USA: +1(669)289-8683
USA: +1(669)289-8683

Homework Questions Section 3

Must show work with some questions

Save Time On Research and Writing
Hire a Pro to Write You a 100% Plagiarism-Free Paper.
Get My Paper

Section 2.2: Circles
KEEP IT 100% MOMENT
It is HIGHLY RECOMMENDED that you either print out
the pdf of this lecture or at least pause the video when
you reach the β€œTRY FOR YOURSELF” examples and try
to solve them for yourself. It reinforces the brain to
think on its own which reinforces the steps and
processes needed in order to solve the problems. YOU
CANNOT BLAME ANYTHING OR ANYONE FOR LACK OF
EFFORT TO LEARN THE MATERIAL ON YOUR END AND
THAT CAN ONLY COME BY WAY OF PRACTICING.
Center-Radius Form Equation of a Circle.
A circle with center 𝒉, π’Œ and radius 𝒓 has an equation
π’™βˆ’π’‰
𝟐
+ π’šβˆ’π’Œ
𝟐
= 𝒓
𝟐
Which is the center-radius form of the equation of the circle. A circle
with center 𝟎, 𝟎 and radius 𝒓 has the equation
𝒙 βˆ’ 𝟎 𝟐 + π’š βˆ’ 𝟎 𝟐 = 𝒓 𝟐 or 𝒙 𝟐 + π’š 𝟐 = 𝒓 𝟐
NOTE: A circle cannot have a negative or imaginary radius!
Center-Radius Form Equation of a Circle.
π’™βˆ’π’‰
𝟐
+ π’šβˆ’π’Œ
𝟐
= 𝒓
𝟐
In CANVAS
π’™βˆ’
𝟐
+ π’šβˆ’
𝟐
=
Assuming you performed your calculations correctly when you
formed your equation, then all you need to do is enter into the
individual boxes what h, k, and 𝐫 𝟐 are.
NOT Center-Radius Form Equation of a Circle.
π’™βˆ’π’‰ πŸβˆ’ π’šβˆ’π’Œ 𝟐 = 𝒓 𝟐
or
π’™βˆ’πŸŽ πŸβˆ’ π’šβˆ’πŸŽ 𝟐 = 𝒓 𝟐
or
𝒙 πŸβˆ’ π’š 𝟐= 𝒓 𝟐
Finding the Radius of a Circle
If the center 𝒉, π’Œ is given and one endpoint π’™πŸ , π’šπŸ is known, then
the radius can be found by using the following:
𝒓=
π’™πŸ βˆ’ 𝒉 𝟐 + π’šπŸ βˆ’ π’Œ 𝟐
If two endpoints π’™πŸ , π’šπŸ and π’™πŸ , π’šπŸ are known, then the radius can be
found by using the following:
𝒓=
π’™πŸ βˆ’ π’™πŸ 𝟐 + π’šπŸ βˆ’ π’šπŸ 𝟐
πŸ’
Finding the Center
If two endpoints π’™πŸ , π’šπŸ and π’™πŸ , π’šπŸ are known, then the center
can be found by using the following:
π’™πŸ + π’™πŸ
π’šπŸ + π’šπŸ
𝒉, π’Œ : 𝒉 =
;π’Œ =
𝟐
𝟐
Ex: Find the center-radius form of the circle.
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’πŸ–, βˆ’πŸ” , π’“π’‚π’…π’Šπ’–π’” πŸπŸ•
𝒉 π’Œ
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’ π‘Ήπ’‚π’…π’Šπ’–π’”: 𝒙 βˆ’ 𝒉 𝟐 + π’š βˆ’ π’Œ 𝟐 = 𝒓 𝟐
𝒙 βˆ’ βˆ’πŸ–
𝟐
+ π’š βˆ’ βˆ’πŸ”
𝒙 + πŸ– 𝟐 + π’š + πŸ” 𝟐 = πŸπŸ•
𝟐
=
πŸπŸ•
𝟐
Ex: Find the center-radius form of the circle.
π‘ͺ𝒆𝒏𝒕𝒆𝒓 πŸ“, πŸ– , π’“π’‚π’…π’Šπ’–π’” πŸ•
Try for yourself!
Ex: Find the center-radius form of the circle.
