Lesson Objectives
• Learn how to write a function using function notation
• Learn how to evaluate a function using function notation
• Learn how to change between the implicit form and explicit form of a function

## Function Notation

Up to this point, we have worked with functions in the following format:
y = 2x - 9
y = x2 + 12
y = x3 - 1
Where y is the dependent variable and x is the independent variable. When we have a function, we normally name the function as: f, g, or h. When working with functions, we will replace our familiar "y" with f(x), g(x), h(x),...etc
f(x) = 2x - 9
f(x) = x2 + 12
f(x) = x3 - 1
f(x) is read as "f of x", which means we have a function "f" that depends on the independent variable "x".
This notation can immediately be used to evaluate a function for a given value of x. Let's suppose we had the following function.
y = x - 3
We have previously seen that we could state that x = some value, and solve for y.
Let's suppose x = 7:
Plug in a 7 for x:
y = 7 - 3
y = 4
y is 4 when x is 7. This can be expressed as the ordered pair: (7,4)
We can show this using function notation by replacing the "x" in "f(x)" with the x-value of 7.
f(x) = x - 3
f(7) = x - 3
This is read as "f of 7"; what is the functions value, when x is 7?
f(7) = 7 - 3
f(7) = 4
The answer is the same either way, we are just learning new notation.
Let's look at a few examples.
Example 1: Find f(0), and f(1)
f(x) = 2x2 - x + 5
f(0) = 2(0)2 - 0 + 5
f(0) = 5
f(1) = 2(1)2 - 1 + 5
f(1) = 2 - 1 + 5
f(1) = 6
Example 2: Find f(-2), and f(5)
f(x) = x3 - 4
f(-2) = (-2)3 - 4
f(-2) = -8 - 4
f(-2) = -12
f(5) = (5)3 - 4
f(5) = 125 - 4
f(5) = 121
In some cases, we will be asked to plug in more complex expressions for our variable. Let's look at a few examples.
Example 3: Find f(a + 7), and f(a - 1)
f(x) = x - 12
Just like when we had a single number, we plug in whatever is inside of the parentheses for x in our function:
f(a + 7) = (a + 7) - 12
f(a + 7) = a - 5
f(a - 1) = (a - 1) - 12
f(a - 1) = a - 13
Example 4: Find f(z + 2), and f(z - 1)
f(x) = x2 + x - 1
f(z + 2) = (z + 2)2 + (z + 2) - 1
f(z + 2) = z2 + 4z + 4 + z + 2 - 1
f(z + 2) = z2 + 5z + 5
f(z - 1) = (z - 1)2 + (z - 1) - 1
f(z - 1) = z2 - 2z + 1 + z - 1 - 1
f(z - 1) = z2 - z - 1

### Functions in Implicit Form

When a function is solved for the dependent variable "y" or "f(x)", it is said to be in "explicit" form. When a function is not solved for the dependent variable "y", it is said to be in "implicit" form. In some cases, we will have a function that is in implicit form and we will be asked to find the value of the function, given a certain input. When this occurs, we will first solve the equation for y. We then will replace y with "f(x)". Lastly, we will evaluate the function for the given value. Let's look at an example.
Example 5: Find f(-3), and f(10)
x2 - y = 20
Let's start by solving for y:
y = x2 - 20
Now we can replace y with f(x):
f(x) = x2 - 20
f(-3) = (-3)2 - 20
f(-3) = 9 - 20
f(-3) = -11
f(10) = (10)2 - 20
f(10) = 100 - 20
f(10) = 80

#### Skills Check:

Example #1

Find f(0). $$f(x)=3x - 1$$

A
$$f(0)=3$$
B
$$f(0)=7$$
C
$$f(0)=5$$
D
$$f(0)=-3$$
E
$$f(0)=-1$$

Example #2

Find f(-2). $$f(x)=\sqrt{2 - x}$$

A
undefined
B
$$f(-2)=8$$
C
$$f(-2)=-6$$
D
$$f(-2)=2$$
E
$$f(-2)=1$$

Example #3

Find f(a + 1). $$f(x)=x^2 - 5$$

A
$$a^2 - 1$$
B
$$a^2 + 2a - 7$$
C
$$a^2 + a - 4$$
D
$$a^2 + 2a - 4$$
E
$$a^2 + 2a - 3$$