TABLE OF CONTENTS:

SESSION 1

SESSION 2

SESSION 3






SESSION 3: CONVERT QUADRATIC FUNCTION TO VERTEX FORM

image

image

image

image






SESSION 2:

A. Odd and Even Function:

image image image image

Your Turn:

image

image

This question is not required. But should give it a try or think about it.

image

  • 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.






Session 1:

Functions:

image

image

image

image

image

image

Now you know more about function, but keep in mind that: 1 INPUT NEVER HAVE MORE THAN 1 OUTPUT

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

Question 1:

image

Question 2:

image

Question 3:

image

Question 4:

image

Question 5:

image

Parents Functions and Transformation:

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

image

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:

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

Now, you can do it in your head:

So vertex is (5,3)

Ex4:

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

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

STEP 3: TRANSFORMATION

Ex1:

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

image

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)

image

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

First Transformation:

image

Second Transformation: (Deal with sign of the linear group)

image

Third Transformation: (Deal with scale and sign of the linear group)

image

Fourth Transformation:

image

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)

image

Ex3:

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

Same things are applied as the above two examples:

image

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.

First Transformation:

image

Second Transformation:

image

Third Transformation:

image

Fourth Transformation:

image

Final Result:

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

image

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:

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

In [ ]: