a) $left{(-3,0), (-1,1), (0,1), (4,5), (0,6) right}$

This relation is not a function,because $x=0$ value of the independent variable corresponds two value of the dependent variable.

Domain of this relation is :$left{ -3,-1,0,4,right}$

Range of this relation is: $left{0,1,5,6 right}$

b) $y=4-x$

This relation is a function, because each $x$-value ( independent variable) corresponds with only one $y$-value (dependent variable)

This is a linear function.

Domain of this function is the set of all real numbers, $(-infty,+infty )$

Range of this function is the set of all real numbers, $(-infty, +infty)$

c) This relation is not a function. If $xgeq-4$ this relation is not pasess vertical-line test.Any vertical line intersects the graph of a relation more two times, so, the relation is not a function.

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{Domain of this relation is: $[-4,+infty)$}$

Range of this relation is: $(-infty, +infty)$

d) This relation is not a function because it is not passes Vertical line test. Vertical line intersects the graph of a relation more than once, so the relation is not a function (see graph)

Domain of this relation is interval $(-4,4)$

Range of this relation is interval $(-4,4)$

begin{table}[]
defarraystretch{1.3}%
begin{tabular}{|l|l|}
hline
a) & begin{tabular}[c]{@{}l@{}}domain = ${ -3,-1,0,4}$\ range = ${ 0,1,5,6}$\ not a function since there are 2 $y$-values for $x=0$end{tabular} \ hline
b) & begin{tabular}[c]{@{}l@{}}domain = ${ xinbold{R}}$\ range = ${ yinbold{R}}$\ function because each $x$-value has only one $y$-valueend{tabular} \ hline
c) & begin{tabular}[c]{@{}l@{}}domain = ${ xinbold{R};|;xgeq-4}$\ range = ${ yinbold{R}}$\ not a function, because at $x>-4$, each has two $y$-valuesend{tabular} \ hline
d) & begin{tabular}[c]{@{}l@{}}domain = ${ xinbold{R};|;-4leq xleq 4}$\ range = ${ yinbold{R};|;-4leq xleq 4}$\ not a function because some $x$-values have two $y$-valuesend{tabular} \ hline
end{tabular}
end{table}

If we want to determine which of relation is a function, we can use

$text{textcolor{#4257b2}{ Vertical line test}}$:,,If any vertical line intersects the graph of a relation more than once, then the relation is not a function”

b) $y=4-3x$

We need use the vertical-line test to see how many points on the graph there were for each value of $x$.

Use graph, the vertical line intersected the graph in only one
place. So, each $x$-value in the domain corresponds with only one
$y$-value in the range.

So, this relation is a function.

c) $y=(x-2)^{2}+4$

Use graph, we can see the vertical line intersected the graph in only one place. So each $x$-value in the domain corresponds with only one $y$-value in the range.

We can conclude: This relation is a function

d) $x^{2}+y^{2}=1$

Using the vertical-line test to see how many points on the graph there were for each value of $x$, we can conclude vertical line intersects the graph of a relation in two points, then the relation is not a function

This relation is not a function.

e) $y=dfrac{1}{x}$

Using vertical line intersects the graph of a relation only once, so the relation is a function.

This relation is a function.

f) $y=sqrt{x}$

e) $y=dfrac{1}{x}$

Using vertical line intersects the graph in only one point, so the relation is a function.

This relation is a function.

We can find many functions whose domain is the set of real numbers and whose range is the set of real numbers less than or equal to 3.

For example,

one of these functions is $y=-x^{2}+3$

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.

