what is function, how to find aroc and iroc

1. A power function is one of the form where
y = xn
Where n is any real number constant.
A power function is the simplest type of polynomial function
Example:
y= x y = x2 y = x3

2. F(X) =3x5 + 2x3 + x2 - 1
Leading
coefficient
Degree
Constant
term ( a0 )
variable
• n must be a whole number

3. ODD-DEGREE POWER FUNCTIONS HAVE POINT SYMMETRY ABOUT THE
ORIGIN, (0,0)

4. Bracket
interval
Inequality Number line In words
(a,b) a <x<b

a

b
X is
greater
than x
than a and
less than
b
(a,b] a < x ≤ b  
a b
X is
greater
than a and
less than
or equal
to b

5. Key features of the
graph
Y=xn ; n =
odd
Y=xn; n = even
Domain {x € R } {x € R }
Range { y € R } { y € R/ y ≥ 0 }
End behaviour as x
∞
y ∞ y
End behaviour as x
∞
y- ∞ y ∞
I
II
III IV

6. Local maximum
Local minimum
VIDEO TIME ! ^_^
Absolute minimum

7. Graphs of polynomial functions with odd degree
🞂 odd-degree polynomials have at least one x - intercepts, up to a max of
nx-intercepts.
🞂 The domain of all odd-degree polynomials is {xε R} and the range is
{ yε R}
🞂 Odd degree functions have NO maximum point * minimum point
🞂 Odd-degree polynomials may have point symmetry
Positive leading
coefficient
Negative leading
coefficient

8. 🞂 Even-degree polynomials may have from zero to a maximum
of n x - intercepts, where n is the degree of the function
🞂 The domain of all even-degree polynomials is {xε R}
🞂 Even-degree polynomials may have line symmetry.
Graphs of Polynomial Functions with Even Degree
Positive leading coefficient Negative leasing coefficient

10. 🞂 Using Y & X –
Intercepts ( Zero‟s )
🞂 The degree of the
function
🞂 Sign of leading coefficient
🞂 Graphic Calculator

11. F ( x ) = ( X – 1 ) ( X + 1 ) ( X – 3 )
Y = 0
X – 1 = 0
= 1
X + 1 = 0
= - 1
X – 3 = 0
= 3
X = 0
Y = X – 1
= - 1
Y = X + 1
= 1
Y = X – 3
= - 3
X intercept(s) : Y = 0 Y intercept(s) : X = 0

12. F ( x ) = ( X – 1 ) ( X + 1 ) ( X – 3 )
X3 – 3X2 – X + 3
(Degree: 3; Three possible
solutions)

13. POSITIVE NEGATIVE
E
O
v
d
e
d
n Degree Polynomials

14. X3 – 3X2 – X + 3
Positive Leading Coefficient
Degree of 3 = Odd

15. 🞂 Clear Your Memory ( Just In Case )
🞂 Press “ ( Y = )“
🞂 Key in formula
🞂 Press “ Graph “
Steps

17. y= a f [k(x+b)] + c

18. „ - ‟ means reflection in the x - axis
a>1
0<a<1
means vertical stretch of factor a
means vertical compression of factor a
k>1
0<k<1
means horizontal compression of factor 1/k
means horizontal stretch of factor 1/k
b>0
b<0
means horizontal shift b unit left
means horizontal shift b unit right
c>0
c<0
means vertical shift c unit up
means vertical shift c unit down
Note: f(-x)
- f(x)
means a reflection in the y-axis
means a reflection in the x - axis

19. Reflection over the x - axis
- f(x) reflects f(x) over the x-axis.

20. Reflection over the y-axis
f(-x) reflects f(x) over the y-axis.

21. a
Vertical Stretch or Compress
a[f(x)] stretches/compresses f(x) vertically

22. k
Horizontal Stretch or Compress
f(kx) stretches/compresses f(x) horizontally

23. b
Slide to the right or left
f(x + / - b) translates f(x) horizontally

24. c
Slide upward or downward
[f(x)+/-] c translates f(x) vertically

25. f(x)=x2
f(x)= 4(1/3 x)2 -
7

28. 🞂 Measure of how quickly one quantity (the
dependent variable) changes with respect to
another quantity (the independent variable)
🞂 Types of rate of change:-
◦ Average
◦ Instantaneous

29. Average Rate Of Change
Instantaneous Rate Of
Change
🞂 Change that takes place over an interval
🞂 Change that takes place in an instant

30. 🞂 Rate of Change
- Average Rate of Change
- Instantaneous Rate of Change
🞂 Slope = Gradient
🞂 Secant = Line that connects two points on a
curve
🞂 Points [Example: P (x,y)]