π‘ͺ𝒆𝒏𝒕𝒆𝒓 𝟏, 𝟏 , π’˜π’Šπ’•π’‰ π’†π’π’…π’‘π’π’Šπ’π’• πŸ“, πŸ’
𝒙 π’š
𝒉 π’Œ
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’ π‘Ήπ’‚π’…π’Šπ’–π’”: 𝒙 βˆ’ 𝒉 𝟐 + π’š βˆ’ π’Œ 𝟐 = 𝒓 𝟐
βˆ’π’‰ 𝟐+
𝒓=
π’™πŸ
𝒓=
πŸ“βˆ’πŸ 𝟐+ πŸ’βˆ’πŸ 𝟐
𝒓=
πŸ’ 𝟐+ πŸ‘ 𝟐
𝒓 = πŸπŸ“
𝒓=πŸ“
π’šπŸ
βˆ’π’Œ 𝟐
π’™βˆ’ 𝟏
𝟐
+ π’šβˆ’ 𝟏
𝟐
= πŸ“ 𝟐
𝒙 βˆ’ 𝟏 𝟐 + π’š βˆ’ 𝟏 𝟐 = πŸπŸ“
Ex: Find the center-radius form of the circle. Try for yourself!
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’πŸ, 𝟐 , π’˜π’Šπ’•π’‰ π’†π’π’…π’‘π’π’Šπ’π’• 𝟎, 𝟐
Ex: Find the center-radius form of the circle.
𝒂 π’„π’Šπ’“π’„π’π’† π’˜π’Šπ’•π’‰ π’†π’π’…π’‘π’π’Šπ’π’•π’” βˆ’πŸ, 𝟐 𝒂𝒏𝒅 𝟏𝟏, πŸ•
π’™πŸ π’šπŸ
π’™πŸ π’šπŸ
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’ π‘Ήπ’‚π’…π’Šπ’–π’”: 𝒙 βˆ’ 𝒉 𝟐 + π’š βˆ’ π’Œ 𝟐 = 𝒓 𝟐
𝒓=
𝒓=
𝒓=
π’™πŸ βˆ’ π’™πŸ 𝟐 + π’šπŸ βˆ’ π’šπŸ 𝟐
πŸ’
𝟏𝟏 βˆ’ βˆ’πŸ
𝟐
+ πŸ•βˆ’ 𝟐
πŸ’
𝟏𝟏 + 𝟏 𝟐 + πŸ• βˆ’ 𝟐 𝟐
πŸ’
𝒓=
𝟏𝟐 𝟐 + πŸ“ 𝟐
πŸ’
π’™πŸ + π’™πŸ
𝒉=
𝟐
π’šπŸ + π’šπŸ
π’Œ=
𝟐
𝒓=
πŸπŸ’πŸ’ + πŸπŸ“
πŸ’
βˆ’πŸ + 𝟏𝟏
𝒉=
𝟐
𝟐+πŸ•
π’Œ=
𝟐
𝒓=
πŸπŸ”πŸ—
πŸ’
𝟏𝟎
𝒉=
𝟐
πŸ—
π’Œ=
𝟐
𝟐
πŸπŸ‘
𝒓=
𝟐
𝒉=πŸ“
Continued→
Ex: Find the center-radius form of the circle.
𝒂 π’„π’Šπ’“π’„π’π’† π’˜π’Šπ’•π’‰ π’†π’π’…π’‘π’π’Šπ’π’•π’” βˆ’πŸ, 𝟐 𝒂𝒏𝒅 𝟏𝟏, πŸ•
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’ π‘Ήπ’‚π’…π’Šπ’–π’”: 𝒙 βˆ’ 𝒉 𝟐 + π’š βˆ’ π’Œ 𝟐 = 𝒓 𝟐
𝒉=πŸ“
πŸ—
π’Œ=
𝟐
πŸπŸ‘
𝒓=
𝟐
πŸ—
𝟐
π’™βˆ’ πŸ“
+ π’šβˆ’
𝟐
πŸ—
𝟐
π’™βˆ’πŸ“ + π’šβˆ’
𝟐
𝟐
𝟐
πŸπŸ‘
=
𝟐
πŸπŸ”πŸ—
=
πŸ’
𝟐
Ex: Find the center-radius form of the circle. Try for yourself!
𝒂 π’„π’Šπ’“π’„π’π’† π’˜π’Šπ’•π’‰ π’†π’π’…π’‘π’π’Šπ’π’•π’” πŸ“, πŸ’ 𝒂𝒏𝒅 βˆ’πŸ‘, βˆ’πŸ
General Form Equation of a Circle.
The equation
𝒙 𝟐 + π’š 𝟐 + 𝒄𝒙 + π’…π’š + 𝒆 = 𝟎
for some real numbers c, d, and e, can have a graph that is a circle or
a point, or is non-existent.
In CANVAS
𝒙^𝟐 + π’š^𝟐 + 𝒄𝒙 + π’…π’š + 𝒆 = 𝟎
Ex: Find the general equation form of a circle.
