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)
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.