Grade 12 – Advanced Functions
(MHF4U)
Unit 1 Polynomial Functions
Topics
Homework
Text book Worksheet
Day 1 Review of Polynomials Products of Polynomials
Factoring Difficult Trinomials
Day 2 Power Functions P. 12 #3-11,14,16-18 Preview of Unit 1
Day 3 Characteristics of Polynomial
Functions
P. 26
#1,2,3i,ii,5,6,7i,ii,8acdef,11-18
Day 4 Graphs of Polynomial functions P. 39 #1,2,6,9-11,13 Graphs of Polynomial functions
#1-3
Day 5 Symmetry P. 39 #3-5,7,8,12,14 Symmetry #1-2
Day 6 Transformation of functions P. 49 #1-7,10,12,15
Day 7 Slope of Secant and Tangent P. 71 #1,4
Day 8 Average and Instantaneous Rates of
Change
P. 63 #9,10,12,13
P. 72 #7,9,10,12
Day 9 Divisions of Polynomial and
Remainder Theorem P. 91 #1-9
Day 10 Factor Theorem P. 102 #1-4,6,9,10,12-15,20
Day 11 Factor Theorem extended P. 102 #5,7,10,11,17-19
Day 12 Solving Polynomial Equations P. 110 #3-5, 6-8,14,15,18,
20, 22
Day 13 Families of Polynomial Functions P. 119 #1-3,
6,7,9,11,13,15,19,22
Day 14 Solving Polynomial Inequalities P. 138 #2,4,6,9,12,13
Day 15 Piecewise functions and Continuity
of Functions
Piecewise functions and
Continuity of Functions
Day 16 Review P. 74 All
P. 140 All
Day 17 Test
MHF4U - Advanced Functions Page 1 of 1 Review: Polynomials Date:
RHHS Mathematics Department © 2014 – E. CHOI
Polynomial Functions The function f is a polynomial if ...)( 01
22
22
11 axaxaxaxaxaxf n
nn
nn
n ++++++= −−
−− .
na are real numbers and n is a whole number. The degree of the function is n if the leading coefficient, .0≠na
Examples 1589)( 21057)( 2523 −+−=+++= xxxxgxxxxf
Example 1:
Expand and Simplify:
a) ( )( )52373 22 +−−+ xxxx b) ( )( ) ( ) ( )xxxxx −+−−− 32525 2
Example 2:
Factor:
a) 672 2 ++ mm b) xxx 42399 23 +−
Example 3:
Evaluate:
a) ( )( ) ( )( )25320 −÷− b) ( )( )( )( )25
320−
−
HW: Product of Polynomials
Factoring Difficult Trinomials
Mathematics Page 1 of 2 Products of Polynomials Date:
RHHS Mathematics Department © 2010 – E. CHOI
Copy the following questions into your notebooks, then expand and simplify.
1. a) )1)(43( 2 +−+ ttt b) )52)(865( 2 −+− bbb
c) )3546)(13( 23 +−−− xxxx d) )52)(38( 2 ++− xxx
e) )6623)(1( 234 −+−− pppp f) )8)(126( 224 −+− ccc
g) )5)(232( 2 −−++ xxx h) )1)(1( 2345 +++++− xxxxxx
i) )132)(45( 22 −−−+ xxxx j) ))(( 2222 yxyxyxyx +−++
2. a) ))(( 42244224 bbaabbaa +++− b) )2)(43( 3223 bababbaa −−+−
c) )13526)(68( 2342 +−+−+− mmmmmm d) )5)(1)(3( ++− xxx e) )5)(4)(3( −−+ xxx f) )4)(13)(72( −−− xxx
g) )3)(3)(3)(3( ++++ xxxx h) )13)(4834( 223 +−−+− ppppp
i) ))(( 222 zyxxzyzxyzyx ++−−−++ j) )122)(433( 232 −−++− xxxxx
3. a) 3)32( −x b) 22 )32( ++ xx
c) )1)(3)(2(5)1()3(2 2 +++−+− xxxxx d) 2)1()3)(1(4 +−+− xxxx
e) 22 )1()1( −−+ xx f) )2)(1()432)(1( 223 −+−−+++ xxxxxx
g) )4)(4(3)1)(1(2 −+−−+ kkkk h) ))(())(( 2222 babababababa ++−+++−
4. a) )52)(74()25)(3(3)52)(43(2 +−−++−−− xxxxxx
b) )35)(35()43(6)25)(2(3 2 bababababa −+−−−−+
c) )832)(2()543)(32( 22 −+−−−++ mmmmmm
d) )435)(3()12)(42( −+−−+−+ yxyxyxyx
e) )52()34()72)(23( 22 babababa −+−+−
f) 22 )23)(32()62)(24(3 −−−++− xxxxx
Mathematics Page 2 of 2 Products of Polynomials Date:
RHHS Mathematics Department © 2010 – E. CHOI
Answers:
1. a) 43 23 +−+ ttt b) 40463710 23 −+− bbb c) 314111818 234 −+−− xxxx
d) 1534112 23 +−− xxx e) 66236623 234567 +−+−−+− ppppppp
f) 817506 246 −+− ccc g) 1017132 23 −−−− xxx h) 16 −x
i) 472472 234 ++−+ xxxx j) 4224 yyxx ++
