EBK PRECALCULUS W/LIMITS
EBK PRECALCULUS W/LIMITS
4th Edition
ISBN: 9781337516853
Author: Larson
Publisher: CENGAGE CO
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Transformation of Graphs
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Question
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Chapter 2.3, Problem 9E

a.

To determine

To graph:The equations y1=x2+2x−1x+3 and y2=x−1+2x+3 in the same viewing window.

a.

Expert Solution
Check Mark

Explanation of Solution

Given:

The given equations are y1=x2+2x−1x+3 and y2=x−1+2x+3

Graph:

The graph of equations y1=x2+2x−1x+3 and y2=x−1+2x+3 are shown in Figure-1 here.

  EBK PRECALCULUS W/LIMITS, Chapter 2.3, Problem 9E , additional homework tip 1

Interpretation:

From the graph of equations y1=x2+2x−1x+3 and y2=x−1+2x+3 , we observe that the shapes of graphs of y1 and y2 are identical.

b.

To determine

To verify: The expressions y1=x2+2x−1x+3 and y2=x−1+2x+3 are equivalent, using their graphs.

Given:

The expressions y1=x2+2x−1x+3 and y2=x−1+2x+3

Concept used:

Two expressions are said to be equivalent if their graphs pass through same points and their geometry is same.

Calculation:

The graphs of the given expressions are

  EBK PRECALCULUS W/LIMITS, Chapter 2.3, Problem 9E , additional homework tip 2

From graphsof equations y1=x2+2x−1x+3 and y1=x2+2x−1x+3 pass through same points and their shapes are same.

Hence, the expressions y1=x2+2x−1x+3 and y2=x−1+2x+3 are equivalent.

b.

Expert Solution
Check Mark

Explanation of Solution

Given:

The expressions y1=x2+2x−1x+3 and y2=x−1+2x+3

Concept used:

Two expressions are said to be equivalent if their graphs pass through same points and their geometry is same.

Calculation:

The graphs of the given expressions are

  EBK PRECALCULUS W/LIMITS, Chapter 2.3, Problem 9E , additional homework tip 3

From graphsof equations y1=x2+2x−1x+3 and y1=x2+2x−1x+3 pass through same points and their shapes are same.

Hence, the expressions y1=x2+2x−1x+3 and y2=x−1+2x+3 are equivalent.

C.

To determine

To verify: The expressions y1=x2+2x−1x+3 and y2=x−1+2x+3 are equivalent; using long division (or algebraic method).

C.

Expert Solution
Check Mark

Explanation of Solution

Given:

Theexpressions y1=x2+2x−1x+3 and y2=x−1+2x+3 .

Method used:

The long- division method.

Calculations:

Consider equation y1=x2+2x−1x+3

  ⇒y1=x+3x−1x2+2x−1x2+3x−x−1x+32

  ⇒y1=x−1+2x+3.........(1)

From result (1) and given expression y2=x−1+2x+3 , it is clear that y1=y2 .

Thus, the expressions y1=x2+2x−1x+3 and y2=x−1+2x+3 are equivalent.

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Chapter 2.3, Problem 8E
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Chapter 2.3, Problem 10E

