Transcript Lesson 2-1

Five-Minute Check (over Chapter 1)
Then/Now
New Vocabulary
Key Concept: Functions
Example 1: Domain and Range
Key Concept: Vertical Line Test
Example 2: Real-World Example
Example 3: Graph a Relation
Example 4: Evaluate a Function
Over Chapter 1
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Name the property illustrated by –15b + 15b = 0.
A. Additive Identity
B. Multiplicative Inverse
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C. Inverse Property of Addition
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Solve 2(c – 5) – 2 = 8 + c.
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Over Chapter 1
Solve |3x – 5| + 4 = 14.
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Solve 2b – 5 ≤ –1. Graph the solution set on a
number line.
A. {b | b ≤ 2}
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Over Chapter 1
Which algebraic equation shows the sentence four
plus a number divided by six is equal to the
product of twelve and the same number?
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You identified domains and ranges for given
situations. (Lesson 0–1)
• Analyze relations and functions.
• Use equations of relations and functions.
• one-to-one function
• onto function
• discrete relation
• continuous relation
• vertical line test
• independent variable
• dependent variable
• function notation
Domain and Range
State the domain and range of
the relation. Then determine
whether the relation is a
function. If it is a function,
determine if it is one-to-one,
onto, both, or neither.
The relation is {(1, 2), (3, 3),
(0, –2), (–4, 0), (–3, 1)}.
Answer: The domain is {–4, –3, 0, 1, 3}. The range
is {–2, 1, 2, 3}. Each member of the domain
is paired with one member of the range, so
this relation is a function. It is onto, but not
one-to-one.
State the domain and range of the
relation shown in the graph. Is the
relation a function?
C.
domain: {–2, –1, 0, 1}
range: {–3, 0, 2, 3}
No, it is not a function.
D.
domain: {–3, 0, 2, 3}
range: {–2, –1, 0, 1}
No, it is not a function.
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domain: {–3, 0, 2, 3}
range: {–2, –1, 0, 1}
Yes, it is a function.
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domain: {–2, –1, 0, 1}
range: {–3, 0, 2, 3}
Yes, it is a function.
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TRANSPORTATION The
table shows the
average fuel efficiency
in miles per gallon for
SUVs for several years.
Graph this information
and determine whether
it represents a function.
Is this relation discrete
or continuous?
Use the vertical line test. Notice that no vertical line can
be drawn that contains more than one of the data
points.
Answer: Yes, this relation is a function. Because the
graph consists of distinct points, the relation
is discrete.
HEALTH The table shows the average weight of a
baby for several months during the first year.
Graph this information and determine whether it
represents a function.
A. Yes, this relation is a
function.
B.
No, this
relation
is not a
function.
C. Yes, this relation is a
function.
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No, this
relation
is not a
function.
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Graph a Relation
Graph y = 3x – 1 and determine the domain and
range. Then determine whether the equation is a
function, is one-to-one, onto, both, or neither.
State whether it is discrete or continuous.
Make a table of values to find
ordered pairs that satisfy the
equation. Choose values for
x and find the corresponding
values for y.
Then graph
the ordered
pairs.
Graph a Relation
Find the domain and range.
Since x can be any real
number, there is an infinite
number of ordered pairs that
can be graphed. All of them
lie on the line shown. Notice
that every real number is the
x-coordinate of some point on
the line. Also, every real
number is the y-coordinate of
some point on the line.
Answer: The domain and range are both all
real numbers.
Graph a Relation
Determine whether the relation is a function and
state whether it is discrete or continuous.
This graph passes the vertical
line test. Every x-value is
paired with exactly one unique
y-value, and every y-value
corresponds to an x-value.
Answer: Yes, the equation
y = 3x – 1 represents a
function. The function is both
one-to-one and onto. Since the
domain and range are both all
real numbers, the relation is
continuous.
Graph y = 2x + 5.
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Evaluate a Function
A. Given f(x) = x3 – 3, find f(–2).
f(x) = x3 – 3
f(–2) = (–2)3 – 3
= –8 – 3 or –11
Answer: f(–2) = –11
Original function
Substitute.
Simplify.
Evaluate a Function
B. Given f(x) = x3 – 3, find f(2t).
f(x) = x3 – 3
f(2t) = (2t)3 – 3
= 8t3 – 3
Answer: f(2t) = 8t3 – 3
Original function
Substitute.
(2t)3 = 8t3
A. Given f(x) = x2 + 5, find f(–1).
A. –4
B. –3
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B. Given f(x) = x2 + 5, find f(3a).
A. 3a2 + 5
B. a2 + 8
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D. 9a2 + 5
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C. 6a2 + 5