2. a) 8448 bbaa ++ b) 432234 29105 babbabaa +−+−
c) 626555557506 23456 +−+−+− mmmmmm d) 15133 23 −−+ xxx
e) 6076 23 +−− xxx f) 2899476 23 −+− xxx g) 811085412 234 ++++ xxxx
h) 4203121154 2345 −+−+− ppppp i) 333 3 zyxyzx ++− j) 486 235 −−+− xxxx
3. a) 2754368 23 −+− xxx b) 912104 234 ++++ xxxx c) 1249403 23 −−−− xxx
d) 11474 23 −−+ xxx e) x4 f) 286 234 ++++ xxxx
g) 462 +− k h) 32a
4. a) 5710311 2 +−− xx b) 22 9314749 baba −+−
c) 3116184 23 −++ mmm d) xyxyx 145413 22 +−−−
e) 3223 5319310820 babbaa −++− f) 2422516 23 −++ xxx
Mathematics Page 1 of 2 Factoring Difficult Trinomials : ax 2+bx +c Date:
RHHS Mathematics Department © 2010 E. Choi
Factor each of the following:
1. a) 672 2 ++ xx b) 12112 2 ++ aa c) 3116 2 ++ yy
d) 132 2 ++ kk e) 143 2 ++ ss f) 3108 2 ++ xx
2. a) 275 2 +− xx b) 6113 2 +− nn c) 31314 2 +− cc
d) 15112 2 +− xx e) 7223 2 +− xx f) 144 2 +− aa
3. a) 673 2 −+ tt b) 456 2 −+ kk c) 328 2 −+ rr
d) 1034 2 −+ mm e) 4195 2 −+ yy f) 1544 2 −+ dd
4. a) 675 2 −− aa b) 10133 2 −− xx c) 212 2 −−mm
d) 994 2 −− kk e) 126 2 −− xx f) 215 2 −− aa
5. a) 4133 2 ++ xx b) 12112 2 +− mm c) 25204 2 +− ss
d) 4129 2 ++ xx e) 12176 2 ++ yy f) 3148 2 +− aa
6. a) 310 2 −− xx b) 56 2 −+ kk c) 2715 2 −− gg
d) 279 2 −+ pp e) 5188 2 −+ cc f) 4415 2 −− xx
7. a) 144 2 +− xx b) 252 2 ++ hh c) 15234 2 +− qq
d) 102910 2 +− uu e) 61710 2 −− mm f) 15148 2 −+ cc
8. a) 12176 2 +− hh b) 31310 2 −+ rr c) 15132 2 ++ ww
d) 31914 2 −− tt e) 73310 2 −− xx f) 16249 2 +− aa
9. a) 607020 2 ++ xx b) 206515 2 +− aa c) 181518 2 −+ aa
d) 101216 2 −− rr e) 547224 2 +− xx f) 405212 2 −− aa
10. a) xxx 45336 23 ++ b) aaa 20266 23 −+ c) yxyyx 45318 2 −−
d) mmm 602510 23 −− e) aaa 42399 23 +− f) aabab 214942 2 −+
11. a) 32032 2 +− xx b) 21324 2 −− ss c) 21194 2 ++ aa
d) 18214 2 −+ xx e) 151910 2 −− aa f) 42521 2 −+ xx
12. a) 301721 2 −+ xx b) 61172 2 −+ xx c) 322815 2 −− xx
d) 22 152248 yxyx −− e) 22 152624 dcdc −+ f) 22 640 zyzy −+
Mathematics Page 2 of 2 Factoring Difficult Trinomials : ax 2+bx +c Date:
RHHS Mathematics Department © 2010 E. Choi
Answers:
1. a) )2)(32( ++ xx b) )4)(32( ++ aa c) )32)(13( ++ yy
d) )1)(12( ++ kk e) )1)(13( ++ ss f) )34)(12( ++ xx
2. a) )1)(25( −− xx b) )3)(23( −− nn c) )37)(12( −− cc
d) )3)(52( −− xx e) )7)(13( −− xx f) 2)12( −a
3. a) )3)(23( +− tt b) )12)(43( −+ kk c) )34)(12( +− rr
d) )2)(54( +− mm e) )4)(15( +− yy f) )32)(52( −+ dd
4. a) )2)(35( −+ aa b) )5)(23( −+ xx c) )3)(72( +− mm
d) )3)(34( −+ kk e) )32)(43( −+ xx f) )13)(25( +− aa
5. a) )4)(13( ++ xx b) )4)(32( −− mm c) 2)52( −s
d) 2)23( +x e) )32)(43( ++ yy f) )32)(14( −− aa
6. a) )12)(35( +− xx b) )1)(56( +− kk c) )23)(15( −+ gg
d) )1)(29( +− pp e) )14)(52( −+ cc f) )23)(25( −+ xx
7. a) 2)12( −x b) )2)(12( ++ hh c) )5)(34( −− qq
d) )52)(25( −− uu e) )2)(310( −+ mm f) )52)(34( +− cc
8. a) )32)(43( −− hh b) )32)(15( +− rr c) )5)(32( ++ ww
d) )32)(17( −+ tt e) )72)(15( −+ xx f) 2)43( −a
9. a) )2)(32(10 ++ xx b) )4)(13(5 −− aa c) )32)(23(3 +− aa
d) )12)(54(2 +− rr e) 2)32(6 −x f) )5)(23(4 −+ aa
10. a) )3)(52(3 ++ xxx b) )5)(23(2 +− aaa c) )32)(53(3 +− xxy
d) )4)(32(5 −+ mmm e) )2)(73(3 −− aaa f) )32)(13(7 +− bba
11. a) )38)(14( −− xx b) )23)(18( −+ ss c) )3)(74( ++ aa
d) )6)(34( +− xx e) )52)(35( −+ aa f) )43)(17( +− xx
12. a) )53)(67( +− xx b) )38)(29( +− xx c) )83)(45( −+ xx
d) )56)(38( yxyx −+ e) )32)(512( dcdc +− f) )38)(25( zyzy −+ .