Domain of this function is interval $(-infty,+infty)$

Range of this function is interval $(-infty,3]$

$$
y=-x^{2}+3
$$

$f(x)=x^{2}+3x-5$ and $g(x)=2x-3$ These are a given functions

a) If we want to determine $f(-1)$, we need substitute $-1$ for $x$ into function $f(x)$

$$
begin{align*}
f(x)&=x^{2}+3x-5\
f(-1)&=(-1)^{2}+3(-1)-5\
&=1-3-5\
&=-7\
end{align*}
$$

So, $f(-1)=-7$

b) If we want to determine $f(0)$, we need substitute $0$ for $x$ into function $f(x)$

$$
begin{align*}
f(x)&=x^{2}+3x-5\
f(0)&=(0)^{2}+3(0)-5\
&=0+0-5\
&=-5\
end{align*}
$$

So, $f(0)=-5$

c)We can evaluate $g(dfrac{1}{2})$ substiting $dfrac{1}{2}$ for $x$ into function $g(x)$

$$
begin{align*}
g(x)&=2x-3\
g(dfrac{1}{2})&=2cdotdfrac{1}{2}-3\
&=1-3\
&=-2\
end{align*}
$$

So, $g(dfrac{1}{2})=-2$

d) We can evaluate $f(2b)$ substiting $2b$ for $x$ into function $f(x)$

$$
begin{align*}
f(x)&=x^{2}+3x-5\
f(2b)&=(2b)^{2}+3(2b)-5\
&=4b^{2}+6b-5\
end{align*}
$$

So is, $f(2b)=4b^{2}+6b-5$

e) We can determine $g(1-4a)$ substiting $1-4a$ for $x$ into function $g(x)$

$$
begin{align*}
g(x)&=2x-3\
g(1-4a)&=2(1-4a)-3\
&=2-8a-3\
&=-1-8a\
end{align*}
$$

So, $g(1-4a)=-1-8a$

f) In this part we need determine $x$ when $f(x)=g(x)$

$$
begin{align*}
f(x)&=g(x)\
x^{2}+3x-5&=2x-3 tag{text{subtract 2x-3 from both sides}}\
x^{2}+3x-5-2x+3&=0\
x^{2}+x-2&=0\
x^{2}-x+2x-2&=0 tag{text{solve equation by factoring}}\
x(x-1)+2(x-1)&=0\
(x-1)(x+2)&=0\
x-1=0 &lor x+2=0\
x=1 & lor x=-2
end{align*}
$$

This means, if $x=1$ or $x=-2$, then $f(x)=g(x)$

a)$f(-1)=-7$

b) $f(0)=-5$

c) $g(dfrac{1}{2})=-2$

d) $f(2b)=4b^{2}+6b-5$

e) $g(1-4a)=-1-8a$

f) $f(x)=g(x)$ when $x=1$ or $x=-2$

a) $f(x)=-2(x-3)^{2}+4$ This is a given function

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.

Use graph, we can determine:

Domain of this function is: $(-infty, +infty)$

Range of this function is : $(-infty,4]$

Graph of the function $f(x)=-2(x-3)^{2}+4$ is:

b) $f(1)$ represents the value or output of the function when the input is $x=1$.

From the graph we can conclude $f(1)$ takes the value -4 at $x=1$,
then $f(1)=-4$ and the point $(1, -4)$ lies on the graph of $f(x)$.

Answer is: $f(1)$ represent $y$-coordinate corresponding to $x=1$.

c)

$text{textcolor{#c34632}{i)}}$ Using the equation of the function we can determine $f(3)-f(2)$

The first we need determine $f(3)$ and $f(2)$:

$$
begin{align*}
f(x)&=-2(x-3)^{2}+4 \
f(3)&=-2(3-3)^{2}+4 tag{text{substitute $3$ for $x$ into function $f(x)$}}\
&=-2cdot0+4\
&=4\
end{align*}
$$

$$
begin{align*}
f(x)&=-2(x-3)^{2}+4 \
f(2)&=-2(2-3)^{2}+4 tag{text{substitute $2$ for $x$ into function $f(x)$}}\
&=-2(-1)^{2}+4\
&=-2+4\
&=2\
end{align*}
$$

