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When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch?

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When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression?

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.

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When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x -axis from a reflection with respect to the y -axis?

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How can you determine whether a function is odd or even from the formula of the function?

For a function f , substitute ( x ) for ( x ) in f ( x ) . Simplify. If the resulting function is the same as the original function, f ( x ) = f ( x ) , then the function is even. If the resulting function is the opposite of the original function, f ( x ) = f ( x ) , then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

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Algebraic

For the following exercises, write a formula for the function obtained when the graph is shifted as described.

f ( x ) = x is shifted up 1 unit and to the left 2 units.

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f ( x ) = | x | is shifted down 3 units and to the right 1 unit.

g ( x ) = | x - 1 | 3

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f ( x ) = 1 x is shifted down 4 units and to the right 3 units.

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f ( x ) = 1 x 2 is shifted up 2 units and to the left 4 units.

g ( x ) = 1 ( x + 4 ) 2 + 2

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For the following exercises, describe how the graph of the function is a transformation of the graph of the original function f .

y = f ( x + 43 )

The graph of f ( x + 43 ) is a horizontal shift to the left 43 units of the graph of f .

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y = f ( x 4 )

The graph of f ( x - 4 ) is a horizontal shift to the right 4 units of the graph of f .

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y = f ( x ) + 8

The graph of f ( x ) + 8 is a vertical shift up 8 units of the graph of f .

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y = f ( x ) 7

The graph of f ( x ) 7 is a vertical shift down 7 units of the graph of f .

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y = f ( x + 4 ) 1

The graph of f ( x + 4 ) 1 is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f .

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For the following exercises, determine the interval(s) on which the function is increasing and decreasing.

f ( x ) = 4 ( x + 1 ) 2 5

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g ( x ) = 5 ( x + 3 ) 2 2

decreasing on ( , 3 ) and increasing on ( 3 , )

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k ( x ) = 3 x 1

decreasing on ( 0 , )

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Graphical

For the following exercises, use the graph of f ( x ) = 2 x shown in [link] to sketch a graph of each transformation of f ( x ) .

Graph of f(x).

For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions.

f ( t ) = ( t + 1 ) 2 3

Graph of f(t).
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h ( x ) = | x 1 | + 4

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k ( x ) = ( x 2 ) 3 1

Graph of k(x).
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Numeric

Tabular representations for the functions f , g , and h are given below. Write g ( x ) and h ( x ) as transformations of f ( x ) .

x −2 −1 0 1 2
f ( x ) −2 −1 −3 1 2
x −1 0 1 2 3
g ( x ) −2 −1 −3 1 2
x −2 −1 0 1 2
h ( x ) −1 0 −2 2 3

g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1

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Tabular representations for the functions f , g , and h are given below. Write g ( x ) and h ( x ) as transformations of f ( x ) .

x −2 −1 0 1 2
f ( x ) −1 −3 4 2 1
x −3 −2 −1 0 1
g ( x ) −1 −3 4 2 1
x −2 −1 0 1 2
h ( x ) −2 −4 3 1 0
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For the following exercises, write an equation for each graphed function by using transformations of the graphs of one of the toolkit functions.

Graph of an absolute function.

f ( x ) = | x - 3 | 2

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Graph of an absolute function.

f ( x ) = | x + 3 | 2

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For the following exercises, use the graphs of transformations of the square root function to find a formula for each of the functions.

For the following exercises, use the graphs of the transformed toolkit functions to write a formula for each of the resulting functions.

Graph of a parabola.

f ( x ) = ( x + 1 ) 2 + 2

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For the following exercises, determine whether the function is odd, even, or neither.

For the following exercises, describe how the graph of each function is a transformation of the graph of the original function f .

g ( x ) = f ( x )

The graph of g is a vertical reflection (across the x -axis) of the graph of f .

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g ( x ) = 4 f ( x )

The graph of g is a vertical stretch by a factor of 4 of the graph of f .

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g ( x ) = f ( 5 x )

The graph of g is a horizontal compression by a factor of 1 5 of the graph of f .

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g ( x ) = f ( 1 3 x )

The graph of g is a horizontal stretch by a factor of 3 of the graph of f .

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g ( x ) = 3 f ( x )

The graph of g is a horizontal reflection across the y -axis and a vertical stretch by a factor of 3 of the graph of f .

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For the following exercises, write a formula for the function g that results when the graph of a given toolkit function is transformed as described.

The graph of f ( x ) = | x | is reflected over the y - axis and horizontally compressed by a factor of 1 4 .

g ( x ) = | 4 x |

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The graph of f ( x ) = x is reflected over the x -axis and horizontally stretched by a factor of 2.

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The graph of f ( x ) = 1 x 2 is vertically compressed by a factor of 1 3 , then shifted to the left 2 units and down 3 units.

g ( x ) = 1 3 ( x + 2 ) 2 3

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The graph of f ( x ) = 1 x is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units.

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The graph of f ( x ) = x 2 is vertically compressed by a factor of 1 2 , then shifted to the right 5 units and up 1 unit.

g ( x ) = 1 2 ( x - 5 ) 2 + 1

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The graph of f ( x ) = x 2 is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units.

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For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation.

g ( x ) = 4 ( x + 1 ) 2 5

The graph of the function f ( x ) = x 2 is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.

Graph of a parabola.
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g ( x ) = 5 ( x + 3 ) 2 2

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h ( x ) = 2 | x 4 | + 3

The graph of f ( x ) = | x | is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.

Graph of an absolute function.
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m ( x ) = 1 2 x 3

The graph of the function f ( x ) = x 3 is compressed vertically by a factor of 1 2 .

Graph of a cubic function.
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n ( x ) = 1 3 | x 2 |

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p ( x ) = ( 1 3 x ) 3 3

The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.

Graph of a cubic function.
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q ( x ) = ( 1 4 x ) 3 + 1

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a ( x ) = x + 4

The graph of f ( x ) = x is shifted right 4 units and then reflected across the vertical line x = 4.

Graph of a square root function.
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For the following exercises, use the graph in [link] to sketch the given transformations.

Graph of a polynomial.

Questions & Answers

An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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