Mathematics 11 Page 1 of 2
Appendix: Factoring Non-Simple Trinomials Date:
RHHS Mathematics Department
A few common techniques to factor Non-simple trinomials cbxax 2
Example: Factor 656 2 xx
Method 1: By Decompositions
Factor 656 2 xx
2 numbers: Product = 3666 & Sum = +5 (They are + 9 and 4 )
656 2 xx
= 26x + 9x - 4x 6 Decompose the middle term
322323 xxx Factoring by groupings
2332 xx Common factoring
Method 2: By Decomposition Formula
Factor 656 2 xx
6
6 6 xx
a
axax
2 numbers: Product = 3666 & Sum = +5 (They are + 9 and 4 )
6
4696
xx Place those 2 numbers inside the brackets
6
232323
xx Common factoring each bracket
6
23326
xx 2332 xx
Method 3: By Trials and Errors (X-Box)
Factor 656 2 xx
2332 xx
3 -2
2 3
6 -2
1 +3
5218
6 2
1 -3
5218
3 2
2 -3
549
3 -2
2 +3
549
3 x 2 = 6 (first term)
(3)(3) + (2)(-2) = 5 (middle term)
(-2)(3)= -6 (last term)
Mathematics 11 Page 2 of 2
Appendix: Factoring Non-Simple Trinomials Date:
RHHS Mathematics Department
Method 4: By Box Method
Factor 656 2 xx
2 numbers: Product = 3666 & Sum = +5 (They are + 9 and 4 )
26x
6
2332 xx
26x +9x
-4x 6
26x +9x
-4x 6
Method 5: By Method
Factor 656 2 xx
2 numbers: Product = 3666 & Sum = +5 (They are + 9 and 4 )
9
a:
3
2
9
6
4
a:
2
3
4
6
32 x 23 x
2332 xx
3x
-2
2x +3
Factor each row
and each column Place the 2 numbers
at each corner
MHF4U - Advanced Functions Page 1 of 2 Power Functions Date:
RHHS Mathematics Department ©2010 – E. CHOI
Polynomial Function The function f is a polynomial if ...)( 01
22
22
11 axaxaxaxaxaxf n
nn
nn
n ++++++= −−
−− .
na are real numbers and n is a whole number. The degree of the function is n if the leading coefficient, .0≠na
Examples 1589)( 21057)( 2523 −+−=+++= xxxxgxxxxf A power function is a polynomial (1 term) of the form naxy = , where n is a whole number.
Examples 2 3 )( 6)( rrAxxf π=−= Power functions (Monomial) have similar characteristics depending on whether their degree is even or odd. Even-degreed Power Function: Even-degree power functions have line symmetry in the y-axis, x = 0.
Odd-degreed Power Function: Odd-degree power functions have point symmetry about the origin, (0, 0)
Power Functions Summary
Function ay = axy = 2axy = 3axy = 4axy =
Example 3=y xy 4= 25xy = 31xy = 42xy = Degree 0 1 2 3 4 Name Constant Linear Quadratic Cubic Quartic Type / Odd-degreed Even-degreed Odd-degreed Even-degreed
Domain Rx∈ Rx∈ Rx∈ Rx∈ Rx∈ Range ay = Ry∈ Ryy ∈≥ ,0 Ry∈ Ryy ∈≥ ,0
End Behaviour (a > 0) Extends from Q2 to Q1 Extends from Q3 to Q1 Extends from Q2 to Q1 Extends from Q3 to Q1 Extends from Q2 to Q1
Symmetry Line Point Line Point Line
Graph
Bracket Interval Inequality Number line In words
),( ba bxa << x is greater than a and less than b ],[ ba bxa ≤≤ x is greater than or equal to a and
less than or equal to b ),[ ∞a ax ≥ x is greater than or equal to a
),( ∞−∞ ∞<<∞− x x is an element of the real numbers Refer to text book P. 8 for more interval summary
MHF4U - Advanced Functions Page 2 of 2 Power Functions Date:
RHHS Mathematics Department ©2010 – E. CHOI
Example 1: Analyzing the Power functions For the following power functions, determine i) the degree and name of the function ii) the type of the function (odd-degreed, even degreed) iii) the domain and range iv) the end behaviour v) the symmetry
a) 34xy −= b) 4xy π= Example 2: Domain and Range of Ordered Pairs For the set of ordered pairs given below, determine i) the domain and range ii) if the ordered pairs represent a function or not
a) ( ) ( ) ( )0,5,4,2,1,3 − b) ( ) ( ) ( )2,2,7,3,5,2
Homework: P.12 # 3 – 11, 14, 16 – 18
MHF4U - Advanced Functions Page 1 of 2 Preview for Unit 1 Date:
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1) Evaluate
a) 40 b)
05 c) 05 d) 15− e)
00 f) 25− g) 30 h) 25−−
2) Label as TRUE or FALSE
a) aa
a 313+=
+ b) 551
5+=
+ aa c) ( ) ( )
rh
rrh 2424 −
⋅=− d) ( ) 3
6332
6+
⋅=+ rr
3) Simplify
a) ab
ba3
15 2 b) 24
14
412
yxxy c) ( )( )
396 yx d)
yxyx
4
346
e) ( )728ab f) ( )6434 25 −− yxyx g)
32
543−
h)
47
253
xx+
−
4) Given: ( ) 542 −+= xxxf , evaluate
a) ( )4f b) ( )hf c) ( )4+af d) ( ) ( )52 ff − e) ( ) ( )bfaf − f) ( ) ( )xfhxf −+
5) Given ( ) 42 23 −+= aaag , evaluate
a) ( )2−g b) ( ) ( )h
xghxg −+
6) Factor the following
a) xx 92 − b) 92 −x c) 5364 1812 −− + baba
d) ( ) ( )524 115125 +−+ xxxx e) 13 −x f) 14 −x
7) Solve for x
a) xax 35 =+ b) 252 =x c) 302 =− xx d) 0187 24 =−− xx
8) Find the slope
a) between the points (4, -2) and (-7, -3) b) perpendicular to the line with slope of 32−
c) between the points (6, -2) and (6, -5) d) between the points (-5, 4) and (7, 4)
MHF4U - Advanced Functions Page 2 of 2 Preview for Unit 1 Date:
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9) Find the equation of the line for each of the following. Write your answer in BOTH the form bmxy += and 0=++ CByAx
a) with slope 7, through ( )5,2 −P
b) through the points ( )8,3− and ( )2,5 −
c) parallel to 0147 =−+ yx and through ( )11 ,1−P
d) with slope 41 and with the same x-intercept as 24 −= xy
10) Sketch the graphs of the following
a) 32 −= xy b) 01643 =−− yx c) ( ) 53 2 −+= xy
Answers
1a) 0 b) undefine c) 1 d) 51 e) indetermine f)
251 g) 0 h)
251− 2a) T b) F c) F d) F
3a) a5 b) 3
123xy c) xy18 d) 26y e) 1472097152 ba f) 3
10y
g) 851− h) ( )
( )28562
+−
xxx
4a) 27 b) 542 −+ hh c) 27122 ++ aa d) -33 e) ( )( )4++− baba f) hhxh 42 2 ++
5a) -16 b) hxhxhx ++++ 2266 22
6a) ( )9−xx b) ( )( )33 +− xx c) ( )6
3 326b
baa + d) ( ) ( )53315 24 −++− xxxx
e) ( )( )11 2 ++− xxx f) ( )( )( )111 2 ++− xxx
7a) 3
5−−
a b) 5± c) 6, -5 d) 3, -3 8a)
111 b)
23 c) undefine d) 0
9a) 0197or 197 =−−−= yxxy b) 01745or 4
1745
=−++−
= yxxy
c) 047or 47 =−++−= yxxy d) 0182or 81
41
=−−−= yxxy
10a)
b) c)
MHF4U - Advanced Functions Page 1 of 2 Characteristics of Polynomial Functions Date:
RHHS Mathematics Department ©2010 – E. CHOI
In general 1) The graph of a polynomial function of degree 1 is a straight line. The graph of a degree-2 polynomial is a U-shaped parabola. The graphs of many degree-3 polynomial are S-shaped parabola. 2) The zeros of a polynomial function f(x) correspond to the x-intercepts in the graph of f(x). Use the x-intercepts and several other points to graph f(x). Example 1 Determine the key features of the graph of each polynomial function. Use these features to match each function with its graph. State the number of x-intercepts, the number of maximum and minimum points, and the number of local maximum and local minimum points for the graph of each function. How are these features related to the degree of the function?