Now we can determine $f(3)-f(2)=4-2=2$

c) $text{textcolor{#c34632}{ii}}$ Using the equation of the function we can determine $2f(5)+7$

$$
begin{align*}
f(x)&=-2(x-3)^{2}+4 \
2f(5)+7&=2cdot(-2(5-3)^{2}+4)+7 tag{text{substitute $5$ for $x$ into function $f(x)$}}\
&=2(-2(2)^{2}+4)+7\
&=2(-8+4)+7\
&=2(-4)+7\
&=-8+7\
&=-1\
end{align*}
$$

c) $text{textcolor{#c34632}{iii)}}$ Using the equation of the function we can determine $f(1-x)$

$$
begin{align*}
f(x)&=-2(x-3)^{2}+4 \
f(1-x)&=-2(1-x-3)^{2}+4 tag{text{substitute $1-x$ for $x$ into function $f(x)$}}\
&=-2(-2-x)^{2}+4\
end{align*}
$$

a) domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;yleq 4}$

b) $f(1)$ corresponds to the $y$-coordinate at $x=1$

c)
text${text{i}}$) 2

ii) $-$1

iii) $-2(-x-2)^2+4$

If $f(x)=x^{2}-4x+3$, determine inputs $x$ for which is $f(x)=8$

$$
begin{align*}
f(x)&=8\
&x^{2}-4x+3=8\
&x^{2}-4x+3-8=0 tag{text{subtract 8 from both sides}}\
&x^{2}-4x-5=0\
&x^{2}+x-5x-5=0 tag{text{set $-4x=x-5x$}}\
&x(x+1)-5(x+1)=0tag{text{factor}}\
&(x+1)(x-5)=0 tag{text{factor}}\
&x+1=0 lor x-5=0 tag{text{set factors equal to 0}}\
&x=-1 lor x=5\
end{align*}
$$

So, this means , if $x=-1$ or $x=5$, then $f(x)=8$

$$
text{$x=-1 lor x=5$}
$$

$bold{Analysis: }$ Since the ball is initially $(t=0)$ at the top of the building 60 m tall, a point on the graph is $(0,60)$. After 2 s, it reached a height of $80$ m above the ground so another point is $(2,80)$ (we may assume that this is the maximum height). Then it hits the ground after $6$ s which means $(6,0)$ is also on the graph.

b) Since the ball reached the ground after 6 s and time cannot be negative, the domain is

${ tinbold{R};|;0leq tleq 6}$

Since the maximum height is $80$ m and height cannot be negative, the range is

${ hinbold{R};|;0leq hleq 80}$

c) We must find the equation of the parabola containing $(0,60)$, (2,80), and $(6,0)$.

Since the maximum height is at $(2,80)$, the vertex form is

$y=a(x-2)^2+80$

Now, we can find the value of $a$ using any of the two points.

$60=a(0-2)^2+80$

$a=dfrac{60-80}{(0-2)^2}=dfrac{-20}{4}=-5$

Therefore, the equation of the parabola is

$h=-5(t-2)^2+80$

$h=-5(t^2-4t+4)+80$

$h=-5t^2+20t-20+80$

$$
h=-5t^2+20t+60
$$

Note that if we do not assume the point $(2,80)$ as the vertex (since it is not explicitly mentioned that it is the maximum height), we can solve it by setting up 3 equations with 3 unknowns.

$60=a(0-h)^2+kimplies 60=ah^2+k$

$80=a(2-h)^2+k$

$0=a(6-h)^2+k$

Subtract 3rd equation from 1st equation

$60-0=a(h^2-(6-h)^2)+(k-k)$

$60=a(h^2-(36-12h+h^2))$

$60=a(h^2-36+12h-h^2)$

$60=a(12h-36)$

$5=a(h-3)$

Subtract 1st equation from 2nd equation

$80-60=a[(2-h)^2-h^2]+(k-k)$

$20=a(4-4h+h^2-h^2)$

$20=a(4-4h)$

$5=a(1-h)$

Thus,

$a(h-3)=a(1-h)$

$h-3=1-h$

$2h=4$

$h=2$

$a=dfrac{5}{1-h}=dfrac{5}{1-2}=-5$

$k=60-ah^2=60-(-5)(2^2)=80$

$$
therefore y=-5(x-2)^2+80
$$

a) The graph has been plotted in the answers.

