2) we need divide the x-coordinate of point (1,1) on $f(x)$ by 3 to compress the graph horizontally by the factor $dfrac{1}{3}$ and then reflect in the y-axis so point is $(-dfrac{1}{3},1)$

3) we need divide the x-coordinate of point (1,1) on $f(x)$ by 3 to compress the graph horizontally by the factor $dfrac{1}{3}$, reflect in the y-axis and vertical stretch by a factor of 5, so point is $(-dfrac{1}{3},5)$

4) The graph $f(x)$ is is translated 2 units right, compressed horizontally by the factor $dfrac{1}{3}$ and reflected in the y-axis, then
graph is stretched vertically by the factor 5, and finally translated 4 units up, so point is $(dfrac{5}{3},9)$

b) translation 2 units to the right and 3 units up

c) horizontal compression by factor $dfrac{1}{2}$ then translation 5 units down

d) reflection in $x$-axis, horizontal stretch by factor 2 and translation 2 units down

e) vertical compression by factor $dfrac{2}{3}$, translation 3 units to the left and 1 unit up

f) vertical stretch by factor 4, reflection in $y$-axis and translation 4 units down

translation 4 units to the left $implies y=fleft(dfrac{1}{3}(x+4)right)$

b) reflection in $y$-axis $implies y=f(-x)$

vertical stretching by factor 2 $implies y=2f(-x)$

translation 3 units to the right and 1 unit up $implies y=2f(-(x-3))+1$

c) horizontal compression by factor $dfrac{1}{2}implies y=f(2x)$

reflection in $x$-axis $implies y=-f(2x)$

vertical stretching by factor 3 $implies y=-3f(2x)$

translation 1 unit to the right and 3 units down $implies y=-3f(2(x-1))-3$

translation 4 units to the left

b) reflection in $y$-axis

vertical stretching by factor 2

translation 3 units to the right and 1 unit up

c) horizontal compression by factor $dfrac{1}{2}$

reflection in $x$-axis

vertical stretching by factor 3

translation 1 unit to the right and 3 units down

$text{textcolor{#4257b2}{Domain}}$ of a function is the set of input values for which the function is defined

$text{textcolor{#4257b2}{The range}}$ is the resulting $y$-values we get after substituting all the possible $x$-values

We can identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $[3,+infty)$

$$
begin{align*}
y&=-f(dfrac{1}{4}(x+1))+2 \
&=-(dfrac{1}{4}(x+1))^{2}+2 tag{text{replace $x$ with $dfrac{1}{4}(x+1)$}}\
&=-dfrac{1}{16}(x+1)^{2}+2 \
end{align*}
$$

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $[2,+infty)$

$$
begin{align*}
y&=0.5f(3(x-4))-1 \
&=0.5(3(x-4))^{2}-1 tag{text{replace $x$ with $3(x-4)$}}\
&=0.5cdot9(x-4)^{2}-1 \
&=4.5(x-4)^{2}-1\
&=dfrac{9}{2}(x-4)^{2}-1\
end{align*}
$$

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $[-1,+infty)$

b) Domain: $(-infty,+infty)$ , Range: $[2,+infty)$

c) Domain: $(-infty,+infty)$ , Range: $[-1,+infty)$

$text{textcolor{#4257b2}{Domain}}$ of a function is the set of input values for which the function is defined

$text{textcolor{#4257b2}{The range}}$ is the resulting $y$-values we get after substituting all the possible $x$-values

We can identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $[1,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $[4,+infty)$

$$
begin{align*}
y&=f(-dfrac{1}{2}(x+4))-3 \
&=sqrt{-dfrac{1}{2}(x+4)}-3 tag{text{replace $x$ with $-dfrac{1}{2}(x+4)$}}\
end{align*}
$$

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,-4]$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $[-3,+infty)$

$$
begin{align*}
y&=-2f(-(x-2))+1 \
&=-2sqrt{-(x-2)}+1 tag{text{replace $x$ with $-(x-2)$}}\
end{align*}
$$

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,2]$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $(-infty,1)$

b) Domain: $(-infty,-4]$, Range:$[-3,+infty)$

c) Domain: $(-infty,2]$, Range:$(-infty,1)$

$text{textcolor{#4257b2}{Domain}}$ of a function is the set of input values for which the function is defined

$text{textcolor{#4257b2}{The range}}$ is the resulting $y$-values we get after substituting all the possible $x$-values

