# Session 3:¶

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

## Parents Functions and Transformation:¶

Here are some parent functions that are introduced by the textbook:

I think you will have no problem dealing with exercises related to constant or linear. So we will try to deal with Absolute Value and Quaratic

Keep in mind:

Step 1: Find the linear group.
Step 2: Find the vertex.(Linear group = 0)
Step 3: Transformation

#### STEP 1: FIND LINEAR GROUP:¶

Let's try some example of identifying the linear group of these functions:

##### Ex1:¶

$g(x) = |x+2| - 1$

Ans: Linear group is x+2

##### Ex2:¶

$f(x) = 2 - \frac{5}{2}|x-7|$

Ans: Linear group is x-7

##### Ex3:¶

$f(x) = (x-5)^2 + 3$

Ans: Linear group is x-5

##### Ex4:¶

$h(x) = 3(2x+3)^2 -2$

Ans: Linear group is 2x+3

#### STEP 2: FIND VERTEX:¶

To find vertex:

a) Let (linear group) = 0
=> Find x-value (This is vertex's x-value)

b) Find y-value of vertex by plugging x-value into a function
##### Ex1:¶

$g(x) = |x+2| - 1$

x + 2 = 0 => x = -2

When x = -2 => g(-2) = -1

So vertex is (-2,-1)

##### Ex2:¶

$f(x) = 2 - \frac{5}{2}|x-7|$

x - 7 = 0 => x = 7

When x = 7 => f(7) = 2

So vertex is (7,2)

##### Ex3:¶

$f(x) = (x-5)^2 + 3$

So vertex is (5,3)

##### Ex4:¶

$h(x) = 3(2x+3)^2 -2$

So vertex is (-3/2, -2)

#### STEP 3: TRANSFORMATION¶

##### Ex1:¶

$g(x) = |x+2| - 1$

Since the vertex that we find is (-2,-1), we can see that the vertext has already tells us how the graph will move:

x = -2 means the original graph (orange color) will move to the left 2 unit y = -1 mneas the orginal graph (orange color) will move down 1 unit

Basically, the original graph move to the left 2 unit and move down 1 unit.

(Do you see the vertex of the original graph which is (0,0) has moved to (-2,-1) ?)

##### Ex2:¶

$f(x) = 2 - \frac{5}{2}|x-7|$

Now we can guess our vertex will move from (0,0) to (7,2)

BE CAREFUL! We have the scale (5/2) in front of linear group. So do it slowly:

###### Final Result:¶

You can compare the final transformation to the original graph.

The whole process could be explained as: The original graph flips upside down, scale with -5/2, then the graph moves along with the vertex from (0,0) to (7,2)

##### Ex3:¶

$f(x) = (x-5)^2 + 3$

Same things are applied as the above two examples:

Notice that: when we don't have the scale factor, the graph just move from the original vertex to the vertex that we find ( In this case, move from (0,0) to (5,3))

##### Ex4:¶

$h(x) = 3(2x+3)^2 -2$

BE CAREFUL! DO YOU NOTICE THE SCALE FACTOR OF 2 INSIDE THE LINEAR GROUP AND THE SCALE OF 3 OUTSIDE THE LINEAR GROUP.

###### Final Result:¶

The original graph is orange. The transformation graph is green.

## HOMEWORK:¶

You do not need to show work, just graph all the transformation and vertex ( I will teach you how to write your results formally later):

Question 1:

$f(x) = |1-x| -5$

Question 2:

$f(x) = |5-\frac{9x}{8}| +4$

Question 3:

$f(x) = 3 -(x +2)^2$

Question 4:

$f(x) = 2 + 3(5x + 1)^2$

Question 5 and 6 are difficult. However, I have shown you how to deal with these types of problems. Try to do these two questions and I will go over these again next week.

Question 5:

$f(x) = x^2 + 2x$

Question 6:

$f(x) = 3x^2 + x + 1$

# SESSION 2¶

### 2. ISOSCELES TRIANGLE ( TWO WAYS OF PROVING THE TRIANGLE AS A ISOSCELES TRIANGLE)¶

#### FIRST WAY:¶

two adjacent sides of a triangle are equal

For Example: In this case, if we are asked to prove that triangle ABC are isosceles, then prove BA = BC

#### SECOND WAY:¶

two adjacent interior angles of a triangle are equal

For Example: In this case, if we are asked to prove that triangle ABC are isosceles, then prove angle A = angle C

### 3. RIGHT TRIANGLE ( PROVING THE TRIANGLE AS A RIGHT TRIANGLE)¶

USE PYTHAGORE's THEOREM TO PROVE THAT THE TRIANGLE IS A RIGHT TRIANGLE

THIS IS PYTHAGORE'S THEOREM:

However, this triangle is so trivial. We need something like this

so c is the distance between point A and point B ( Take a look at part 1. above to review about calculating distance).

b is the distance between point A and point C.

a is the distance between point B and point C.

### 4. REVIEW ABOUT SYMMETRY: ( LOOK AT SESSION 1)¶

Note:

When there is any question about SYMMETRY, you should always think of ODD or EVEN function

# Tyler 1st Session¶

## A. INTERVAL NOTATION:¶

### II. "One side" case:¶

Let's try with some exercises. Convert these set-builder notations to Interval Notation:

a) $x > 1$

b) $x \le -2$

c) $\frac{3}{2} \le x < 5$

#### Instructions:¶

• Step 1: Convert all the problem into the standard case: < x < but not this > x > or this x >

Example: x >1 will becomes x < 1

• Step 2: Now, there are only two situations, we need to deal with. First is a < x or x < b

With a < x. The Interval Notation will be (a, $\infty$)

With x < b. The Interval Notation will be (-$\infty$,b)

### III. "Or" Case¶

• Example:

Given $f(x) = \frac{x^2 + 5x}{1-x}$ . Answer these:

a) $f(5-x) =$

b) $f(x^2 + 4x)$

c) $f(-x)$

d) $f(\frac{1}{x})$

#### Instructions:¶

* Step 1: Replace x by a. Write f(a)
* Step 2: Replace a by x_SOMETHING. Write f(x_SOMETHING)

a) Find f(5-x) = ?

We have $f(x) = \frac{x^2 + 5x}{1-x}$

• Step 1: Replace x by a. Then write f(a) =

$f(a) = \frac{a^2 + 5a}{1-a}$

• Step 2: Replace a by x_SOMETHING. Then write f(x_SOMETHING) =

In this case x_SOMETHING = 5-x

So, it becomes

$f(a) = \frac{a^2 + 5a}{1-a} => f(5-x) = \frac{(5-x)^2 + 5(5-x)}{1-(5-x)}$

Now, your turn with b) c) and d)