a) 142)( 23 ++−= xxxxf b) 4510)( 24 −++−= xxxxg
c) xxxxh −+−= 35 52)( d) 316)( 26 +−= xxxp i) ii) iii) iv)
142)( 23 ++−= xxxxf 4510)( 24 −++−= xxxxg xxxxh −+−= 35 52)( 316)( 26 +−= xxxp
Name
Sign of leading coefficient
End behaviours
y - intercept
Graph
# of x - intercept(s)
Degree
# of local max point
# of local min point
# of max point
# of min point
MHF4U - Advanced Functions Page 2 of 2 Characteristics of Polynomial Functions Date:
RHHS Mathematics Department ©2010 – E. CHOI
Key features of Graphs of polynomial functions • A polynomial function of degree n, where n is a whole number greater than 1, may have at most n – 1 local minimum and local
maximum points. • For any polynomial function of degree n, the nth differences - are equal (or constant) - have the same sign as the leading coefficient - are equal to [ ] !)1)(2)(3)...(2)(1( annnna =−− , where a is the leading coefficient.
Example 2 Each table of values represents a polynomial function. Use finite differences to determine i) the degree of the polynomial function ii) the sign of the leading coefficient iii) the value of the leading coefficient a)
x y 1st differences
2nd differences
3rd differences
4th differences
-3 -36 -2 -12 -1 -2 0 0 1 0 2 4 3 18 4 48
b)
x y 1st differences
2nd differences
3rd differences
4th differences
-2 -54 -1 -8 0 0 1 6 2 22 3 36 4 12 5 -110
Homework P. 26 #1,2,3i,ii, 5,6,7i,ii, 8acdef,11-18
MHF4U - Advanced Functions Page 1 of 4
Graphs of Polynomial Functions Date:
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Name of Polynomial Functions
For Odd-Degree Polynomial Functions
If positive leading coefficient ( 132 23 xxxy ), the graph extends from Q3 to Q1.
If negative leading coefficient ( 4623 xxxy ), the graph extends from Q2 to Q4.
For Even-Degree Polynomial Functions
If positive leading coefficient ( 2069 234 xxxxy ), the graph extends from Q2 to Q1.
If negative leading coefficient ( 146 24 xxxy ), the graph extends from Q3 to Q4.
In general:
For a polynomial ....1 nn bxaxy
There are at most 1n concavities to the function.
n is ODD n is EVEN
a > 0 End behaviors are uphill End behaviors are positive
a < 0 End behaviors are downhill End behaviors are negative
0 1 2 3 4 5 6 7 8 9 10 Constant Linear Quadratic Cubic Quartic Quintic Sextic Septic Octic Nonic Decic
Quadratic: Degree = 2, cbxaxy 2
)3)(4(
122
xxy
xxy
25.125.02 xy
2
2
)1(
12
xy
xxy 43
136
2
2
xy
xxy
2 different real roots 2 equal real roots
Order 2
no real root
(2 imaginary/complex conjugate roots)
(always 2 1,12 ii )
Cubic: Degree = 3, dcxbxaxy 23
)2)(1)(1(
22 23
xxxy
xxxy 3xy 2
23
)3)(5(
4521
xxy
xxxy
)64)(4(
2410
2
3
xxxy
xxy
3 different real roots 3 equal real roots
Order 3 2 equal real roots (Order 2)
& 1 different real root 1 real root
& 2 imaginary roots
MHF4U - Advanced Functions Page 2 of 4
Graphs of Polynomial Functions Date:
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Quartic: Degree =4, edxcxbxaxy 234
)4)(1)(2)(3(
242636 234
xxxxy
xxxxy 4)2( xy 2
234
)2)(3)(1(
12496
xxxy
xxxxy 22
234
)2()3(
3612112
xxy
xxxxy
4 different real roots 4 equal real roots
(Order 4)
2 equal real roots (Order 2)
& 2 different real roots
2 pairs of equal real roots
(Order 2)
)136()4(
2088192
22
234
xxxy
xxxxy )136)(5)(4(
2603313
2
234
xxxxy
xxxxy 54 xy 43243 24 xxxy
2 equal real roots (Order 2)
& 2 imaginary roots
2 different real roots
& 2 imaginary roots 4 equal imaginary roots 2 pairs of 2 imaginary roots
Example 1
Given the following functions,
i) State the x-intercepts (i.e. zeros of the function.) ii) State the y-intercept.
iii) State the degree of the functions iv) End behaviors
v) Sketch the graph. vi) Describe the roots
a) ( ) ( 1)( 2)( 3)f x x x x
b) 2( ) 3( 1)( 2)f x x x
MHF4U - Advanced Functions Page 3 of 4
Graphs of Polynomial Functions Date:
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c) ( ) ( 3)( 2)( 1)( 4)f x x x x x
d) 22 )12()3()( xxxxf
Example 2
From the graph below, determine the equation f(x). Express the equation in standard form.
