Transcript Glencoe Precalculus

Find the value of x 2 + 4x + 4 if x = –2.
A. –8
B. 0
C. 4
D. 16
Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7.
A. –9
B. 9
C. 19
D. 41
Factor 8xy 2 – 4xy.
A. 2x(4xy 2 – y)
B. 4xy(2y – 1)
C. 4xy(y 2 – 1)
D. 4y 2(2x – 1)
A.
B.
C.
D.
1.1 Functions
Objectives
• Use Set Notation
• Use Interval Notation
Set – a collection of objects
Example: Colors, Cars
Element – are the objects that belong to a set.
Example: red, orange, blue, ….
Nissan, Audi, Jeep, …
Infinite Set
A set that has an unending list of elements
Countable – a collection of objects
Uncountable – are the objects that belong to a set.
Use Set-Builder Notation
A. Describe {2, 3, 4, 5, 6, 7} using set-builder
notation.
The set includes natural numbers greater than or
equal to 2 and less than or equal to 7.
This is read as the set of all
x such that 2 is less than or
equal to x and x is less than
or equal to 7 and x is an
element of the set of natural
numbers.
Use Set-Builder Notation
B. Describe x > –17 using set-builder notation.
The set includes all real numbers greater than –17.
Use Set-Builder Notation
C. Describe all multiples of seven using set-builder
notation.
The set includes all integers that are multiples of 7.
Describe {6, 7, 8, 9, 10, …} using set-builder
notation.
A.
B.
C.
D.
Interval Notation
Is a method of writing numbers in a set.
Recall the Number Line
Use Interval Notation
A. Write –2 ≤ x ≤ 12 using interval notation.
The set includes all real numbers greater than or
equal to –2 and less than or equal to 12.
Answer: [–2, 12]
Use Interval Notation
B. Write x > –4 using interval notation.
The set includes all real numbers greater than –4.
Answer:
(–4,
)
Use Interval Notation
C. Write x < 3 or x ≥ 54 using interval notation.
The set includes all real numbers less than 3 and all
real numbers greater than or equal to 54.
Answer:
Write x > 5 or x < –1 using interval notation.
A.
B.
C. (–1, 5)
D.
Review
Homework
1.1 Functions Continued
• What is a function?
• How do we use the Vertical Line Test?
x represents the domain
y represents the range
Turn your pencil vertically.
Does you pencil pass through the graph
more than once?
Identify Relations that are Functions
B. Determine whether the table
represents y as a function of x.
Answer: No; there is more than one y-value for an
x-value.
Identify Relations that are Functions
C. Determine whether the graph
represents y as a function of x.
Answer: Yes; there is exactly one y-value for each xvalue. Any vertical line will intersect the
graph at only one point. Therefore, the
graph represents y as a function of x.
Practice
WKST
Review
Homework
And
Worksheet
Quiz on Section 1.1
Tuesday, September 16
Day 4
Extra Help – Second half of Lunch
1.1 – Functions
Objectives
• Determine if the equation is a function
• Find function values
• Find the domain of the function
Identify Relations that are Functions
D. Determine whether x = 3y 2 represents y as a
function of x.
To determine whether this equation represents y as a
function of x, solve the equation for y.
x = 3y 2
Original equation
Divide each side by 3.
Take the square root of each side.
Identify Relations that are Functions
This equation does not represent y as a function of
x because there will be two corresponding y-values,
one positive and one negative, for any x-value greater
than 0.
Let x = 12.
Answer: No; there is more than one y-value for an
x-value.
Determine whether 12x 2 + 4y = 8 represents y as a
function of x.
A. Yes; there is exactly one y-value for each
x-value.
B. No; there is more than one y-value for an
x-value.
Find Function Values
A. If f(x) = x 2 – 2x – 8, find f(3).
To find f(3), replace x with 3 in f(x) = x 2 – 2x – 8.
f(x) = x 2 – 2x – 8
Original function
f(3) = 3 2 – 2(3) – 8
Substitute 3 for x.
=9–6–8
Simplify.
= –5
Subtract.
Answer: –5
Find Function Values
B. If f(x) = x 2 – 2x – 8, find f(–3d).
To find f(–3d), replace x with –3d in f(x) = x 2 – 2x – 8.
f(x) = x 2 – 2x – 8
f(–3d) = (–3d)2 – 2(–3d) – 8
= 9d 2 + 6d – 8
Answer: 9d 2 + 6d – 8
Original function
Substitute –3d for x.
Simplify.
Find Function Values
C. If f(x) = x2 – 2x – 8, find f(2a – 1).
To find f(2a – 1), replace x with 2a – 1 in f(x) = x 2 – 2x – 8.
f(x) = x 2 – 2x – 8
f(2a – 1) = (2a – 1)2 – 2(2a – 1) – 8
Original function
Substitute
2a – 1 for x.
= 4a 2 – 4a + 1 – 4a + 2 – 8 Expand
(2a – 1)2 and
2(2a – 1).
= 4a 2 – 8a – 5
Answer: 4a 2 – 8a – 5
Simplify.