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’πŸ–, βˆ’πŸ” , π’“π’‚π’…π’Šπ’–π’” πŸ‘
𝒉 π’Œ
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’ π‘Ήπ’‚π’…π’Šπ’–π’”: 𝒙 βˆ’ 𝒉 𝟐 + π’š βˆ’ π’Œ 𝟐 = 𝒓 𝟐
𝒙 βˆ’ βˆ’πŸ–
𝟐
+ π’š βˆ’ βˆ’πŸ”
𝟐
= πŸ‘ 𝟐
𝒙+πŸ– 𝟐+ π’š+πŸ” 𝟐 =πŸ—
𝒙+πŸ– 𝒙+πŸ– + π’š+πŸ” π’š+πŸ” =πŸ—
π’™πŸ + πŸ–π’™ + πŸ–π’™ + πŸ”πŸ’ + π’šπŸ + πŸ”π’š + πŸ”π’š + πŸ‘πŸ” = πŸ—
π’™πŸ + πŸπŸ”π’™ + πŸ”πŸ’ + π’šπŸ + πŸπŸπ’š + πŸ‘πŸ” = πŸ—
π’™πŸ + πŸπŸ”π’™ + π’šπŸ + πŸπŸπ’š + πŸ”πŸ’ + πŸ‘πŸ” βˆ’ πŸ— = 𝟎 β†’ π’™πŸ + π’šπŸ + πŸπŸ”π’™ + πŸπŸπ’š + πŸ—πŸ = 𝟎
In CANVAS = 𝒙^𝟐 + π’š^𝟐 + πŸπŸ”π’™ + πŸπŸπ’š + πŸ—πŸ = 𝟎
Ex: Find the general equation form of a circle.
π‘ͺ𝒆𝒏𝒕𝒆𝒓 βˆ’πŸ’, 𝟏 , π’“π’‚π’…π’Šπ’–π’” πŸ“
Try for yourself!
Ex: Find the center-radius of the circle by completing the square.
π’™πŸ + π’šπŸ + πŸπŸπ’™ βˆ’ πŸπ’š + πŸπŸ– = 𝟎
π’™βˆ’π’‰ 𝟐+ π’šβˆ’π’Œ 𝟐 = 𝒓 𝟐
π’™πŸ + πŸπŸπ’™ + π’šπŸ βˆ’ πŸπ’š + πŸπŸ– = 𝟎
π‘ͺ𝒆𝒏𝒕𝒆𝒓 = 𝒉, π’Œ
π’™πŸ + πŸπŸπ’™ + π’šπŸ βˆ’ πŸπ’š = βˆ’πŸπŸ–
π‘Ήπ’‚π’…π’Šπ’–π’” =
𝒓 𝟐
π’™πŸ + πŸπŸπ’™ + ? 𝟐 + π’šπŸ βˆ’ πŸπ’š + ? 𝟐 = βˆ’πŸπŸ– + ? 𝟐 + ? 𝟐
π’™πŸ + πŸπŸπ’™ + πŸ” 𝟐 + π’šπŸ βˆ’ πŸπ’š + βˆ’πŸ 𝟐 = βˆ’πŸπŸ– + πŸ” 𝟐 + βˆ’πŸ 𝟐
𝒙+πŸ” 𝟐+ π’šβˆ’πŸ 𝟐 =πŸ—
π‘ͺ𝒆𝒏𝒕𝒆𝒓 = βˆ’πŸ”, 𝟏
π‘Ήπ’‚π’…π’Šπ’–π’” = πŸ— = πŸ‘
Ex: Find the center-radius of the circle by completing the square.
π’™πŸ + π’šπŸ βˆ’ πŸπŸπ’™ + πŸπŸŽπ’š + πŸπŸ“ = 𝟎
Try for yourself!
Ex: Find the center-radius of the circle by completing the square.
πŸ’π’™πŸ + πŸ’π’šπŸ + πŸ’π’™ βˆ’ πŸπŸ”π’š βˆ’ πŸπŸ— = 𝟎
π’™βˆ’π’‰ 𝟐+ π’šβˆ’π’Œ 𝟐 = 𝒓 𝟐
πŸ’ 𝟐 πŸ’ 𝟐 πŸ’
πŸπŸ”
πŸπŸ— 𝟎
𝒙 + π’š + π’™βˆ’
π’šβˆ’
=
π‘ͺ𝒆𝒏𝒕𝒆𝒓 = 𝒉, π’Œ
πŸ’
πŸ’
πŸ’
πŸ’
πŸ’
πŸ’
πŸπŸ—
𝟐
π‘Ήπ’‚π’…π’Šπ’–π’”
=
𝒓
𝟐
𝟐
𝒙 + π’š + 𝒙 βˆ’ πŸ’π’š βˆ’
=𝟎
πŸ’
πŸπŸ—
𝟐
𝟐
𝒙 + 𝒙 + π’š βˆ’ πŸ’π’š =
πŸ’
πŸπŸ—
𝟐
𝟐
𝟐
𝟐
𝒙 + 𝒙 + ? + π’š βˆ’ πŸ’π’š + ? =
+ ? 𝟐+ ? 𝟐
πŸ’
𝟐
𝟐
𝟏
πŸπŸ—
𝟏
𝟐
𝟐
𝟐
𝒙 +𝒙+
+ π’š βˆ’ πŸ’π’š + βˆ’πŸ =
+
+ βˆ’πŸ 𝟐
𝟐
πŸ’
𝟐
𝟐
𝟏
𝒙+
+ π’šβˆ’πŸ 𝟐 =πŸ—
𝟐
𝟏
π‘Ήπ’‚π’…π’Šπ’–π’” = πŸ— = πŸ‘
π‘ͺ𝒆𝒏𝒕𝒆𝒓 = βˆ’ , 𝟐
𝟐
Ex: Find the center-radius of the circle by completing the square.
