So, Let's try some exercises to test your understanding about function:

__Question 1:__

__Question 2:__

__Question 3:__

__Question 4:__

__Question 5:__

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
```

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

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

Ans: Linear group is x+2

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

Ans: Linear group is x-7

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

Ans: Linear group is x-5

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

Ans: Linear group is 2x+3

```
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
```

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

x + 2 = 0 => x = -2

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

So vertex is (-2,-1)

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

x - 7 = 0 => x = 7

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

So vertex is (7,2)

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

Now, you can do it in your head:

So vertex is (5,3)

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

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

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

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) ?)

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

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:

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)

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

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))

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

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

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

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:__

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

`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`

`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`

`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.

**Note:**

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

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

a) $x > 1$

b) $x \le -2$

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

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)

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

```
* 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

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

Simplify this fraction. Your task!!!

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

if f(x) is an even function, then f(x) is symmetric with respect to y-axis

if f(x) is odd function, then f(x) is symmetric with respect to x-axis

Now, if you have done PART C. Tell me which one is symmetric with respect to y-axis. Which one is symmetric with respect to x-axis.