We can identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $[0,+infty)$

$$
begin{align*}
y&=4f(2(x-1))-2 \
&=4|2(x-1)|-2 tag{text{replace $x$ with $2(x-1)$}}\
end{align*}
$$

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $[-2,+infty)$

$$
begin{align*}
y&=-dfrac{1}{2}f(3(x+2))+4 \
&=-dfrac{1}{2}|3(x+2)|+4 tag{text{replace $x$ with $3(x+2)$}}\
end{align*}
$$

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $(-infty,4]$

b) domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq -2}$

c) domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq 4}$

The graph of $f(x)$ will horizontally shift to the right 2 units

So, there are used translation right 2 units

The graph of $f(x)$ will vertically shift upward by 2 units

So, there are used translation up 2 units

The graph of $f(x)$ will compresses vertically by factor of 0.5.

So, there are used vertical compression, by factor of 0.5

$y=2cdotdfrac{1}{x}$

The graph of $f(x)$ will stretches vertically by factor of 2.

So, there are used vertical stretch, by factor of 2

The graph of $f(x)$ will compresses horizontaly by factor of $dfrac{1}{2}=0.5$.

So, there are used horizontal compression, by factor of 0.5

The graph of $f(x)$ is reflected across the x-axis.

So, there is used reflection across the x-axis.

b) vertical translation 2 units upward

c) vertical compression by factor $dfrac{1}{2}$

d) vertical stretching by factor 2

e) horizontal compression by factor $dfrac{1}{2}$

f) reflection in $x$-axis

$$
begin{align*}
g(x)&=f(2x+6) \
&=(2x+6)^{2} tag{text{replace $x$ with $2x+6$}}\
end{align*}
$$

$$
begin{align*}
h(x)&=f(-3x-12) \
&=sqrt{-3x-12} tag{text{replace $x$ with $-3x-12$}}\
end{align*}
$$

$$
begin{align*}
p(x)&=f(4x+8)\
&=|4x+8| tag{text{replace $x$ with $4x+8$}}\
end{align*}
$$

$T(s)=dfrac{20}{s}implies$ vertical stretching by factor 20

$T(s)=dfrac{15}{s-3}implies$ vertical stretching by factor 15 then translation 3 units to the right

$T(s)=dfrac{15}{s-3}implies$ vertical stretching by factor 15 then translation 3 units to the right

then, stretched vertically by the factor 3 $longrightarrow$ $g_{2}(x)=3sqrt{-x}$

then translated 5 units right $longrightarrow$ $g_{3}(x)=3sqrt{-(x-5)}$

then translated 2 units down $longrightarrow$ $g_{4}(x)=3sqrt{-(x-5)}-2$

Use these steps, we get new function $textcolor{#4257b2}{h(x)=3g(-(x-5))-2=3sqrt{-(x-5)}-2}$

See graph

then, stretched vertically by the factor 3 $longrightarrow$ $y_{2}=3f(-x)$

then translated up 2 units $longrightarrow$ $y_{3}=3f(-x)+2$

then translated 1 units left $longrightarrow$ $y_{4}=3f(-(x+1))+2$

$text{Use these steps we get new function: $textcolor{#4257b2}{h(x)=3f(-(x+1))+2}$}$

Parent function of this given function is $y=dfrac{1}{x}$

The graph of a function may be transformed using:

horizontal shift which shifts the graph 2 units to the right and verticaly shift 1 unit up, vertical stretch of 3, reflection in the x-axis.

Use these steps we get given function $y=dfrac{3}{-(x-2)}+1$, it corresponds to $text{textcolor{#4257b2}{Graph C}}$

Parent function of this given function is $y=|x|$.

The graph of $y=2|x-3|-2$ is stretched vertically by the factor 2, and then translated 2 units down units and 3 unit right.

Using these transformation we get function $y=2|x-3|-2$, it corresponds to $text{textcolor{#4257b2}{Graph E}}$.

Parent function of this given function is $y=sqrt{x}$.

The graph of $y=sqrt{x}$ is reflected across the y-axis, stretched vertically by the factor 2, and then translated 3 units left and 2 units down

It corresponds to $text{textcolor{#4257b2}{graph A}}$.

Parent function of this given function is $y=x^{2}$

The graph of $y=x^{2}$ is translated down 3 units and 2 units right.