πŸ—π’™πŸ + πŸ—π’šπŸ + πŸπŸπ’™ βˆ’ πŸπŸ–π’š βˆ’ πŸπŸ‘ = 𝟎
Try for yourself!
Ex: Graph the circle.
π’™βˆ’πŸ 𝟐+ π’šβˆ’πŸ 𝟐 =πŸ—
π‘ͺ𝒆𝒏𝒕𝒆𝒓 = 𝟏, 𝟐
π‘Ήπ’‚π’…π’Šπ’–π’” = πŸ— = πŸ‘
Ex: Graph the circle.
Try for yourself!
𝒙+𝟐 𝟐+ π’š+𝟐 𝟐 =πŸ’
Ex: Given the graph of a circle, find the equation of the circle.
Where is the center?
π‘ͺ𝒆𝒏𝒕𝒆𝒓 = βˆ’πŸ, 𝟎
What is the radius?
π‘Ήπ’‚π’…π’Šπ’–π’” = πŸ‘
Why? The radius is 3 because in each
direction (left, right, up, down) it
takes exactly 3 units to touch the
circle.
𝒙 βˆ’ βˆ’πŸ
𝟐
+ π’šβˆ’ 𝟎
𝒙+𝟐 𝟐+ π’šβˆ’πŸŽ 𝟐 =πŸ—
𝟐
= πŸ‘ 𝟐
Ex: Given the graph of a circle, find the equation of the circle.
Try for yourself!
Section 2.3: Functions
KEEP IT 100% MOMENT
It is HIGHLY RECOMMENDED that you either print out
the pdf of this lecture or at least pause the video when
you reach the β€œTRY FOR YOURSELF” examples and try
to solve them for yourself. It reinforces the brain to
think on its own which reinforces the steps and
processes needed in order to solve the problems. YOU
CANNOT BLAME ANYTHING OR ANYONE FOR LACK OF
EFFORT TO LEARN THE MATERIAL ON YOUR END AND
THAT CAN ONLY COME BY WAY OF PRACTICING.
Section 2.3: Functions
Relation – A relation is a set of ordered pairs.
Function – A relation in which, for each distinct value of the first
component or (x), of the ordered pairs, there is exactly one value
of the second component or (y).
Domain – A domain is the set of all values of the independent
variable or (x) and that value must not repeat. YOU SHOULD LIST
THE VALUES IN ASCENDING ORDER!
Range – A range is the set of all values that are dependent or (y).
YOU SHOULD LIST THE VALUES IN ASCENDING ORDER!
Common Ways to Represent Function Notation.
The following are ways to represent function notation:
𝒇 𝒙 or π’ˆ 𝒙 or 𝒉 𝒙 and so on…
Also the most common way to represent a function is by…
π’š=
Section 2.3: Functions
Ex: Give the domain and range of the relation and determine if the
relation is a function.
πŸ‘, βˆ’πŸ , πŸ’, 𝟐 , πŸ’, πŸ“ , πŸ”, πŸ–
π‘«π’π’Žπ’‚π’Šπ’: πŸ‘, πŸ’, πŸ’, πŸ”
π‘Ήπ’‚π’π’ˆπ’†: βˆ’πŸ, 𝟐, πŸ“, πŸ–
π‘­π’–π’π’„π’•π’Šπ’π’: 𝑡𝒐! Why? Because there is a value
in the domain that repeats.
Section 2.3: Functions
Ex: Give the domain and range of the relation and determine if the
relation is function. Try for yourself!
𝟐, πŸ‘ , πŸ’, πŸ“ , πŸ”, πŸ• , πŸ–, πŸ— , 𝟏𝟎, 𝟏𝟏
π‘«π’π’Žπ’‚π’Šπ’:
π‘Ήπ’‚π’π’ˆπ’†:
π‘­π’–π’π’„π’•π’Šπ’π’:
Section 2.3: Functions
Ex: Determine if the relation is function. Give the domain and range of
each relation.
