Transcript Lesson 1 - WCHS Math

Five-Minute Check
Then/Now
New Vocabulary
Key Concept: Real Numbers
Example 1: Use Set-Builder Notation
Example 2: Use Interval Notation
Key Concept: Function
Key Concept: Vertical Line Test
Example 3: Identify Relations that are Functions
Example 4: Find Function Values
Example 5: Find Domains Algebraically
Example 6: Real-World Example: Evaluate a Piecewise-Defined
Function
Find the value of x 2 + 4x + 4 if x = –2.
A. –8
B. 0
C. 4
D. 16
Solve 5n + 6 = –3n – 10.
A. –8
B. –2
C.
D. 2
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.
You used set notation to denote elements, subsets,
and complements. (Lesson 0-1)
• Describe subsets of real numbers.
• Identify and evaluate functions and state their
domains.
• set-builder notation
• interval notation
• function
• function notation
• independent variable
• dependent variable
• implied domain
• piecewise-defined
function
• relevant domain
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.
Answer:
Use Set-Builder Notation
B. Describe x > –17 using set-builder notation.
The set includes all real numbers greater than –17.
Answer:
Use Set-Builder Notation
C. Describe all multiples of seven using set-builder
notation.
The set includes all integers that are multiples of 7.
Answer:
Describe {6, 7, 8, 9, 10, …} using set-builder
notation.
A.
B.
C.
D.
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.
Identify Relations that are Functions
A. Determine whether the relation represents y as
a function of x.
The input value x is the height of a student in
inches, and the output value y is the number of
books that the student owns.
Answer: No; there is more than one y-value for an
x-value.
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.
Identify Relations that are Functions
D. Determine whether x =3y 2 represents y as a
function of x.
No; there is more than one y-value for an x-value. 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 = 2.
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 the function value
for 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 the function value
for 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 the function value
for 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.
If
A.
B.
C.
D.
, find f(6).
Find Domains Algebraically
A. State the domain of the function
.
Because the square root of a negative number cannot
be real, 4x – 1 ≥ 0. Therefore, the domain of g(x) is all
real numbers x such that x ≥
, or
Answer: all real numbers x such that x ≥
or
.
,
Find Domains Algebraically
B. State the domain of the function
When the denominator of
.
is zero, the expression
is undefined. Solving t 2 – 1 = 0, the excluded values in
the domain of this function are t = 1 and t = –1. The
domain of this function is all real numbers except
t = 1 and t = –1, or
Answer:
.
Find Domains Algebraically
C. State the domain of the function
This function is defined only when 2x – 3 > 0.
Therefore, the domain of f(x) is
or
Answer:
.
or
.
State the domain of g(x) =
A.
B.
C.
D.
or [4, ∞)
or [–4, 4]
or (− , −4)
.
Evaluate a Piecewise-Defined
Function
A. FINANCE Realtors in a
metropolitan area studied
the average home price per
square foot as a function
of total square footage.
Their evaluation yielded
the following piecewisedefined function. Find the
average price per square
foot for a home with the
square footage of
1400 square feet.
Evaluate a Piecewise-Defined
Function
Because 1400 is between 1000 and 2600,
use
to find p(1400).
Function for 1000 ≤ a < 2600
Substitute 1400 for a.
Subtract.
= 85
Simplify.
Evaluate a Piecewise-Defined
Function
According to this model, the average price per square
foot for a home with a square footage of 1400 square
feet is $85.
Answer: $85 per square foot
Evaluate a Piecewise-Defined
Function
B. FINANCE Realtors in a
metropolitan area studied
the average home price
per square foot as a
function of total square
footage. Their evaluation
yielded the following
piecewise-defined
function. Find the average
price per square foot for a
home with the square
footage of 3200 square
feet.
Evaluate a Piecewise-Defined
Function
Because 3200 is between 2600 and 4000, use
to find p(3200).
Function for
2600 ≤ a < 4000.
Substitute 3200 for a.
Simplify.
Evaluate a Piecewise-Defined
Function
According to this model, the average price per square
foot for a home with a square footage of 3200 square
feet is $104.
Answer: $104 per square foot
ENERGY The cost of residential electricity use
can be represented by the following piecewise
function, where k is the number of kilowatts. Find
the cost of electricity for 950 kilowatts.
A. $47.50
B. $48.00
C. $57.50
D. $76.50