Explanation Given: The function is f ( x ) = | x | + 1 . a. y = f ( x − 1 ) b. y = − f ( x ) c. y = − f ( x ) + 2 d. y = f ( x − 1 ) − 2 e. y = f ( x + 1 ) Calculation: Part a: Find the points to draw the graph of the function y = f ( x − 1 ) . The given function is f ( x ) = | x | + 1 . Substitute x − 1 in place of x . f ( x − 1 ) = | x − 1 | + 1 We can rewrite it as: y = | x − 1 | + 1 Put x = − 2 in the given function. y = | ( − 2 ) − 1 | + 1 = 3 + 1 = 4 Put x = − 1 in the given function. y = | ( − 1 ) − 1 | + 1 = 2 + 1 = 3 Put x = 0 in the given function. y = | 0 − 1 | + 1 = 1 + 1 = 2 Put x = 1 in the given function. y = | 1 − 1 | + 1 = 0 + 1 = 1 Put x = 2 in the given function. y = | 2 − 1 | + 1 = 1 + 1 = 2 The points are: x − 2 − 1 0 1 2 y 4 3 2 1 2 The graph is: Hence, the graph of option (a) is (v). Part b: Find the points to draw the graph of the function y = − f ( x ) . The given function is f ( x ) = | x | + 1 . Multiply with ( − ) of the above function. − f ( x ) = − ( | x | + 1 ) We can rewrite it as: y = − ( | x | + 1 ) Put x = − 2 in the given function. y = − ( | − 2 | + 1 ) = − ( 3 ) = − 3 Put x = − 1 in the given function. y = − ( | − 1 | + 1 ) = − ( 2 ) = − 2 Put x = 0 in the given function. y = − ( | 0 | + 1 ) = − ( 1 ) = − 1 Put x = 1 in the given function. y = − ( | 1 | + 1 ) = − ( 2 ) = − 2 Put x = 2 in the given function. y = − ( | 2 | + 1 ) = − ( 3 ) = − 3 The points are: x − 2 − 1 0 1 2 y − 3 − 2 − 1 − 2 − 3 The graph is: Hence, the graph of option (b) is (iv). Part c: Find the points to draw the graph of the function y = − f ( x ) + 2 . The given function is f ( x ) = | x | + 1 . Multiply with ( − ) of the above function. − f ( x ) = − ( | x | + 1 ) Add 2 on both sides. − f ( x ) + 2 = − | x | − 1 + 2 = − | x | + 1 We can rewrite it as: y = − | x | + 1 Put x = − 2 in the given function. y = − | x | + 1 = − | − 2 | + 1 = − 1 Put x = − 1 in the given function

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Question

Chapter 1.1, Problem 80E

To determine

Match the functions (a)-(e) with their graphs (i)-(v).

a. y=f(x−1)

b. y=−f(x)

c. y=−f(x)+2

d. y=f(x−1)−2

e. y=f(x+1)

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Chapter 1.1, Problem 79E

Chapter 1.1, Problem 81E

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Chapter 1 Solutions

CALCULUS (CLOTH)

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Match the functions (a)-(e) with their graphs (i)-(v). a. y = f ( x − 1 ) b. y = − f ( x ) c. y = − f ( x ) + 2 d. y = f ( x − 1 ) − 2 e. y = f ( x + 1 )

Solution Summary: The author explains how to draw the graph of the function y=f(x-1).

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Match the functions (a)-(e) with their graphs (i)-(v).

a. y=f(x−1)

b. y=−f(x)

c. y=−f(x)+2

d. y=f(x−1)−2

e. y=f(x+1)

Want to see the full answer?

Chapter 1 Solutions

Match the functions (a)-(e) with their graphs (i)-(v). a. y = f ( x − 1 ) b. y = − f ( x ) c. y = − f ( x ) + 2 d. y = f ( x − 1 ) − 2 e. y = f ( x + 1 )

Solution Summary: The author explains how to draw the graph of the function y=f(x-1).

Concept explainers

Videos

Match the functions (a)-(e) with their graphs (i)-(v). a. y=f(x−1) b. y=−f(x) c. y=−f(x)+2 d. y=f(x−1)−2 e. y=f(x+1)

Want to see the full answer?

Chapter 1 Solutions

Match the functions (a)-(e) with their graphs (i)-(v).

a. y=f(x−1)

b. y=−f(x)

c. y=−f(x)+2

d. y=f(x−1)−2

e. y=f(x+1)