Summary of Lesson (2-4-13) (Ryan)

LESSON 1

Addition of Case 7 and 8

Case 7

y=3e^x

y=-e^x

y=e^0 (This is not exponential)
y=3e^x + 4
y= -e^x -4
y=e^-x

y = | x |

if x>0, y=x
if x<0, y=-x
if x=0, y=0

Case 8

y= 3lgx

y= -lgx (Reflection about the y-axis)

y= lnx

y= 3lgx + 4

y= -lgx - 4

y= log2 x

Case 2a

y= |-3/4x|

or

y=abs (-3/4 x)

y=|3/4x|

or

y=abs (3/4 x)

Hint: For EOY Mr Johari can convert normal graph problems to a trigonometry problem or add in different topics

Let y = f(x)

Case 1: y = -f(x)

this would be a reflection about the x-axis

Case 2: y = f(x) + c

this would be a vertical shift (moving up and down the graph)

Case 3: y = f(-x)

this would be a reflection about the y-axis

case 4: y = absolute f(x)

when x > 0 , y = f(x)

when x < 0 , y = -f(x)


LESSON 2

Prove that

1                              1
_______     +  _________ = 1
log a ab           log b ab

Answer (Courtesy of Justin) :

log a a               log b b
_______     +  _________ =  log ab a (change to base a )+ log ab b = log ab ab = 1
log a ab           log b ab

Lesson summary 26/3 Shaun Ng

Quadratic inequalities: 

Step 1: Find roots (by factorising)

Step 2: Sketch and include x-intercept (roots)

Step 3: Check sign, <0 answer is part below x-axis

           >0 above x axis

Practice QNs Answers p26 - p29 below

Lesson Summary 13/3/13 Mason Sim_12

Practice 10 Page 158
Show that the equation x^2+(k+1)x+k=0 has real roots for all values of k .

Practice 12 Page 159
Show that the roots of the equation x^2-2x-p+2=0 are real and distinct if p>1 

Practice 15 Page 162

That should be all for todays lesson 

12/3 Math Scribe Post – Justin (11)

What we did today ~(•0•)~ 
- Quiz on laws of surds, logs, indices, and some log questions below
We did this to revise our concepts as it is important to understand the laws before we can manipulate them. Also because logs are fun. 

- Diagnostic Test on Quadratic Equations
To revise and re-emphasize on what are the key concepts in this topic. Also because quadratic is fun. 

- Peer Assessment 
This taught us how to improve our answers, in terms of mathematical concepts, explanations and presentation.
Presentation will ultimately affect your explanation, which then affects your mathematical concept which is a shown in your presentation. It's a vicious cycle. 
Through this cycle, we learn that we communicate with the reader when writing out math equations. We must explain what we're doing in the question well, or else it won't make any sense. This form of communication is key in exams. 

Mathematical concepts we learnt today ~(•0•)~ 
MISCONCEPTIONS: (reasons for our tests)
a^(-n) = a^(1/n) 
This is wrong as when the power is negative, your base is now a denominator, not your power. 
Make sure to get your concepts down for the exam, for they are our foundation! ^^

CONCEPTS: 
x + 1 = 0 
This is an equation. It shows that both sides are of the same value.
x + 1 > 0 
This is an inequality, it shows in the inequality between both sides in terms of value.
x + 1 
This is an expression, of not needed to be solved but rather simplified. 

Quadratic ... in a nutshell

by Mr Johari
Source: http://perfectmaths.com/quadratic-equations/

Lesson Summary 5 March

Steps (shortcut) to sketching ><

1) if a is positive, the graph has a min turning point
 if a is negative, the graph has a max turning point

2) c refers to the y intercept of the graph

3) D, the discriminant, affects the nature of the roots
if D is greater than zero, the roots are real and different
if D is equal to zero, the roots are real and similar
if D is less than zero, the roots are imaginary

4) in completing the square [ (x+h)^2+k ], the turning point is (h,k)

eg. y=x^2-3x+5
1) is a graph with a min turning point
2) has a y intercept of 5
3) is imaginary (D=-11)
4) has a turning point of (1.5, 2.75)

___________________________________________________________
Difference between plotting and sketching

                     Drawing/Plotting                 SKetching
Scale                       Y                               N
Accuracy                  Y                               N
Plotting Points            Y                               N
Label                       Y                               Y
Construction lines       Y                               N

Lesson Summary

Quadratic

-graph(parabola)
one turning point
line of symmetry( (X1+X2)/2 )
intercept to y ( c of y=ax^2+bx+c)
intercept to x  (X1 X2)
(a determine its shape)