b) domain = ${ tinbold{R};|;0leq tleq 6}$ , range = ${ hinbold{R};|;0leq hleq 80}$

c) $h=-5t^2+20t+60$

a) domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq 3}$

b) domain = ${ xinbold{R};|;xgeq -2}$ , range= ${ yinbold{R};|;ygeq 0}$

a) Let be $A$ area , $w$ width, and $l$ length of the rectangular.

Use the given section of the rectangular we can determine lenth $l$ as a function of the area $A$ and width $w$:

$$
begin{align*}
&540=3w+2l\
&540-3w=2l
tag{text{subtract 3w from the both sides}}\
&l=dfrac{540-3w}{2} tag{text{divide both sides by 2}}\
end{align*}
$$

We know, the tota area of the rectangular is equal $A=length cdot width$

Substitute equation of the length $l$ into equation of the area $A$ we get:

$$
begin{align*}
A&=lcdot w\
&=dfrac{540-3w}{2}cdot w\
&=dfrac{w(540-3w)}{2}\
end{align*}
$$

b) $w$ is a width of the rectangular, so $w$ must be greater than zero

$A$ is area of the rectangular, so $A$ must be greater than zero, but if we look a equation of the area $A=dfrac{w(540-3w)}{2}$, we can conclude that A is greater than zero if $540-3w>0$ , so, this means 3w must be less than 540:

$$
begin{align*}
540-3w>0\
3w<540\
w< dfrac{540}{3}\
w<180\
end{align*}
$$

This means, domain of the function $A=dfrac{w(540-3w)}{2}$ is:

$Domain=left{ w in R| 0<w<180 right}$

b) $A=dfrac{w(540-3w)}{2}$

This is a quadratic function that opens down It has two zeros, at 0 and 180. The vertex lies halfway in between the
zeros, above the x-axis, so the numbers
in the domain have to be between 0
and 180.

The vertex is halfway between $w=0$ and $w=180$, so,the $x$-coordinate of the vertex is: $dfrac{0+180}{2}=90$

Substitute $w=90$ into the area function $A$ we will find the $y$-coordinate of the vertex:

$$
begin{align*}
A(w)&=dfrac{w(540-3w)}{2}\
A(90)&=dfrac{90(540-3cdot90)}{2}\
&=dfrac{90cdot 270}{2}\
&=12150\
end{align*}
$$

So, vertex is $(90,12150)$

This means that: Range od the function is $left{ Ain R| 0<Aleq12150right}$

c) Area is a quadratic function with vertex $(90,12150)$, so, maximum area we get if width $w=90$ and length $l=3w=3cdot90=270$

a) $A(w)=dfrac{(540-3w)w}{2}$

b) domain = ${ xinbold{R};|;0leq wleq 180}$

range = ${ Ainbold{R};|;0<Aleq 12;150}$

c) $l=270$ m , $w=90$ m

a) $f(x)=2x-5$ this is a given linear function

one of the methods for determining the inverse of a linear function is draw a graph of linear function and then determine its reflection in the line $y=x$

Use graph to determine the slope-intercept form of inverse :slope is 0.5 and $y$-intercept is 2.5.

Inverse function is: $f^{-1}(x)=0.5 x +2.5$

b) Second method is reverse operations:

Linear function is: $f(x)=dfrac{x+3}{7}$

Use reverse operations we get:

$$
begin{align*}
f(x)&=dfrac{x+3}{7}
7f(x)&=x+3 tag{text{multiply both sides by 7}}\
7f(x)-3&=x tag{text{subtract 3 from both sides}}
end{align*}
$$

