November 14th, 2013

11/14/2013

tt2_ch.2_revision_1214-11-2013.pdf
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25 Comments

11/14/2013 11:42:23 am

sir, could you please do out the full Q. 12 on 2.6 because I dont understand it at all?

Mr R

11/14/2013 12:40:39 pm

The full solution to Q.12 using the quicker method (method 1) is uploaded. Using the alternative method the steps are identical to Curve D solutions:
Roots: x= -2 and x = 6
Factors: (x+2)(x-6)
Function in terms of factors: h(x) = k(x+2)(x-6)
h(2) = k(2+2)(2-6)=4
-16k=4
k=-1/4
Expand:h(x)=(-1/4)(x+2)(x-6)
h(x)= (-1/4)(x^2 -4x -12)
Complete square form: h(x) = (-1/4)(x^2 -4x +4 -4 -12)
h(x) = (-1/4)( (x-2)^2 -16)
h(x) = -1/4(x-2)^2 +4

ie: a=-1/4, p=4, q=2

Mr R

11/14/2013 01:52:42 pm

I mean a = 1/4 (in last line)

11/14/2013 11:54:23 am

sir, what's the answer to that question? The back of the book says p=16 for curve C

Mr R

11/14/2013 12:30:40 pm

Thank you for this.

The book answer is incorrect for curve C. It has multiplied the correct answer by 4 which doesn't affect the solutions but does change the function. The test is to input x = 2 which should give an output of y = 4. This is true in my solution (uploaded) but not in book solution.

fiona

11/14/2013 12:19:32 pm

sir will we need to know quadratic and linear questions in context?

Mr R

11/14/2013 12:41:36 pm

Yes

Anon123

11/14/2013 12:40:26 pm

sir could you help me with 2.10 Q 1 i am really stuck

Mr R

11/14/2013 12:44:32 pm

The solutions to this question are included in the 2.10 questions uploaded from class:
Link below
http://colaistebride.weebly.com/uploads/1/0/7/1/10716199/tt6_2.10_cubic_graphs_23-10-2013.pdf

Unknown

11/14/2013 12:41:46 pm

sir could you do 2.1 question 6 please

Mr R

11/14/2013 12:56:25 pm

Solution uploaded in latest blog

abc

11/14/2013 12:44:45 pm

is there any topic we dont need to do

Mr R

11/14/2013 12:51:23 pm

Chapter 1!

abc

11/14/2013 12:53:11 pm

i mean any execise in chapter 2 thats not on the test

Mr R

11/14/2013 12:57:40 pm

I would like students to be familiar with everything in Chapter 2 at this stage.

Anon123

11/14/2013 12:52:07 pm

in Q 1 2.10 i dont understand where the K came from

Anon123

11/14/2013 01:05:12 pm

never mind got it

Mr R

11/14/2013 01:05:28 pm

k is the "multiplier". There are an infinite number of cubic equations that have the roots x = -1, x = 1 and x = 3 as in part (i). The specific function is the factors multiplied by the specific multiplier. The value of k is found by the constraint that is contains another specific point other than the roots in part (i) we know that it contains (0, 3) i.e. if we input 0 it will give us an output of 0. This means that k=1 so now we know that this curve is 1 multiplied by the factors.

Siobhan

11/14/2013 12:52:30 pm

sir would you be able to do Q.17 page 88 advanced revision exercise please

11/14/2013 01:04:14 pm

Mr R

11/14/2013 01:16:27 pm

This is the same as Q5 in the first test as in Blog 6-11-2013 . (linked below)

http://colaistebride.weebly.com/uploads/1/0/7/1/10716199/algebra_2_test_solutions_part_1.pdf

Part (iv) the image of the curve in the x-axis is the negative of the first answer: i.e. f(x) = -2x^3 + 24 x^2 -90x + 100

Part (v) is found by noticing that the reflected roots in the y-axis would be x= -2, x= -5 and x= -5 giving factors (x + 2)(x +5)(x +5) with the multiplier remaining n=2 since the reflection doesn't rescale of stretch the curve....

siobhan

11/14/2013 01:28:58 pm

okay thanks

Mr R

11/14/2013 01:31:44 pm

You're very welcome

11/14/2013 01:00:07 pm

Is there anything on chapter 2 we dont need to know? In the last test we didnt need to know section 2.4 and 2.3. Do we need to know them for this test?

Mr R

11/14/2013 01:06:46 pm

Test is on whole chapter

November 14th, 2013

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