## Thursday, 7 June 2012

### Math Assignment!

Here's a link to my independent math assignment! :)

http://www.glogster.com/tay-math/video-glog-by-tay-math-independent-math-assignment/g-6lgcls32o6sgts052t023a0

I used the Glogster tool and it was really cool!

I decided to make a video glog so that i could play around with iMovie at the same time.

Check it out !

## Thursday, 9 February 2012

### Square Root Function

* Not finished

Square Root Function
$\small \fn_jvn a\sqrt{b(x-h)}+k$

Basic Properties:
$\small \fn_jvn \sqrt{x}$ : Means the positive square root of X, which is only possible if X > 0

$\small \fn_jvn -\sqrt{x}$: Means the negative square root of X, which is only possible if X > 0

$\small \fn_jvn \sqrt{-x}$: Means the positive square root of -X, which is only possible if X< 0

$\small \fn_jvn -\sqrt{-x}$: Means the negative square root of -X, which is only possible if X < 0

On the graph :

Transformed function: $\fn_jvn a\sqrt{b(x-h)}+k$

Role of h and k is to shift the function according to their values. (h,k) is the point where the function begins.

Examples:

$\small \fn_jvn \sqrt{x-1}+2$                                                                                          $\small \fn_jvn -\sqrt{x+2}-4$

$\small \fn_jvn \sqrt{-(x-2)}-4$                                $\small \fn_jvn -\sqrt{-(x-2)}+4$

Simplifying out parameter b:

Factor out parameter b and factor out of parameter h.  $\small \fn_jvn \sqrt{b} \cdot\sqrt{x\pm h}$
Sometimes, b can only simplified to -1.

Examples:
G(x) = $\small \fn_jvn \sqrt{-x+2}+4$                H(x) = $\small \fn_jvn \sqrt{10-4x}-7$

G(x) = $\small \fn_jvn \sqrt{-(x-)2}+4$                H(x) = $\small \fn_jvn 2\sqrt{-(x+\frac{10}{4})}-7$

## Tuesday, 7 February 2012

### Absolute Value Function

I've been condensing my notes into a couple pages per topi. I plan to do this kind of post for all of the functions, so I hope you like it and hope it helps with anyone trying to review the basics :)

Absolute Value Function
Y= a |bx-h| + k

Basic absolute values:
|4| =  4         |1-5| = 4            |n • m| = |n| • |m|
|-3| = 3         |-6 x 2| = 12     |n/m| = |n| / |m|

If |?| = a and a > 0, there are 2 possible answers : -a and a
Example: |a| = 66, a can equal 66 or - 66

|?| = -1  IS IMPOSSIBLE
Simplifying parameter b :
To simplify parameter b, simply make it into parameter a by multiplying the absolute value of b with the value of a. If  parameter h is involved, factor out parameter b.

Examples:

f'(x) = 2|-3x|        g(x)= 4 |2x-2| +3            h(x)= |2x|
f(x)= 6 |x|            g(x)= 4 |2 (x-1)| + 3       h(x)= 2 |x|
g(x) = 8 |x-1| + 3

TO FIND THE VERTEX WE MUST FIRST SIMPLIFY PARAMETER B

Properties of f(x)= a |x-h|+ k

If a > 0
Domain = R
Range = [k, ∞
Vertex (h,k)
Min: k
Slopes of sides: +a, -a

If a< o

Domain = R
Range = -∞, k]
Vertex (h,k)
Max: k
Slopes of sides: +a, -a

Finding the rule:

1. Write template : y= a |x-h|+ k
2. Fill in given parameters
3. Substitute given point
4. Solve for missing parameters

Example 1: Given point (9, 17) and vertex (3,5)
1. y= a |x-h|+ k
2. y= a |x-3|+ 5
3. 17= a |9-3|+ 5
4. 17= 6a +5
a=2
f(x)= 2 |x-3|+ 5

Example 2: Given zeroes -5 and 3, and minimum value of -2
* Because the vertex h is between the 2 zeroes, add and divide by 2 : (-5+3)/ 2 = -1
vertex (-1, -2)
1. y= a |x-h|+ k
2. y= a |x+1|- 2
3. 17= a |3+1|- 2
4. 2= 4a
a= 0.5
f(x)= 0.5 |x+1|- 2

Finding the zero(es)
In an absolute value function, there can be 0, 1 or 2 zeroes.
* change result of absolute value so you have both a negative and positive answer before you solve for x
Example:  f(x)= |x+2| - 6
0= |x+2| -6
6=x+2
AND
-6= x+2
x= 4 or x= -8

Solving inequalities
In an absolute value function, there are 4 different possibilities for the answer of an inequality, they are:
a) 1 interval   X E [ 1st zero, 2nd zero]
b) 2 intervals   X E -∞, 1st zero] U [ 2nd zero, ∞
c) The entire X axis
d) The empty set X E ∅

Example: Rule < 0

a)

b)

c)

d)