Try for yourself!
π‘­π’–π’π’„π’•π’Šπ’π’:
π‘«π’π’Žπ’‚π’Šπ’:
π‘Ήπ’‚π’π’ˆπ’†:
Section 2.3: Functions
Domain & Range of a Linear Equation of the Forms
𝑨𝒙 + π‘©π’š = π‘ͺ
π‘«π’π’Žπ’‚π’Šπ’: βˆ’βˆž, ∞
π’š = π’Žπ’™ + 𝒃
π‘Ήπ’‚π’π’ˆπ’†: βˆ’βˆž, ∞
NOTE: It does not matter in which direction
the line maybe going, if the equation is a
linear equation, then the domain & range
follows as previously stated. Linear equations
ARE FUNCTIONS.
Section 2.3: Functions
Domain & Range of a Quadratic Equation of the Forms
π’‚π’™πŸ + 𝒃𝒙 + 𝒄 = 𝟎
π’š = π’‚π’™πŸ + 𝒃𝒙 + 𝒄
π’š=𝒂 π’™βˆ’π’‰ 𝟐+π’Œ
π‘«π’π’Žπ’‚π’Šπ’: βˆ’βˆž, ∞
Assuming the graphs
are continuous!
NOTE: The RANGE of a quadratic
π‘Ήπ’‚π’π’ˆπ’†: βˆ’βˆž, 𝟎
π‘Ήπ’‚π’π’ˆπ’†: 𝟎, ∞
equation is the interval to which
the values for (y) increase instead
of decrease. Quadratic equations
ARE FUNCTIONS.
Section 2.3: Functions
Domain & Range of a Circle of the Form
𝒙 βˆ’ 𝒉 𝟐 + π’š βˆ’ π’Œ 𝟐 = π’“πŸ
π‘«π’π’Žπ’‚π’Šπ’: βˆ’πŸ‘, πŸ‘
NOTE: The DOMAIN of a circle equation is the
exact interval to which the circle crosses the
x-axis. The RANGE of a circle equation is the
exact interval to which the circles crosses the
y-axis. Circle equations ARE NOT FUNCTIONS.
π‘Ήπ’‚π’π’ˆπ’†: βˆ’πŸ‘, πŸ‘
Section 2.3: Functions
Domain & Range of a Square Root Equation of the Form
π’š= 𝒙
𝒇 𝒙 = 𝒙
π‘«π’π’Žπ’‚π’Šπ’: ሾ𝟎, ∞ሻ
π‘Ήπ’‚π’π’ˆπ’†: ሾ𝟎, ∞ሻ
NOTE: The DOMAIN of a square root equation is the
exact interval to which the graph either touches,
crosses, or hovers over or under the x-axis.
Sometimes you may have to solve the square root
equation in order to determine that exactly. The
RANGE of a square root equation is the exact interval
to which the graph crosses or hovers beside the yaxis. Square root equations ARE FUNCTIONS.
Section 2.3: Functions
Domain & Range of a Square Root Equation of the Form
π’š= 𝒙+𝟏
𝒇 𝒙 = 𝒙+𝟏
π‘«π’π’Žπ’‚π’Šπ’: αˆΎβˆ’πŸ, ∞ሻ
π‘Ήπ’‚π’π’ˆπ’†: ሾ𝟎, ∞ሻ
NOTE: The DOMAIN of a square root equation is the
exact interval to which the graph either touches,
crosses, or hovers over or under the x-axis.
Sometimes you may have to solve the square root
equation in order to determine that exactly. The
RANGE of a square root equation is the exact interval
to which the graph crosses or hovers beside the yaxis. Square root equations ARE FUNCTIONS.
Section 2.3: Functions
Domain & Range of a Rational Equation of the Form
𝟏
π’š=
𝒙
π‘«π’π’Žπ’‚π’Šπ’: βˆ’βˆž, 0 βˆͺ 0, ∞
π‘Ήπ’‚π’π’ˆπ’†: βˆ’βˆž, 0 βˆͺ 0, ∞
NOTE: The DOMAIN of a rational equation is the exact
interval to which the graph will not cross the x-axis known
as the vertical asymptote. Sometimes you may have to
solve the rational equation in order to determine that
exactly. The RANGE of a rational equation is the exact
interval to which the graph hovers or runs along the y-axis
known as the horizontal asymptote. Rational equations
ARE FUNCTIONS.
