at which there are **two y values for one x value** in IB mathematics.

**Example 4**

Identify the **horizontal and vertical asymptotes** for these functions if they exist.

a) y=2^x

b) y=2x/(x+1)

c) y=(x+2)/(x+1)(x-2)

**Answers**

This can be shown both **graphically and by calculation**. Graphically an asymptote is shown by the line which the curve tends towards but never reaches. Below I have shown these by calculation.

(a) Horizontal asymptote:

y=2

(b) Horixontal asymptote:

2x/x=2

y=2

Vertical asymptote:

0=x+1

x=-1

(c) Horizontal asymptote:

y=(0+2)/(0+1)(0-2)

y=0

Vertical asymptote:

0=(x+1)(x-2)

x=-1 and x=2

- To find the vertical asymptote take the
**donominator**and make it**equal to 0**

End of this part!

Drafted by Gina