-
(a+b)^2=a^2+2ab+b^2
a^2-b^2=(a+b)(a-b)


-function
(base on a vertical line test) <many to one>
basic form: ax^2+bx+c=0
roots:
    discriminant: D=b^2-4a
    D>0, real & different roots
    D=0, real & equal
    D<0, imaginary(complex root)
solving:
    cross method
    formula: x= ( -b+(-)√D )/ (2a)
    


sum of roots: -b/a
product of roots: c/a

x^2-(-b/a)x+c/a=0
x^2-(X1+X2)x+X1X2=0
x^2-(sum of roots)x+product of roots=0

Lesson summary 3.1

Discriminant table

Sorry for posting late:P
 Crystal

Lesson Summary (February 27, 2013)

Worksheet A02d
6) log₅(log₃x)=log₁₀₀100
log₅(log₃x)=1
log₃x=5¹
x=3⁵
  =243

7) log₅x=a            log√5y=b
express xy² as a power of 5
x=5ª                      y=√5ᵇ
                               =5^b/2

xy²=5ª x 5^2b/2
     =5^a+2b/2
     =5^a+b

*Don't cancel your answer even if it seems wrong (sometimes it might be correct) only cancel after realising the mistake and redoing.

Worksheet A02e
9) 6ⁿ + 6^n+1 + 6^+2
= 6ⁿ+6ⁿ6+6ⁿ6²
= 6ⁿ(1+6+36)
= 6ⁿ(43)
∴divisible by 43 for all natural values of n.

11) log₅(5-4x) = log√5(2-x)
                        = log₅(2-x)/log₅5^1/2
                        = log₅(2-x)/(1/2)
                        = log₅(2-x)²

5-4x = (2-x)²
        = 4 - 4x + x²
x²=1
x=±1

* Sorry for the late post

Lesson Summary (February 26th 2013)

Morning Lesson:

We learnt how to solve logarithms using different methods.

Method 1: Covert log form to exponential form.
Method 2: Use change of base formula first.

For example:

Afternoon Lesson:

From the tables we have calculated out, we can conclude that:

- Power:   log_a (x^p) = p log_a x

-  Product: log_a (xy) = log_a x + log_a y

- Quotient:  

A formula can also be inputed into your graphic calculator to help change base:

Sorry I will input the formula when I get the paper back ><

Otherwise, you can use this log rule to change base too:

- Change of base formula: 

http://www.mathwords.com/c/change_of_base_formula.htm

Things to take note:

log_a (x + y) ≠ log_a x + log_a y

log_a (x – y) ≠ log_a x – log_a y

Lesson Summary 19/2/2013: Wee Chang Han

Logarithm 

Laws of Logarithm:

1.

a^x = y

is equals to:

log a(y) = x

2. 

log10 = lg

log e = l(n), n = natural log.

3.

lg 1 = 0

log10 (1) = 0

10^0 = 1

Because:

loga(1) = 0

a ^0 = 1

where a ≠0.

4.

loga(a) = 1

where a ≠ 0, a >0

log10(10) = 1

because:

log10(10) = 1

10^1 = 10

Note:

5.

log10(-5) = error?

why?

-5 = 10^?

there is no value for the power of 10 to become -5, therefore log a base b = c, where a ≠ a negative integer or  < 0.

6.

loga(b) = log10(b) / log10(a)

            = ln(b) / ln(a)

            

caution!:

lg(b)/lg(a) ≠ b/a (do not cancel!!)

Correct:

lg(b)/ lg(a) = calculate both the top and the bottom.

7.

logc(ab) = logc(a) + logc(b)

8.

logc (a/b) = logc (a) - logc (b)

9.

log10(100) = log10(10)^2

                  = 2log10(10)

                  = 2

note:

loga(b)^n = n loga(b)

15/02/13 Lesson summary

Lesson Summary 14/2/13 Ryan Tan

Surds

Important = *

*

(2√3)^2

=2√3 x 2√3

=4√3√3

=4(√3)^2

=4(3)

=12

Conjugate Surds

*

(a+b)(a-b) = a^2 - b^2

(√a+√b)(√a-√b)=(√a)^2-(√b)^2

(a^1/2 + b^1/2) (a^1/2 - b^1/2) = (a^1/2)^2 - (b^1/2)^2

2√3^2=

a) 2√3 √3 (correct)

b)2(√3)^2 (correct)

c)2 . 3^1/2x2 (correct)

d) 4(3) (wrong)

e) √4(3) ^2 (wrong)

f) √4 √9 (wrong)

To convert irrational surds into rational surds

By similar surd

1/√2 x √2(to both numerator and denominator)