(2, -45)
Homework: P. 39 #1, 2, 6, 9, 10, 11, 13
MHF4U - Advanced Functions Page 4 of 4
Graphs of Polynomial Functions Date:
RHHS Mathematics Department ©2013 – E. CHOI
Exercise:
1) Given the following functions,
i) State the x-intercepts ii) State the y-intercept.
iii) State the degree of the functions iv) Sketch the graph.
v) Describe the roots
a) )2)(1( xxy b) )3)(1)(2( xxxy c) )4)(1)(3)(2( xxxxy
d) 2)2)(1( xxy e)
2)3( xxy f) 2)1)(2)(1( xxxy
g) 22 )2()2( xxy h)
2)3)(2)(13)(32( xxxxy i) )49)(1)(1( 22 xxxxy
2) Sketch a possible general shape for the graphs of the following:
A sixth-degree function that has a coefficient of x6 that is
(i) positive: (ii) negative.
3) From the graph below, determine the equation f(x). Leave the answer in i) factor form, ii) standard form.
Answers
3i) 2)2)(1)(1()( xxxxf ii) 4434)( 234 xxxxxf
1) a) b) c) d) e) f) g) h) i)
i) -1, 2 -2, 1, -3
2, -3,
-1, 4 1, -2 0, 3 1, -2, -1 -2, 2
-3/2,
1/3,-2,3
1,
2/3, -2/3
ii) -2 -6 -24 -4 0 -2 16 -54 -4
iii) 2nd
3rd
4th
3rd
3rd
4th
4th
5th
5th
v) 2 diff real 3 diff real 4 diff real
1 real & 2
equal real
1 real & 2
equal real
2 real & 2
equal real
2 pair of 2
equal real
3 real and 2
equal real
3 real or 3
real & 2
complex
(3, -8)
MHF4U - Advanced Functions Page 1 of 2
Symmetry Date:
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Symmetry
To determine the symmetry, replace –x with x in the relation, then determine the symmetry according to the
following.
Even Function:
An even function satisfies
f(-x)=f(x) for all x in its
domain. The graph of an
even function is
symmetric about the y-
axis.
Odd Function:
An odd function satisfies
f(-x)=-f(x) for all x in its
domain. The graph of an
odd function is symmetric
about the origin.
For Polynomial functions,
An even-degree polynomial function is an even function if the exponent of each term is even.
An odd-degree polynomial function is an odd function if the exponent of each term is odd.
Example 1
Determine whether the function is even, odd, or neither. Check the symmetries from the screen captures.
(a) 2
2( )
9
xf x
x
(b)
9
4)(
2
3
x
xxxf
(c) 1 3
( )1
xf x
x
(d) xxxxf 293)( 35
(e) 452)( 24 xxxf (f) 464)( 236 xxxxf (g) 212)( xxxxf
MHF4U - Advanced Functions Page 2 of 2
Symmetry Date:
RHHS Mathematics Department ©2010 – E. CHOI
Exercise:
1. State whether the functions whose graphs are given are even, odd, or neither.
a) b) c)
d) e) f)
2. Determine whether each of the following functions is even, odd, or neither.
a) 52)( xxf b) xxxf 32)( 2 c) xxxf 33)(
d) 1
1)(
2
xxf e)
1
1)(
3
xxg f)
24 32)( xxxh
g) xxxxF 35)( h) 1
)(2
x
xxG i)
5
3
1)(
xxxG
j) 1
)(2
3
x
xxf k)
4
1)(
2
2
x
xxd l) 31243)( 234 xxxxk
Answers:
1a) odd b) neither c) even d) odd e) neither f) even
2a) neither b) neither c) odd d) even e) neither f) even g) odd h) odd i) odd j) odd k) even l) neither
Homework: P. 39 #3-5,7,8,12,14
MHF4U - Advanced Functions Page 1 of 2
Transformations of Functions: qpxkafy +−= )]([ Date:
RHHS Mathematics Department ©2012 – E. CHOI
0 3 6-3-6 x
y
3
6
-3
-6
Example 2: Translation of Cubic function Rewrite 5)22(3 3 ++−= xy into qpxkafy +−= )]([ ,state the values of a, k, p and q and sketch the relation and state the domain and range.
Example 1: Translations of functions
Given )(xfy = , sketch 52212 +
−−= xfy
State the values of a, k, p and q and sketch the relation.
)(xfy =
ORDER Stretches Reflections Translations
MHF4U - Advanced Functions Page 2 of 2
Transformations of Functions: qpxkafy +−= )]([ Date:
RHHS Mathematics Department ©2012 – E. CHOI
Example 3: Describing Transformations from an equation a) State the a, k, p & q, and then describe the function and its transformations. b) State the domain and range. c) State the vertex and axis of symmetry (for functions that are even only)
i) 4
)5(312)(
−= xxf ii) 4)]3(2[
41 5 +−−= xy
Homework: P. 49 #1-7,10,12,15
MHF4U - Advanced Functions Page 1 of 3 The Slope of a Secant and Slope of a Tangent Date:
RHHS Mathematics Department ©2013 – E. CHOI
Slopes of Secants and Average Rates of Change A rate of change is a measure of how quickly one quantity (the dependent variable) changes with respect to another quantity (the independent variable). Average rates if change
• Represent the rate of change over a specific interval • Correspond to the slope of a secant between two points P1( 11 , yx ) and P2( 22 , yx ) on a curve.
• Average rate of change 12
12
xxyy
xy
−−
=∆∆
= (Slope of secant!!)