Section 2.3: Functions
Domain & Range of an Absolute Value Equation of the Form
π’š= 𝒙
π‘«π’π’Žπ’‚π’Šπ’: αˆΊβˆ’βˆž, 𝟎ሿ βˆͺ ሾ𝟎, ∞ሻ
π‘Ήπ’‚π’π’ˆπ’†: ሾ𝟎, ∞ሻ
NOTE: The DOMAIN of an absolute value equation is the
exact interval to which the graph will cross, touch, or hover
above or below the x-axis. Sometimes you may have to
solve the absolute value equation in order to determine
that exactly. The RANGE of an absolute equation is the
exact interval to which the graph crosses, hovers or runs
along the y-axis. Absolute Value Equations ARE FUNCTIONS.
Section 2.3: Functions
Domain & Range of an Absolute Value Equation of the Form
π’š= 𝒙+𝟏
π‘«π’π’Žπ’‚π’Šπ’: αˆΊβˆ’βˆž, βˆ’πŸαˆΏ βˆͺ αˆΎβˆ’πŸ, ∞ሻ
π‘Ήπ’‚π’π’ˆπ’†: ሾ𝟎, ∞ሻ
NOTE: The DOMAIN of an absolute value equation is the
exact interval to which the graph will cross, touch, or hover
above or below the x-axis. Sometimes you may have to
solve the absolute value equation in order to determine
that exactly. The RANGE of an absolute equation is the
exact interval to which the graph crosses, hovers or runs
along the y-axis. Absolute Value Equations ARE FUNCTIONS.
Vertical Line Test for Functionality.
The vertical line test is a quick technique used to determine if a
graph is a function. A vertical or straight line is drawn through the
graph and if the lines touches the graph more than once then the
graph is NOT A FUNCTION.
π‘­π’–π’π’„π’•π’Šπ’π’
π‘­π’–π’π’„π’•π’Šπ’π’
𝑡𝒐𝒕 𝒂 π‘­π’–π’π’„π’•π’Šπ’π’
π‘­π’–π’π’„π’•π’Šπ’π’
Vertical Line Test for Functionality.
The vertical line test is a quick technique used to determine if a
graph is a function. A vertical or straight line is drawn through the
graph and if the lines touches the graph more than once then the
graph is NOT A FUNCTION.
π‘­π’–π’π’„π’•π’Šπ’π’
π‘­π’–π’π’„π’•π’Šπ’π’
π‘­π’–π’π’„π’•π’Šπ’π’
Section 2.3: Functions
𝑬 𝒙 : Decide whether the equation is a function and then give
the domain and range.
𝒂. π’š = 𝒙 + πŸ’
π‘­π’–π’π’„π’•π’Šπ’π’
π‘­π’–π’π’„π’•π’Šπ’π’: 𝒀𝒆𝒔! It’s Linear.
π‘«π’π’Žπ’‚π’Šπ’: βˆ’βˆž, ∞
π‘Ήπ’‚π’π’ˆπ’†: βˆ’βˆž, ∞
Section 2.3: Functions
𝑬 𝒙 : Decide whether the equation is a function and then give
the domain and range.
𝒃. π’š = πŸπ’™ βˆ’ 𝟏
π‘­π’–π’π’„π’•π’Šπ’π’: 𝒀𝒆𝒔!
𝟏
π‘«π’π’Žπ’‚π’Šπ’: ΰ΅€ , ∞ሻ
𝟐
πŸπ’™ βˆ’ 𝟏 = 𝟎
+𝟏 +𝟏
πŸπ’™ = 𝟏
𝟐
𝟐
𝟏
𝒙=
𝟐
π‘Ήπ’‚π’π’ˆπ’†: ሾ𝟎, ∞ሻ
π’š = πŸπ’™ βˆ’ 𝟏
π’š=
𝟐
𝟏
βˆ’πŸ
𝟐
π’š= πŸβˆ’πŸ = 𝟎 =𝟎
π‘­π’–π’π’„π’•π’Šπ’π’
Section 2.3: Functions
𝑬 𝒙 : Decide whether the equation is a function and then give
the domain and range.
𝟏
𝒄. π’š =
π’™βˆ’πŸ
π‘­π’–π’π’„π’•π’Šπ’π’: 𝒀𝒆𝒔!
π‘«π’π’Žπ’‚π’Šπ’:
π’™βˆ’πŸ=𝟎
+𝟏 +𝟏
π‘Ήπ’‚π’π’ˆπ’†: βˆ’βˆž, 0 βˆͺ 0, ∞
𝟏
𝟎=
π’™βˆ’πŸ
𝟎=𝟏
π‘­π’–π’π’„π’•π’Šπ’π’
𝒙=𝟏
βˆ’βˆž, 1 βˆͺ 1, ∞
Notice here that if π’š = 𝟎, we cannot solve
for 𝒙. The 𝒙 would cancel out. Therefore,
π’š = 𝟎 cannot be in the range but all other
values for π’š are in the range.