It corresponds to $text{textcolor{#4257b2}{graph G}}$

Parent function of this function is $y=dfrac{1}{x}$

The graph of $y=dfrac{1}{x}$ is reflected in one of the axes and translated 3 units down.

It corresponds to $text{textcolor{#4257b2}{graph F}}$.

Parent function of this function is $y=|x|$

The graph of $y=|x|$ is reflected acros the y-axis, translated 4 units left and 2 units down.

It corresponds to $text{textcolor{#4257b2}{Graph D}}$

Parent function of this function is $y=sqrt{x}$

The graph of $y=sqrt{x}$ is reflected across x-axis and y-axis, stretched vertically by the factor 0.5, translated 1 unit right and 1 unit up.

It corresponds to $text{textcolor{#4257b2}{Graph H}}$

$text{Parent function of this function is $y=x^{2}$}$

The graph of $y=x^{2}$ is reflected in the y-axis, stretched vertically by the factor $dfrac{1}{2}$ ,and then translated 4 units left and 1 unit up.

It corresponds to $text{textcolor{#4257b2}{Graph B}}$

b) $longrightarrow$ Graph E

c) $longrightarrow$ Graph A

d) $longrightarrow$ Graph G

e) $longrightarrow$ Graph F

f) $longrightarrow$ Graph D

g) $longrightarrow$ Graph H

h) $longrightarrow$ Graph B

We need determine $a, k, c$ and $d$.

The graph of $f(x)=sqrt{x}$ is stretched vertically by the factor 2 $longrightarrow$ $y=2sqrt{x}$

then translated 4 units right $longrightarrow$ $y=2sqrt{x-4}$

and then reflected in the x-axis $longrightarrow$ $y=-2sqrt{x-4}$

Use these steps, we get new function $textcolor{#4257b2}{y=-2g(x-4)=-2sqrt{x-4}}$

If we compare this function with $y=af[k(x-d)]+c$, we can conclude that:

$a=-2, k=1, c=0, d=4$

We can identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $[4,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $(-infty,0)$

We need determine $a, k, c$ and $d$.

The graph of $f(x)=dfrac{1}{x}$ is compressed vertically by the factor 2 $longrightarrow$ $y=dfrac{1}{2x}$

then reflecred in the y-axis $longrightarrow$ $y=dfrac{1}{2}(dfrac{1}{-x})$

then translated 3 units left $longrightarrow$ $y=dfrac{1}{2}(dfrac{1}{-(x+3)})$

and thentranslated 4 units down $longrightarrow$ $y=dfrac{1}{2}(dfrac{1}{-(x+3)})-4$

Use these steps, we get new function $y=dfrac{1}{2}f(-(x+3))-4$

If we compare this function with $y=af[k(x-d)]+c$, we can conclude that:

$a=dfrac{1}{2}, k=-1, c=-4, d=-3$

We can identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval $left{xin R|xne-3 right}$:

$text{textcolor{#4257b2}{Range}}$ of the function is interval:$left{yin R|yne-4 right}$

We need determine $a, k, c$ and $d$.

The graph of $f(x)=|x|$ is compressed horizontaly by the factor $dfrac{1}{3}$ $longrightarrow$ $y=dfrac{1}{3}|x|$

then, verticaly streched by the factor 3 $longrightarrow$ $y=3(dfrac{1}{3}|x|)$

then translated 1 units right $longrightarrow$ $y=3(dfrac{1}{3}|x-1|)$

and then translated 6 units down $longrightarrow$ $y=3(dfrac{1}{3}|x-1|)-6$

Use these steps, we get new function $y=3f(dfrac{1}{3}|x-1|)-6$

If we compare this function with $y=af[k(x-d)]+c$, we can conclude that:

$a=3, k=dfrac{1}{3}, c=-6, d=1$

We can identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

From the graph we can conclude that:

$text{textcolor{#4257b2}{Domain}}$ of the function is interval : $(-infty,+infty)$

$text{textcolor{#4257b2}{Range}}$ of the function is interval: $[-6,+infty)$

Algebraic, an x-intercept is a point on the graph where y is zero,

$text{If $f(x)=(x-2)(x+5)$, we need determine x-intercepts}$.