Recall: Equation of Lines Example 1: Average Rate of Change The table shows the amount of money in a bank saving account over 6 months. a) Draw a time-amount graph. b) Calculate the average rate of change over the 5 months period. Example 2: Equation of line given two points Find a linear equation passing through the points (-1, -1) and (2, 5).
Time (months) Amount ($)
0 500 1 700 2 900 3 900 4 600 5 1200
(2) Equation of a straight line Given slope m and point (x1, y1), we can write equation of a line by point-slope form: 1 1( )y y m x x− = −
(1) Slope of a straight line As in the figure,
12
12
12
22
,,
xxyy
xym
yyyxxx
AB −−
=∆∆
=
−=∆−=∆
0 x
y ),( 22 yxB
),( 11 yxA
)( 12 xxx −=∆
)( 12 yyy −=∆
Time (months)
Am
ount
($)
Amount of money in the bank
MHF4U - Advanced Functions Page 2 of 3 The Slope of a Secant and Slope of a Tangent Date:
RHHS Mathematics Department ©2013 – E. CHOI
Example 3 Find the slope of the tangent to the curve f (x) = x2 at the point P(1, 1). Take a point Q(x, f (x)) on the curve, then PQ is a secant
(1) Tangent and secant i) ii) Tangents Secants i) Tangent is a straight line touching the curve at one and only one point. ii) Secant is a straight line cutting the curve at two distinct points. (2) Slope of a tangent We need two points to find the slope of a straight line. How can we find the slope of the tangent at one point only. Note the following example.
t t s s
P(1, f(1))
MHF4U - Advanced Functions Page 3 of 3 The Slope of a Secant and Slope of a Tangent Date:
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From Example 3, the steps of finding the slope of the tangent can be summarized as follows: Slope of Tangent to a Curve (Without using limits) To find the slope of tangent to the function y = f (x) at x = a.
(1) Find slope of secant: ( ) ( )s
y f x f amx x a
∆ −= =∆ −
(in simplified form)
(2) Let 0.001x a= ± , find 1 2 1 2and , ( )s s s sm m m m<
(3) Then 1 2s t sm m m< < .
mt can be approximated. Example 4 Find the equation of the tangent line to the function 3( ) 3 2 4f x x x= + − at the point (-1, -9).
Homework: P. 62 #1-5,7 P. 70 #2,3,5
MHF4U - Advanced Functions Page 1 of 1 Slope of Tangents Date:
RHHS Mathematics Department ©2013 – E. CHOI
Exercise 1) Find the slope of tangent to the curve at the given point.
a) 22 += xy at (2, 6). b) ( ) 962 +−= xxxf at (1, 4) c) ( ) 3−= xxg at (7, 2)
d) x
y 1= at
21,2
e) ( )2
1−
=x
xh at (3, 1)
f) 2
1x
y = at (1, 1)
g) x
xy−
=1
at (2, -2)
h) ( )x
xf 1= at (1, 1)
2) i) Find the equation of the tangent at the given point
ii) Graph the curve and the tangent. a) 92 −= xy at 3=x b) ( ) 142 −+= xxxg at (-2, -5) c) 24 xy −= at 2=x d) ) e)d) ( ) 23 −= xxg at (2, 6) f)e) ( ) 38 xxf −= at (2, 0)
g)f)x
y−
=1
1 at (2, -1)
h)g) xy += 16 at 5=y Answers
1a) 4 b) -4 c) 41 d)
41− e) -1 f) -2 g) 1 h)
21−
2a) 186,6 −= xy b) 5,0 −=y c) 84,4 +−=− xy d) 1812,12 −= xy e) 2412,12 +−=− xy
f) 3,1 −= xy g) 1041
101
+= xy
MHF4U - Advanced Functions Page 1 of 4 Average and Instantaneous Rates of Change Date:
RHHS Mathematics Department ©2012 – E. CHOI
Introduction to Limits Example 1 Determine the equation of the normal to the curve 3( ) 3 2f x x x= − − at the point (-1, 5)
(x, f(x))
Let h be the horizontal displacement between P & Q on x-axis. • As Q P, • Secant PQ is getting closer to
become Tangent at P • h 0 • MPQ MP •
hafhafM PQ
)()( −+=
• h
afhafMhP
)()(lim0
−+=
→
y=f(x)
y
x
p
Q1
Tangent Line Q3 Q2
a a + h h
(a+h, f(a+h))
(a, f(a))
y=f(x)
• As Q P, • Secant PQ is getting closer
to bacome Tangent at P • x a • MPQ MP •
axafxfM PQ −
−=
)()(
• ax
afxfMaxP −
−=
→
)()(lim
y
x
p Q1
Tangent Line Q3 Q2
a x3 x2 x1
(a, f(a))
MHF4U - Advanced Functions Page 2 of 4 Average and Instantaneous Rates of Change Date:
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Example 2 A pebble is dropped from a cliff with height of 80 m. After t seconds, it is s metres above the ground, where
2( ) 80 5 ,0 4s t t t= − ≤ ≤ . (a) Find the average velocity of the pebble between the times t=1 s and t =3 s. (b) Find the average velocity of the pebble during the 2nd second.
(1) Distance and velocity Distance with respect to time is usually denoted as: )(ts Velocity with respect to time is usually denoted as: )(tv (2) Average velocity (va) In this case, you are always given two different times (i.e. .2 1 and tt )
The formula is 12
12 )()(tt
tststsva −
−=
∆∆
= or h
tshts )()( −+=
(3) Instantaneous velocity (vI) It is the velocity at a particular point of time (i.e. t=a). We say “change of distance s with respect to time t at t=a”.
(1) Find the slope of secant: ( ) ( )s
s s t s amt t a
∆ −= =∆ −
(in simplified form)
(2) Let 0.001t a= ± , find 1 2 1 2and , ( )s s s sm m m m< .