Section 2.3: Functions
𝑬 𝒙 : Decide whether the equation is a function and then give
the domain and range. Try for yourself!
π’š = πŸ• βˆ’ πŸπ’™
π‘­π’–π’π’„π’•π’Šπ’π’:
π‘«π’π’Žπ’‚π’Šπ’:
π‘Ήπ’‚π’π’ˆπ’†:
Section 2.3: Performing
Operations on Functions
Ex: Find 𝒇 βˆ’π’™ when 𝒇 𝒙
𝟐
= βˆ’πŸπ’™ βˆ’ πŸ‘π’™ βˆ’ πŸ“
𝒇 βˆ’π’™ = βˆ’πŸπ’™πŸ βˆ’ πŸ‘π’™ βˆ’ πŸ“
𝒇 βˆ’π’™ = βˆ’πŸ βˆ’π’™ 𝟐 βˆ’ πŸ‘ βˆ’π’™ βˆ’ πŸ“
𝒇 βˆ’π’™ = βˆ’πŸπ’™πŸ + πŸ‘π’™ βˆ’ πŸ“
Ex: Find 𝒇 π’˜ when 𝒇 𝒙
𝟐
= 𝒙 + πŸπ’™ + πŸ–
Try for yourself!
Ex: Find 𝒇 π’Œ βˆ’ 𝟏 when 𝒇 𝒙
𝟐
= πŸ’π’™ + πŸ‘π’™ + πŸ’
𝒇 π’Œβˆ’πŸ =πŸ’ π’Œβˆ’πŸ 𝟐+πŸ‘ π’Œβˆ’πŸ +πŸ’
𝒇 π’Œ βˆ’ 𝟏 = πŸ’ π’ŒπŸ βˆ’ π’Œ βˆ’ π’Œ + 𝟏 + πŸ‘ π’Œ βˆ’ 𝟏 + πŸ’
𝒇 π’Œ βˆ’ 𝟏 = πŸ’ π’ŒπŸ βˆ’ πŸπ’Œ + 𝟏 + πŸ‘ π’Œ βˆ’ 𝟏 + πŸ’
𝟐
𝒇 π’Œ βˆ’ 𝟏 = πŸ’π’Œ βˆ’ πŸ–π’Œ + πŸ’ + πŸ‘π’Œ βˆ’ πŸ‘ + πŸ’
𝟐
𝒇 π’Œ βˆ’ 𝟏 = πŸ’π’Œ βˆ’ πŸ“π’Œ + πŸ“
Try for yourself!
𝟐
Ex: Find 𝒇 𝒅 + 𝟐 when 𝒇 𝒙 = 𝒙 βˆ’ πŸπ’™
Ex: Find π’ˆ 𝒂 βˆ’ 𝟏 when π’ˆ 𝒙
𝟏
= 𝒙+πŸ‘
πŸ“
𝟏
π’ˆ π’‚βˆ’πŸ = π’‚βˆ’πŸ +πŸ‘
πŸ“
𝟏
𝟏
π’ˆ π’‚βˆ’πŸ = π’‚βˆ’ +πŸ‘
πŸ“
πŸ“
𝟏
πŸπŸ’
𝟏
𝟏 πŸπŸ“
= 𝒂+
π’ˆ π’‚βˆ’πŸ = π’‚βˆ’ +
πŸ“
πŸ“
πŸ“
πŸ“ πŸ“
Try for yourself!
Ex: Find 𝒉 𝒛 + 𝟐 when 𝒉 𝒙
𝟐
𝟏
=βˆ’ 𝒙+
πŸ‘
𝟐
Determining if a graph is defined or not along the y-axis
for a given or known value of x.
1) If for a given value x there is a hole in the graph along the
y-axis, then x is not defined or does not exist (DNE) at that
location.
2) If for a given value x there is not a hole in the graph along
the y-axis, then x is defined at that location.
Ex: Find 𝒇 𝟎 , 𝒇 πŸ’ and 𝒇 βˆ’πŸ if they exist.
𝒇 𝟎 = πŸ’
𝒇 πŸ’ = πŸ’
𝒇 βˆ’πŸ = 𝟐
Ex: Find 𝑯 𝟎 , 𝑯 πŸ‘ and 𝑯 βˆ’πŸ if they exist.
𝑯 𝟎 =πŸ‘
𝑯 πŸ‘
Does Not Exist
or
=
DNE
𝑯 βˆ’πŸ = 𝟐
Ex: Find 𝒇 𝟎 , 𝒇 βˆ’πŸ and 𝒇 βˆ’πŸ if they exist.
Try for yourself!
Section 2.3: Increasing,
Decreasing, Constant Functions
Increasing, Decreasing, Constant Functions.