Using the definitions of the intercepts, we will proceed as follows:

$text{$y = 0$ for the x-intercepts, so}$

$$
begin{align*}
y&=0\
(x-2)&(x+5)=0\
x-2=0 &lor x+5=0\
x=2 &lor x=-5\
end{align*}
$$

$text{Then the x-intercepts are the points $(2,0)$ and $(-5,0)$}$

$text{$y = 0$ for the x-intercepts, so}$

$$
begin{align*}
y&=0\
-4(x-2)&(x+5)=0\
x-2=0 &lor x+5=0\
x=2 &lor x=-5\
end{align*}
$$

$text{Then the x-intercepts are the points $(2,0)$ and $(-5,0)$}$

We can conclude that: the x-intercepts are the same as in part a)

$y=f(-dfrac{1}{3}x)=(-dfrac{1}{3}x-2)(-dfrac{1}{3}x+5)$ ($text{replace $x$ with $-dfrac{1}{3}x$}$)

$text{$y = 0$ for the x-intercepts, so}$

$$
begin{align*}
y&=0\
(-dfrac{1}{3}x-2)&(-dfrac{1}{3}x+5)=0\
-dfrac{1}{3}-2=0 &lor -dfrac{1}{3}x+5=0\
-dfrac{1}{3}x=2 &lor -dfrac{1}{3}x=-5 tag{text{multiply both sides of equation by -3}}\
x=-6 & lor x=15\
end{align*}
$$

$text{Then the x-intercepts are the points $(-6,0)$ and $(15,0)$}$

$$
begin{align*}\
y&=f(-(x+2))\
&=(-(x+2)-2)(-(x+2)+5)\
&=(-x-2-2)(-x-2+5)\
&=(-x-4)(-x+3) (text{replace $x$ with $-(x+2)$})\
end{align*}
$$

$text{$y = 0$ for the x-intercepts, so}$

$$
begin{align*}
y&=0\
(-x-4)&(-x+3)=0\
-x-4=0 &lor -x+3=0\
-x=4 &lor -x=-3 tag{text{multiply both sides of equation by -1}}\
x=-4 & lor x=3\
end{align*}
$$

$text{Then the x-intercepts are the points $(-4,0)$ and $(3,0)$}$

b) $(2,0)$ and $(-5,0)$

c) $(-6,0)$ and $(10,0)$

a) $(-4,0)$ and $(3,0)$

(1) horizontally compress by a factor of $dfrac{1}{|k|}implies y=f(kx)$

if $k<0$, also reflect in the $y$-axis

(2) vertically stretch by a factor of $aimplies y=acdot f(kx)$

if $a<0$, also reflect in the $x$-axis

(3) horizontally translate $d$ units to the right $implies y=acdot f(k(x-d))$

if $d<0$, translate to the left

(4) vertically translate $c$ units up $implies y=acdot f(k(x-d))+c$

if $c<0$, translate downwards

if $k<0$, also reflect in the $y$-axis

(2) vertically stretch by a factor of $aimplies y=acdot f(kx)$

if $a<0$, also reflect in the $x$-axis

(3) horizontally translate $d$ units to the right $implies y=acdot f(k(x-d))$

if $d<0$, translate to the left

(4) vertically translate $c$ units up $implies y=acdot f(k(x-d))+c$

if $c<0$, translate downwards

$text{then, scompressed vertically by the factor $dfrac{1}{4}$ $longrightarrow$ $y=-dfrac{1}{4}x^{2}$}$

$text{then translated 6 units left $longrightarrow$ $y=-dfrac{1}{4}(x+6)^{2}$}$

$text{then translated 2 units up $longrightarrow$ $y=-dfrac{1}{4}(x+6)^{2}+2$}$

$$
text{Use these steps, we get new function $textcolor{#4257b2}{y=-dfrac{1}{4}(x+6)^{2}+2}$}
$$

$text{The domain of a function is the set of all real values of $x$ that will give real values for $y$, domain is interval $(-infty, +infty)$}$

$text{The range of a function is the set of all real values of $y$ that we can get by plugging real numbers into $x$, so, range is interval $[0,+infty)$}$

We can identify the domain and range of functions is by using graphs.

Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis

The range is the set of possible output values, which are shown on the y-axis.

$text{The domain of a function is the set of nonegative real values of $x$ that will give real values for $y$, domain is interval $[0, +infty)$}$

$text{The range of a function is the set of all real values of $y$ that we can get by plugging real numbers into $x$, so, range is interval $[0,+infty)$}$

Reflect the right-half of $f(x)$ with respect to the line $y=x$