31. 🞂 Refers to the slope of the secant between the
points.
🞂 Average rate of change = ∆ y / ∆ x
= change in
y/change in x
= y2-y1
x2-x1

32. Time (t) in hours ; x 0 1 2 3 3.5
Distance d(t) in km; y 50 98 156 200 256
• Average velocity= slope/gradient=m= ∆ y / ∆ x
Average rate of change = (y2-y1)/(x2-x1)
Time ∆d ∆ t ∆ d / ∆ t = Average
ROC
0<t<1 d(1) – d(o) = 48km 1 - 0=1 48 km/h
1<t<2 d(2) – d(1) = 58km 2 - 1=1 58km/h
2<t<3 d(3) – d(2) = 44km 3 - 2=1 44km/h
3<t<4 d(3.5) – d(3) =
56km
3.5-3=0.5 112km/h

33. 🞂 John drops a ball from the cliff of a hill of 150m.
After t seconds it is d meters above the ground,
where d(t) = 90– 4t2; 0<t<5
a) Calculate the average rate of change of the distance
of the ball above the ground between the times t =
1 and t = 4
Average velocity = d(4) – d(1) / 4 - 1
=2 6 - 86 / 4 - 1
= - 20m/s
b) Find the average rate of change of the distance of
the ball above the ground between t=1 and t=2.5
Average velocity = d(2.5) – d(1) / 2.5 – 1
= 65-86/2.5-1
= - 14m/s

34. 200
0
400
1000
800
600
2 4 6 8 10 12
Amount Of Money
Amount Of Money

35. amount of
🞂 Find the average rate of change of the
money in the account:
i) Month 2 to month 4
ii) Month 4 to month 8
iii) Month 8 to month 10
iv) Month 8 to month 12
i) A.R.O.C = 400-200/ 4-
= 50
ii) A.R.OC = 800-400/ 8 - 4
= 100
iii) A.R.O.C = 800-800/10-8
= 0
iv) A.R.O.C = 500-800/ 12-8
= - 75

36. 🞂 How can you tell that the A.R.O.C is
positive/negative by examining
a) the table of values ?
- POSITIVE because as x increases, y increases
- NEGATIVE because as x increases, y
decreases
b) the graph ?
- POSITIVE because the graph is increasing
from left to right
- NEGATIVE because the graph is decreasing
from left to right

37. 🞂 How can you tell the rate of change is constant or
non-constant by examining
a) The table of values ?
- CONSTANT= the change in the dependent variables
is the same for each one unit increase in the
independent variable.
- NON- CONSTANT = the change in the dependent
variables varies for each one unit increase in the
independent variable.
b) The graph ?
- CONSTANT = Graph is linear
- NON- CONSTANT= Graph is not linear
c) The average rate of change ?
- CONSTANT = A.R.O.C is the same
- NON- CONSTANT = A.R.O.C varies

39. - As a point Y becomes very close to a tangent point
X, the slope of the secant line becomes closer to
(approaches) the slope of the tangent line.
is used to denote the word
- An arrow
“approaches”.
- Example: As Y X, the slope of secant XY the
slope of the tangent at X.
- Thus, the A.R.O.C between X and Y becomes closer to
the value of the instantaneous rate of change at X.

40. 🞂 From a graph
- Draw a tangent line on the graph and
estimating the slope of that tangent from the
graph

41. x y
0.6 5.0
0.8 5.3
1.0 4.9
1.3 3.0
1.5 3.53
🞂 From the table of values method
Slope of secant between X= 0.8 &
X= 1.0
= 4.9-5.3/1.0-0.8
= - 2
Slope of secant between X= 1.0 &
X= 1.3
= 3.0- 4.9/1.3-1.0
= - 6.3
Instantaneous rate of change at
X= 1.0= a+b/2
Therefore,
= - 2+(-6.3)/2
= - 4.15

42. 🞂 From an equation method
- Looking for a trend as the slopes of secants
get closer and closer to the slope of the
tangent.
- Example:
A ball is tossed up in the air so that it‟s position s in meters and
at time t in second, is given by:
s(t) = - 5t2 + 30t + 2
Interval ∆s ∆ t Average rate of change
1<t<2 S(2) – s(1) = 15 2- 1 = 1 15
1<t<1.5 S(1.5) –s(1) = 1.75 1.5 - 1 = 0.5 17.5
1<t<1.1 S(1.1) – s(1) =
1.95
1.1-1 = 0.1 19.5
1<t<1.0
1
S(1.01) – s(1) =
1.995
1.01 – 1 =
0.01
19.95
Therefore, the I.R.O.C at t=1, approximately 20m/s.

43. 🞂 Using the graphing calculator (Tangent
Operation)
- Page 515 in the text book