Chapter 2 Solutions

EBK PRECALCULUS W/LIMITS

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Prob. 14ECh. 2.7 - Prob. 15ECh. 2.7 - Prob. 16ECh. 2.7 - Prob. 17ECh. 2.7 - Prob. 18ECh. 2.7 - Prob. 19ECh. 2.7 - Prob. 20ECh. 2.7 - Prob. 21ECh. 2.7 - Prob. 22ECh. 2.7 - Prob. 23ECh. 2.7 - Prob. 24ECh. 2.7 - Prob. 25ECh. 2.7 - Prob. 26ECh. 2.7 - Prob. 27ECh. 2.7 - Prob. 28ECh. 2.7 - Prob. 29ECh. 2.7 - Prob. 30ECh. 2.7 - Prob. 31ECh. 2.7 - Prob. 32ECh. 2.7 - Prob. 33ECh. 2.7 - Prob. 34ECh. 2.7 - Prob. 35ECh. 2.7 - Prob. 36ECh. 2.7 - Prob. 37ECh. 2.7 - Prob. 38ECh. 2.7 - Prob. 39ECh. 2.7 - Prob. 40ECh. 2.7 - Prob. 41ECh. 2.7 - Prob. 42ECh. 2.7 - Prob. 43ECh. 2.7 - Prob. 44ECh. 2.7 - Prob. 45ECh. 2.7 - Prob. 46ECh. 2.7 - Prob. 47ECh. 2.7 - Prob. 48ECh. 2.7 - Prob. 49ECh. 2.7 - Prob. 50ECh. 2.7 - Prob. 51ECh. 2.7 - Prob. 52ECh. 2.7 - Prob. 53ECh. 2.7 - Prob. 54ECh. 2.7 - Prob. 55ECh. 2.7 - Prob. 56ECh. 2.7 - Prob. 57ECh. 2.7 - Prob. 58ECh. 2.7 - Prob. 59ECh. 2.7 - Prob. 60ECh. 2.7 - Prob. 61ECh. 2.7 - Prob. 62ECh. 2.7 - Prob. 63ECh. 2.7 - Prob. 64ECh. 2.7 - Prob. 65ECh. 2.7 - Prob. 66ECh. 2.7 - Prob. 67ECh. 2.7 - Prob. 68ECh. 2.7 - Prob. 69ECh. 2.7 - Prob. 70ECh. 2.7 - Prob. 71ECh. 2.7 - Prob. 72ECh. 2.7 - Prob. 73ECh. 2.7 - Prob. 74ECh. 2.7 - Prob. 75ECh. 2.7 - Prob. 76ECh. 2.7 - Prob. 77ECh. 2.7 - Prob. 78ECh. 2.7 - Prob. 79ECh. 2.7 - Prob. 80ECh. 2.7 - Prob. 81ECh. 2.7 - Prob. 82ECh. 2.7 - Prob. 83ECh. 2.7 - Prob. 84ECh. 2.7 - Prob. 85ECh. 2.7 - Prob. 86ECh. 2.7 - Prob. 87ECh. 2.7 - Prob. 88ECh. 2.7 - Prob. 89ECh. 2.7 - Prob. 90ECh. 2 - Prob. 1RECh. 2 - Prob. 2RECh. 2 - Prob. 3RECh. 2 - Prob. 4RECh. 2 - Prob. 5RECh. 2 - Prob. 6RECh. 2 - Prob. 7RECh. 2 - Prob. 8RECh. 2 - Prob. 9RECh. 2 - Prob. 10RECh. 2 - Prob. 11RECh. 2 - Prob. 12RECh. 2 - Prob. 13RECh. 2 - Prob. 14RECh. 2 - Prob. 15RECh. 2 - Prob. 16RECh. 2 - Prob. 17RECh. 2 - Prob. 18RECh. 2 - Prob. 19RECh. 2 - Prob. 20RECh. 2 - Prob. 21RECh. 2 - Prob. 22RECh. 2 - Prob. 23RECh. 2 - Prob. 24RECh. 2 - Prob. 25RECh. 2 - Prob. 26RECh. 2 - Prob. 27RECh. 2 - Prob. 28RECh. 2 - Prob. 29RECh. 2 - Prob. 30RECh. 2 - Prob. 31RECh. 2 - Prob. 32RECh. 2 - Prob. 33RECh. 2 - Prob. 34RECh. 2 - Prob. 35RECh. 2 - Prob. 36RECh. 2 - Prob. 37RECh. 2 - Prob. 38RECh. 2 - Prob. 39RECh. 2 - Prob. 40RECh. 2 - Prob. 41RECh. 2 - Prob. 42RECh. 2 - Prob. 43RECh. 2 - Prob. 44RECh. 2 - Prob. 45RECh. 2 - Prob. 46RECh. 2 - Prob. 47RECh. 2 - Prob. 48RECh. 2 - Prob. 49RECh. 2 - Prob. 50RECh. 2 - Prob. 51RECh. 2 - Prob. 52RECh. 2 - Prob. 53RECh. 2 - Prob. 54RECh. 2 - Prob. 55RECh. 2 - Prob. 56RECh. 2 - Prob. 57RECh. 2 - Prob. 58RECh. 2 - Prob. 59RECh. 2 - Prob. 60RECh. 2 - Prob. 61RECh. 2 - Prob. 62RECh. 2 - Prob. 63RECh. 2 - Prob. 64RECh. 2 - Prob. 65RECh. 2 - Prob. 66RECh. 2 - Prob. 67RECh. 2 - Prob. 68RECh. 2 - Prob. 1CTCh. 2 - Prob. 2CTCh. 2 - Prob. 3CTCh. 2 - Prob. 4CTCh. 2 - Prob. 5CTCh. 2 - Prob. 6CTCh. 2 - Prob. 7CTCh. 2 - Prob. 8CTCh. 2 - Prob. 9CTCh. 2 - Prob. 10CTCh. 2 - Prob. 11CTCh. 2 - Prob. 12CTCh. 2 - Prob. 13CTCh. 2 - Prob. 14CTCh. 2 - Prob. 15CTCh. 2 - Prob. 16CTCh. 2 - Prob. 17CTCh. 2 - Prob. 18CTCh. 2 - Prob. 1PSCh. 2 - Prob. 2PSCh. 2 - Prob. 3PSCh. 2 - Prob. 4PSCh. 2 - Prob. 5PSCh. 2 - Prob. 6PSCh. 2 - Prob. 7PSCh. 2 - Prob. 8PSCh. 2 - Prob. 9PSCh. 2 - Prob. 10PSCh. 2 - Prob. 11PSCh. 2 - Prob. 12PSCh. 2 - Prob. 13PS
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