= √2 /2

By conjugate surd

(1/ √3 - √2) x √3 + √2 (to both numerator and denominator)

=√3 + √2/ 3-2

=√3 + √2 / 1

Lesson Summary 7/2/13 Shaun Ng

Level Test

Elementary math

Duration: 45 mins

30 marks

Topics: 

1. Algebraic manipulation 

• Factorisation
• Expansion

1. Algebraic Fraction       

3  Indices (no surds + logs) (quiz c)

1. Quadratic Function/graph plotting  (U shaped and n shaped)

Additional math

Duration: 45 mins

30 marks

Topics:

1. Polynomials
2. Remainder Factor Theorem
3. Cubic expression/equation
4. Partial fraction

(All Jumbled up)

(Start doing Exam Prep Questions)

Question for fun

Given that the roots of a quadratic function is -3 and 1. Find the function if x=-3 (x+3)=0

1. Case 1 

      the coefficient of x^2 is 2     (3)

1. Case 2

      y intercept is -6.              (3) 

Sketch the above function(s), showing clearly the x and y intercepts, and the turning point. State also the nature of turning point. (2)

y = f(x) = ax^2+bx+c

   = A(x+d)(x+e)

   = A(x_3)(x-1)

   = A(x^2+2x-3)

   =Ax^2+2Ax-3A

1. f(x) = 2(x+3)(x-1)

            = 2x^2+4x+6

1. f(x) = A(x+3)(x-6)

    x=0, f(0) = A(3)(-1) = -6

    A = 2

Hence, f(x) = 2(x+3)(x-1)

= 2x^2+4x-6

Nature of turning point = Minimum tp

Turning point = Two x intercepts added/2

Sub x value into answer for a.]]]

Lesson Summary 6th Feb - Wednesday ; Kaelan

Practice 1,2,3,4 Answers:

Practice 1:

1 = A(x+2) + B(x+1)

1 = (A+B)x + 2A + B

Compare Co-efficients. A+B = 0

    2A+B=1

Therefore:  A = 1; B = -1

   1             1

------   -   ------

(x+1)      (x+2)

b) Simple:

    9x+9                 9(x+1)                 9

--------------- =  -----------------  =  --------

(x+1)(x-2)          (x+1)(x-2)          (x-2)

c) 3x+5 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)

Sub x = -1

  2 = 2A

 A = 1

Sub x = -2, -1 = -B, B = 1

Sub x = -3, -4 = 2C, C = -2

Therefore :    1                      1                    2

                -------------   +  -------------   -  ------------

                  (x+1)               (x+2)              (x+3)

Practice 2 :

1 = Ax(x-1) + B(x-1) + Cx^2 OR        1=(Ax+B)(x-1)+Cx^2

1 = (A+C)x^2 + (B-A)x -B OR     x=1, C=1

Compare Coefficients OR       x=0, B=-1

B=-1 OR        x=-1, A = -1

A =-1    

C = 1

A question : 

        x^2

------------------

       9-x^2

Long divide to give :  -1 + 9/9-x^2

                                  = -1  + 9/(3+x)(3-x)

9 = A(3-x) + B(3+x)

x=3 , B=1.5

x=-3, A=1.5

 -1 + 3/(6-2x) + 3/(6+2x)

NOTE: ALWAYS FACTORISE, A NON FACTORISED POLYNOMIAL WILL NOT WORK. have fun in a loop.

Lesson Summary 5th February

Homework :

Linoit

Comment on the posts on the math blog about the error 

Prepare a thing plastic red file for filing . 

Corrections

Remember :

Rules of index 

Thanks to Justin 

For Comparison

In order to rid the denominator of the surd/square root  , we rationalize it . 

Rationalizing is the process whereby we multiply the denominator by another fraction with the numerator and denominator like it . We make use of the identities such as : (a+b)^2 = a^2 + 2ab + b^2 ,

(a-b)^2 = a^2-2ab+b^2 or (a+b)(a-b) = a^2-b^2 

Recall : Prime factorization 

It is a way of breaking down a compound number into  prime factors 

Partial Fraction Summary (Denzel)

Lesson Summary ~(•O•)~ 4/2/13

RECALL: 

~(•o•)~ Partial Fractions! Read through the summaries posted by yourself/others
~(•o•)~ Level Tests are on Week 7 & 8, starting mugging. 


HOMEWORK: 
~(•o•)~ Check what you haven't done on the homework spreadsheet. 
~(•o•)~ AceLearning for those who did badly on the various quizzes and tests. 
~(•o•)~ Assignment 01 and 02 are now due on Friday~!