(3) Then 1 2s t sm m m< < .
vI=mt can be determined. Using Limits
htshtsV
hI)()(lim
0
−+=
→
s
t
y = s(t)
t1
s(t)
s
t
y = s(t)
t2 t1
s(t2)
s(t1)
MHF4U - Advanced Functions Page 3 of 4 Average and Instantaneous Rates of Change Date:
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Example 3 A ball is tossed straight up so that its position s, in metres, at time t, in seconds, is given by
2305)( 2 ++−= ttts . What is the velocity of the ball at 4=t ? (4) Rates of change (R) Similarly, we can write the “Rate of Change, R” as shown in the following examples. (a) Rate of change of volume V with respect to time t at t=a is:
(1) Find the slope of secant: ( ) ( )s
V V t V amt t a
∆ −= =∆ −
(in simplified form)
(2) Let 0.001t a= ± , find 1 2 1 2and , ( )s s s sm m m m< .
(3) Then 1 2s t sm m m< < .
R=mt can be determined. (b) Rate of change of surface area of sphere A with respect to radius r at r=a is:
(1) Find the slope of secant: ( ) ( )ar
aArArAms −
−=
∆∆
= (in simplified form)
(2) Let 0.001r a= ± , find 1 2 1 2and , ( )s s s sm m m m< .
(3) Then 1 2s t sm m m< < .
R=mt can be determined.
MHF4U - Advanced Functions Page 4 of 4 Average and Instantaneous Rates of Change Date:
RHHS Mathematics Department ©2012 – E. CHOI
Example 4: David drains the water from a hot tub. The tub holds 1600 L of water. It takes 2 h for the water to drain
completely. The volume of water in the tub is modeled by ( ) ( )212091 ttV −= , where V is the volume in litres at
t minutes and 1200 ≤≤ t . Find the instantaneous rate of change exactly 60 minutes. Example 5 Estimate the instantaneous rate of change of the surface area of a sphere with respect to its radius when the radius is 10 cm.
Homework: P. 63 #9,10,12,13 P. 72 #7,9,10,12
MHF4U - Advanced Functions Page 1 of 3
Division of Polynomials & Remainder Theorem Date:
RHHS Mathematics Department ©2014 – E. CHOI
Division of polynomials can be done using a method similar to that used to divide whole numbers.
Remainder Theorem: If f(x) is divided by (x-p) giving a quotient q(x) and a remainder r, then ).( pfr
Example 1
i) Perform the following divisions and express the answers in the form f(x)=d(x)q(x)+r(x).
ii) Use the remainder theorem to compare the remainder in i) for a and b only.
a) )3()3432( 23 xxxx b) )2()4( 23 xxx
c) )31()47423( 2234 xxxxxx
)()()()( xrxqxdxf
1430 = ______________
However, if you are dividing a polynomial by something
more complicated than just a simple monomial, you'll need a
different method for simplification. The method is called
"polynomial long division" or “polynomial synthetic
division” method.
?31430
476
3 ) 1430
12
23
21
20
18
2
divisor
d(x)
quotient
q(x)
dividend
f(x)
remainder
r(x)
3
1430
xd
xrxq
xd
xf
MHF4U - Advanced Functions Page 2 of 3
Division of Polynomials & Remainder Theorem Date:
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Example 2
When a certain polynomial is divided by 3x , its quotient is 752 xx and its remainder is -1. What is the
polynomial?
Example 3:
Use synthetic division to divide
a) 3 2(4 5 3 7) ( 2)x x x x b)
3 2(12 2 11 16) (3 2)x x x x
Example 4: Synthetic Division (Divisor is not linear)
Perform the following divisions and express the answers in the form f(x)=d(x)q(x)+r(x).
a) )2()8254( 2234 xxxxxx
Definition of Synthetic division
Synthetic division is a shortcut method of polynomial division in the special case of
dividing by a linear factor )( bax . It is generally used, however, not for dividing
out factors but for finding zeroes (or roots) of polynomials
3 1 4 3 0
4 7 6
2 2 2
MHF4U - Advanced Functions Page 3 of 3
Division of Polynomials & Remainder Theorem Date:
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b) )2()781132( 2234 xxxxxx
Example 5
When 2x is divided into f(x), the remainder is 3. Determine the remainder when 2x is divided into each of
the following:
a) 1)( xf ; b) )74()( xxf
Example 6
Use the remainder theorem to determine the remainder when 2452 23 xxx is divided by )1)(2( xx .
Example 7
If 3)( 23 xgxmxxf is divided by 1x , the remainder is 3. If f(x) is divided by 2x , the remainder is
–7. What are the values of m and g?
Homework:
P. 91 #1-9
32 x
22 xx ) 2452 23 xxx
xxx 422 23
23 2 x
633 2 xx
83 x
From division
MHF4U - Advanced Functions Page 1 of 2 Factor Theorem Date:
RHHS Mathematics Department ©2012 – E. CHOI
Example 1 Factor 352 2 −− xx Example 3 Factor 241423 +−− xxx Example 4 Factor
(a) 33 yx − b) 6427 3 +y c) 33 )(405 yxu +−
The Remainder Theorem tells us that if the remainder is zero on division by )( px − , then 0)( =pf . If the remainder is zero, then )( px − divides evenly into )(xf , and )( px − is a factor of )(xf . Conversely, if px − is a factor of )(xf , then the remainder )( pf must equal zero. These two statements give us the Factor Theorem, which is an extension of the Remainder Theorem. Definition
)( ax − is a factor of )(xf if and only if 0)( =af .