1) 𝒇 increases on an interval when π’™πŸ < π’™πŸ & 𝑓 π’™πŸ < 𝑓 π’™πŸ 2) 𝒇 decreases on an interval when π’™πŸ < π’™πŸ & 𝑓 π’™πŸ > 𝑓 π’™πŸ
3) 𝒇 constant on an interval when π’™πŸ π‘Žπ‘›π‘‘ π’™πŸ & 𝑓 π’™πŸ = 𝑓 π’™πŸ
For graphing using open interval notation purposes, parentheses ( )
are used.
For graphing using closed interval notation purposes, brackets [ ]
are used.
Ex: Find increasing, decreasing, and constant intervals
of the function below if they exist.
Increasing, Decreasing, Constant Functions.
1) 𝒇 increases on an interval when π’™πŸ < π’™πŸ & 𝑓 π’™πŸ < 𝑓 π’™πŸ 2) 𝒇 decreases on an interval when π’™πŸ < π’™πŸ & 𝑓 π’™πŸ > 𝑓 π’™πŸ
3) 𝒇 constant on an interval when π’™πŸ π‘Žπ‘›π‘‘ π’™πŸ & 𝑓 π’™πŸ = 𝑓 π’™πŸ
𝒇 𝒙 = π’™πŸ‘ βˆ’ 𝟏 on the interval (-4,4).
π‘°π’π’„π’“π’†π’‚π’”π’Šπ’π’ˆ: βˆ’πŸ’, πŸ’
𝒇 βˆ’πŸ’ = βˆ’πŸ’ πŸ‘ βˆ’ 𝟏 𝒇 πŸ’ = πŸ’ πŸ‘ βˆ’ 𝟏
𝒇 πŸ’ = πŸ”πŸ’ βˆ’ 𝟏
𝒇 βˆ’πŸ’ = βˆ’πŸ”πŸ’ βˆ’ 𝟏
π‘«π’†π’„π’“π’†π’‚π’”π’Šπ’π’ˆ: 𝑡𝒐𝒏𝒆
𝒇 βˆ’πŸ’ = βˆ’πŸ”πŸ“
𝒇 πŸ’ = πŸ”πŸ‘
π‘ͺ𝒐𝒏𝒔𝒕𝒂𝒏𝒕: 𝑡𝒐𝒏𝒆
Ex: Find increasing, decreasing, and constant intervals
of the function below if they exist.
𝒇 𝒙 = βˆ’ 𝒙 βˆ’ πŸ’ on the interval (4,7).
Try for yourself!
π‘°π’π’„π’“π’†π’‚π’”π’Šπ’π’ˆ:
π‘«π’†π’„π’“π’†π’‚π’”π’Šπ’π’ˆ:
π‘ͺ𝒐𝒏𝒔𝒕𝒂𝒏𝒕:
Ex: Find increasing, decreasing, and constant intervals
of the graph if they exist.
π‘°π’π’„π’“π’†π’‚π’”π’Šπ’π’ˆ: ሾ𝟏, ∞ሻ
π‘«π’†π’„π’“π’†π’‚π’”π’Šπ’π’ˆ: αˆΊβˆ’βˆž, βˆ’πŸαˆΏ
π‘ͺ𝒐𝒏𝒔𝒕𝒂𝒏𝒕:
βˆ’πŸ, 𝟏
Ex: Find increasing, decreasing, and constant intervals
of the graph if they exist.
π‘°π’π’„π’“π’†π’‚π’”π’Šπ’π’ˆ: 𝑡𝒐𝒏𝒆
π‘«π’†π’„π’“π’†π’‚π’”π’Šπ’π’ˆ: αˆΊβˆ’βˆž, 𝟎ሿ βˆͺ ሾ𝟎, ∞ሻ
π‘ͺ𝒐𝒏𝒔𝒕𝒂𝒏𝒕:
𝑡𝒐𝒏𝒆
Ex: Find increasing, decreasing, and constant intervals
of the graph if they exist.
Try for yourself!
π‘°π’π’„π’“π’†π’‚π’”π’Šπ’π’ˆ:
π‘«π’†π’„π’“π’†π’‚π’”π’Šπ’π’ˆ:
π‘ͺ𝒐𝒏𝒔𝒕𝒂𝒏𝒕:
Ex: Find increasing, decreasing, and constant intervals
of the graph if they exist.
Try for yourself!
π‘°π’π’„π’“π’†π’‚π’”π’Šπ’π’ˆ:
π‘«π’†π’„π’“π’†π’‚π’”π’Šπ’π’ˆ:
π‘ͺ𝒐𝒏𝒔𝒕𝒂𝒏𝒕:

Are you stuck with your online class?
Get help from our team of writers!
Order Today

Order your essay today and save 20% with the discount code RAPID

Order Now