REVISE: 
~(•o•)~ The 9 laws of indices: (here's a helpful table)

Please remember to fully revise all laws as they will be used in Indices and Surds. 


LESSON: 

~(•o•)~ Indices and Surds: Forms

Index Form: When there is an index for the base. (Eg. 2^6)
Surd Form: When the base has a squareroot with it (√5)
Radical Form: When the base has an fractional index (2^(2/123))

~(•o•)~ Indices and Surds: Rationalization 

Rationalization is the act of eliminating the squareroot (√x) from the denominator. 

BACK TO BASICS– "  1/2 VS 1/3"
How do we determine which is bigger? 
We change make them have an equal denominator. Thus, the LCM between 2 and 3 is 6. Therefore, we multiply 1/2 with 3/3 (Note how this is 1 whole), and 1/3 with 2/2. Eventually, we get "3/6 > 2/6"

ON TO SURDS! – "1/2 VS 1/√2"

How do we determine which is bigger? 

Once again, we change them to make them have an equal denominator. In this case, we change the denominator to 2. As the first fraction already has a denominator of 2, we will work with 1/√2. 

2 divided by √2 is √2. Thus, we multiply 1/√2 by √2/√2 (1 whole), and get √2/2. 

Therefore, we would get "1/2 < √2/2"

Rationalizing with a^2 - b^2 – "1/(3 - 2) VS 1/(√3 - √2)"

To rid of the surd denominator, we use the 'a^2  b^2 = (a + b)(a - b)' concept. 

In this scenario, our denominator is (√3 - √2) which is our (a - b). Thus, to rid of it, we multiply it by (√3 + √2) which represents our (a + b).

Thus, we multiply 1/(√3 - √2) by (√3 + √2)/(√3 - √2) <-- which is one whole! 

This gives us (√3 + √2)/(3 - 2) = (√3 + √2)/1 

Therefore, "1 < (√3 + √2)"

LESSON SUMMARY :D:D:D:D:DD:D

Heres what we've covered in class today...

1): Please do remember to organize your workings and provide necessary explanation otherwise the marker would not understand your concept.

2) FRACTIONS: What is a fraction?

A fraction is numerical quantity that is not a whole number 

Numerator/ Denominator

Whole Number- Both values of the numerator and the denominator are equal.

Improper Fraction-  The degree of the unknown in the numerator is higher than the denominator it is said to be a improper fraction 

Proper Fraction– The degree of the unknown in the numerator is lower than the denominator it is said to be a proper fraction.However, if the both the degrees are the same, it would NOT be a proper fraction

3)PARTIAL FRACTIONS

Definition: Partial fraction is breaking apart the final expression into its initial polynomial fractions.

How to solve a Partial Fraction:

Step 1: Factorising the denominator

Step 2: Separate

Step 3: Multiply through by the bottom so there no longer will be fractions

Step 4: Find the constant

Express it as a partial fraction

23 Jan Lesson Summary

Mr Johari's Interesting question...


(a) Solve the equation 2x^3 - 7x^2 - 7x + 30 = 0

2x^3 - 7x^2 - 7x + 30 = 0

(x + 2)(2x - 5)(x - 3) = 0

x = -2 or 3 or 2.5


(b) When the expression x^2 + bx + c is divided by x-2, the remainder is R. When the expression is divided by x+1, the remainder is also R

(i) Find the value of b

let f(x) = x^2 + bx + c

f(2) = f(-1)

4 + 2b + c = 1 - b + c

b = -1


(ii) When the expression is divided by x-4, the remainder is 2R. Find the value of c and R

f(4) = 2f(2)

16 + 4b + c = 2(2 + 2b + c)

sub b

16 - 4 + c = 2(4 - 2 + c)

c = 8

R = f(2) = (4 - 2 + 8) = 10

(iii) When the equation is divided by x-t, the remainder is 5R. Find the two possible values of t

f(t) = 5 x 10

t^2 + tb + c = 50

t^2 - t + 8 = 50

t = 7 or -6

(c)

the sketch shows part of the graph y = x^3+px^2+qx+r where p, q, and r are constants.
The points A, B, and C have co-ordinates (-2,0), (2,0) and (4,0) respectively.
Find p, q, r

let f(x) = x^3 + px^2 + qx + r

f(-2) = 0

f(2) = 0

f(4) = 0

Therefore: f(x) = (x + 2)(x - 2)(x - 4)
f(x) = x^3 - 4x^2 - 4x + 16

comparing coef.

-4x^2 = px^2
p = -4

-4x = qx
q = -4

r = 16