The Sum and Difference of Cubes
))((
))((2233
2233
yxyxyxyx
yxyxyxyx
+−+=+
++−=−
Example 2 a) Show that 2−x is a factor of 653 23 −+− xxx . b) Factor 653 23 −+− xxx
MHF4U - Advanced Functions Page 2 of 2 Factor Theorem Date:
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Example 5 Prove that )( ax − is a factor of abcxcabcabxcbax −+++++− )()( 23
Example 6
Given the function 22 23 +−−= xxxy
i) State the intercepts (if any)
ii) State the degree
iii) Sketch and describe the roots.
iv) Determine the symmetry
0 x
y
Homework: p.102 #1 – 4, 6, 9, 10, 12 – 15, 20
Advanced Functions Page 1 of 1
Division of Polynomials 2 Date:
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Example 1: Synthetic Division (Divisor is not linear)
Perform the following divisions and express the answers in the form f(x)=d(x)q(x)+r(x).
a) )2()8254( 2234 xxxxxx
b) )2()781132( 2234 xxxxxx
MHF4U - Advanced Functions Page 1 of 2 Factor Theorem Extended Date:
RHHS Mathematics Department ©2012 – E. CHOI
Example 1 Factor 1096 23 −−− xxx Example 2 Factor 107176 234 −−+− xxxx
If the coefficient of the highest order term in the polynomial is not one, there could be a factor of the form
)( pqx − or )( pqx + where q is not equal to one and thus qp or
qp
− will not be an integer.
MHF4U - Advanced Functions Page 2 of 2 Factor Theorem Extended Date:
RHHS Mathematics Department ©2012 – E. CHOI
Example 3 A function g(x) with integer coefficients has the following properties:
0)3( =g , 043
=
−g , )2( +x is a factor of g(x), 84)1( −=g .
Determine g(x) if it is a) cubic b) quartic Example 4 Factor fully bxbaxbaax 2)2()( 23 +−−+−+ .
Homework P.102 #5,7,10,11,17–19
MHF4U - Advanced Functions Page 1 of 2 Solving Polynomial Equations Date:
RHHS Mathematics Department ©2012 – E. CHOI
Example 1 Solve 04123 23 =−−+ xxx Example 2 Solve a) 05139 23 =+++ xxx b) 0543 =−+ xx
Recall: Quadratic Formula, a
acbbx2
42 −±−= . Any equation of the form 0)( =xf can be solved if
)(xf can be expressed as a combination of linear and quadratic factors.
MHF4U - Advanced Functions Page 2 of 2 Solving Polynomial Equations Date:
RHHS Mathematics Department ©2012 – E. CHOI
Example 3 Solve 02136 23 =++− xxx Example 4 Solve 02524 24 =−− xx Example 5 Solve 9)35)(55( 22 =+−−− xxxx
Homework P.110 #3 – 5, 6 – 8, 14, 15, 18, 20, 22
Advanced Functions Page 1 of 2 Families of Polynomial Functions Date:
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A family of functions is a set of functions that have the same characteristics. Polynomial functions with the same zeros are said to belong to the same family. The graphs of polynomial functions that belong to the same family have the same x-intercept but have different y-intercepts (Unless zero is one of the x-intercept). An equation for the family of polynomial functions with zeros naaaa ,...,,, 321 is
),)...()()(( 321 naxaxaxaxky −−−−= where k 0, ≠∈ kR . Example 1: The zero of a family of quadratic functions are 2 and -3. a) Determine an equation for this family functions. b) Write equations for two functions that belong to this family. c) Determine an equation for the member of the family that passes through the point (1, 4). Example 2: The zero of a family of cubic functions are -2, 1 and 3. a) Determine an equation for this family functions. b) Write equations for two functions that belong to this family. c) Determine an equation for the member of the family whose graph has a y-intercept of -15. d) Sketch graph of the function in part c).
Advanced Functions Page 2 of 2 Families of Polynomial Functions Date:
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Example 3: a) Determine a simplified equation for the family of quartic functions with zeros 1± and 32± . b) Determine an equation for the member of the family whose graph passes through the point (2, 18). Example 4: Determine an equation for the quartic function in factors form represented by this graph. Homework
p.119 # 1 – 3, 6, 7, 9, 11, 13, 15, 19 ,22
MHF4U - Advanced Functions Page 1 of 2 Solving Polynomial Inequalities Date:
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When the equal sign in an equation is replaced by any of ≤≥<> or ,,, , the equation becomes an inequality. 1) 2) 3) 4) Example 1 Solve and show on a number line a) 062 <−− xx b) 0352 2 ≥−+ xx Example 2 Solve and show on a number line
09923 <+−− xxx
3 x
3<x
3 x
3≥x -2 5
x
5or 2 >−< xx -2 5
x
52 ≤≤− x
> or <
≥ or ≤
x x
x
MHF4U - Advanced Functions Page 2 of 2 Solving Polynomial Inequalities Date:
RHHS Mathematics Department
A polynomial inequality with a double root Example 3 Solve 2 2( 1)( 2) 0x x x− + − ≥ algebraically. Check graphically.
A polynomial inequality with imaginary roots Example 4 Solve 2 2(4 )( 2 2) 0,x x x− − + < and check by graphical calculator.
x
x
Homework p.138 # 2, 4, 6, 9, 12, 13
Advanced Functions Page 1 of 2
Piecewise Functions and Continuity of Functions Date:
RHHS Mathematics Department
Piecewise Function: A function that uses different formulas for different parts of its domain.
Continuous Functions: When a function is being continuous at a point, it means that the graph passes
through the point without a break.
Example 1:
Consider
1
3)(
2
x
xxf
1 if
1 if
x
x
Determine the continuities of the functions.
Example 2:
Consider
4
1
)( 2x
x
xf
2 xif
22 if
2 if
x
x
Determine the continuities of the functions.
Advanced Functions Page 2 of 2
Piecewise Functions and Continuity of Functions Date:
RHHS Mathematics Department
Example 3:
Consider
2)4(
2)4(
)(2
2
x
x
x
xf
2 xif
22 if
2 if
x
x
Determine the continuities of the functions.
Functions continuous in all real numbers
Exercise
1) Find the value(s) of x at which the functions are discontinuous.
a) 3
9)(
2
x
xxf b)
x
xxg
47)(
c)
3
2 1)(
x
xxh
d) 9
4)(
2
x
xxf e)
6
13)(
2
xx
xxg f)
3,1
3,)(
xx
xxxh
2) Sketch a graph of
3,5
3,1)(
xx
xxxh
3) Sketch a graph of
.0,3
.0,)(
2
x
xxxf Is the function continuous?
4) Examine the continuity of the function
.2 if ,2
.21 if ,1
1 if ,
)(
x
x
xx
xf