BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

15th Edition

ISBN: 9781608408405

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Cengage Learning

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Question

Chapter 4, Problem 5CA

(a)

To determine

To Find:The model for each transformation of the graph $f(x)=x$ .

(a)

Expert Solution

Answer to Problem 5CA

$g(x)=1(x−(−3))+2$

Explanation of Solution

Given:The graph is a translation $2 units$ up and $3 units$ left.

$g(x)=....(x−....)+....$ .

The translation $2 units$ up involves adding $2 units$ to the function and translation $3 units$ left involves subtracting $3 units$ .

Thus

$f(x)=xg(x)=f(x)+2g(x)=x+2g(x)=(x−(−3))+2thusg(x)=1(x−(−3))+2$

Hence, $g(x)=1(x−(−3))+2$ .

(b)

To determine

To Find: The model for each transformation of the graph $f(x)=x$ .

(b)

Expert Solution

Answer to Problem 5CA

$g(x)=1(x−2)+(−3)$

Explanation of Solution

Given: The graph is a translation $2 units$ right and $3 units$ down.

$g(x)=....(x−....)+....$ .

The translation $2 units$ right involves subtracting $2 units$ to the variable and translation $3 units$ down involves subtracting $3 units$ to the function value.

Thus

$g(x)=f(x−2)g(x)=x−2g(x)=(x−2)−3thusg(x)=1(x−2)+(−3)$

Hence, $g(x)=1(x−2)+(−3)$ .

(c)

To determine

To Find: The model for each transformation of the graph $f(x)=x$ .

(c)

Expert Solution

Answer to Problem 5CA

$g(x)=2(x−0)+2$

Explanation of Solution

Given: The graph is vertical stretch by a factor of $2 units$ followed by the translation $2 units$ up.

$g(x)=....(x−....)+....$ .

Vertical stretch by two means mean multiply the function by two and translation of $2 units$ up means adding $2 units$ to the function.

$g(x)=2f(x)g(x)=2xg(x)=2x+2thusg(x)=2(x−0)+2$

Hence, $g(x)=2(x−0)+2$ .

(d)

To determine

To Find: The model for each transformation of the graph $f(x)=x$ .

(d)

Expert Solution

Answer to Problem 5CA

$g(x)=12(x−3)+(−4)$

Explanation of Solution

Given: The graph is translation $3 units$ right and vertical shrink by a factor of $12$ followed by the translation $4 units$ down.

$g(x)=....(x−....)+....$ .

$g(x)=f(x−3)g(x)=x−3g(x)=12(x−3)−4thusg(x)=12(x−3)+(−4)$

Hence, $g(x)=12(x−3)+(−4)$ .

Chapter 4, Problem 4CA

Chapter 4, Problem 6CA

Chapter 4 Solutions

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

Ch. 4.1 - Prob. 11E Ch. 4.1 - Prob. 12E Ch. 4.1 - Prob. 13E Ch. 4.1 - Prob. 14E Ch. 4.1 - Prob. 15E Ch. 4.1 - Prob. 16E Ch. 4.1 - Prob. 17E Ch. 4.1 - Prob. 18E Ch. 4.1 - Prob. 19E Ch. 4.1 - Prob. 20E Ch. 4.1 - Prob. 21E Ch. 4.1 - Prob. 22E Ch. 4.1 - Prob. 23E Ch. 4.1 - Prob. 24E Ch. 4.1 - Prob. 25E Ch. 4.1 - Prob. 26E Ch. 4.1 - Prob. 27E Ch. 4.1 - Prob. 28E Ch. 4.1 - Prob. 29E Ch. 4.1 - Prob. 30E Ch. 4.1 - Prob. 31E Ch. 4.1 - Prob. 32E Ch. 4.1 - Prob. 33E Ch. 4.1 - Prob. 34E Ch. 4.1 - Prob. 35E Ch. 4.1 - Prob. 36E Ch. 4.1 - Prob. 37E Ch. 4.1 - Prob. 38E Ch. 4.1 - Prob. 39E Ch. 4.1 - Prob. 40E Ch. 4.1 - Prob. 41E Ch. 4.1 - Prob. 42E Ch. 4.1 - Prob. 43E Ch. 4.1 - Prob. 44E Ch. 4.1 - Prob. 45E Ch. 4.1 - Prob. 46E Ch. 4.1 - Prob. 47E Ch. 4.1 - Prob. 48E Ch. 4.1 - Prob. 49E Ch. 4.1 - Prob. 50E Ch. 4.1 - Prob. 51E Ch. 4.1 - Prob. 52E Ch. 4.1 - Prob. 53E Ch. 4.1 - Prob. 54E Ch. 4.1 - Prob. 55E Ch. 4.1 - Prob. 56E Ch. 4.2 - Prob. 1E Ch. 4.2 - Prob. 2E Ch. 4.2 - Prob. 3E Ch. 4.2 - Prob. 4E Ch. 4.2 - 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