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