The inverse function of $f(x)=dfrac{x+3}{7}$ is

$$
f^{-1}(x)=7x-3
$$

c) $f(x)=4-dfrac{1}{2}x$ This is a given function

Switch $x$ and $f(x)$, then solve for $f(x)$:

$$
begin{align*}
x&=4-dfrac{1}{2}f(x)\
x-4&=-dfrac{1}{2} tag{text{subtract 4 from both sides}}\
-2(x-4)=f(x) tag{text{multiply both sides by -2}}\
f(x)=-2x+8
end{align*}
$$

So, inverse function of $f(x)=4-dfrac{1}{2}x$ is

$f^{-1}(x)=-2x+8$

a) $f^{-1}(x)=0.5x+2.5$

b) $f^{-1}(x)=7x-3$

c) $f^{-1}(x)=-2x+8$

a) Let $x$ be the number of people who will attend the end. The 15 000 from the sponsors is fixed regardless of the number of people who will attend so it must be our $y$-intercept.

$f(x)=30x+15;000$

b) There is no upper limit in the number of people who could join while the number of people can’t be negative.

domain = ${ xinbold{R};|;xgeq 0}$

Regardless of the number of people who would join, the income could not be less than the fixed $$15;000$ from the sponsors.

range = ${ yinbold{R};|;ygeq 15;000}$

c) The number of people as a function of income is the inverse function of $f(x)$.

To find the inverse, swap the positions of the variable and solve for $y$

$y=30x+15000$

$x=30y+15000$

$30y=x-15000$

$y=f^{-1}(x)=dfrac{x-15000}{30}$

The domain of the inverse is the range of the original function.

domain = ${ xinbold{R};|;xgeq 15000}$

a) $f(x)=30x+15000$

b) domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq 15000}$

c) $f^{-1}(x)=dfrac{x-15000}{30}$ , domain = ${ xinbold{R};|;xgeq 15;000}$

The given graphs have undergone transformations of the form $f(kx)$. We know that this is horizontal compression by a factor of $dfrac{1}{|k|}$ if $|k|>1$ and horizontal stretching by $dfrac{1}{|k|}$ if $0<|k|<1$.

If $k<0$, then it is also reflected in the $y$-axis.

a) The graph looks like upper half-parabola that opens to the right and it passes through $(4,2)$. Thus, we can assume the parent function as $y=sqrt{x}$.

Now, the transformed graph (in red) appears to be compressed. Notice that it apparently passes through $(4,4)$. (though not exactly, this might be graphic error)

$y=sqrt{kx}$

$4=sqrt{k(4)}$

$16=4k$

$k=4$

Therefore, the equation is $y=sqrt{4x}$

b) The graph looks like a hyperbola so we can assume the parent function as $y=dfrac{1}{x}$. Observe that the red graph is reflected in the $y$-axis $f(x)to f(-x)$, horizontally stretched and passes through $left(5,1right)$.

$y=dfrac{1}{-kx}$

$1=dfrac{1}{-k(5)}$

$k=-dfrac{1}{5}$

Therefore, the equation is $y=dfrac{1}{-frac{1}{5}x}$

a) $y=sqrt{4x}$

b) $y=dfrac{1}{-frac{1}{5}x}$

a) In this part, the given functions are:

$f(x)=x^{2}$

$g(x)=(dfrac{1}{2}x)^{2}$

$h(x)=-(2x)^{2}$

the function $f(x)$ is a parent function of the functions $g(x)$ and $h(x)$.

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.

$f(x)$ : Domain=${xin R| -inftyleq xleq +infty}$ , Range=$left{ yin R| 0leq y<+infty right}$

$g(x)$ : Domain=${xin R | -inftyleq xleq +infty}$ , Range=$left{ yin R | 0leq y<+infty right}$

$h(x)$ : Domain=${xin R| -inftyleq xleq +infty}$ , Range=$left{ yin R| -infty<yleq 0 right}$

b) The given functions are:

$f(x)=|x|$

$g(x)=|-4x|$

$h(x)=|dfrac{1}{4}x|$

The function $f(x)=|x|$ is a parent function of the functions $g(x)$ and $h(x)$.

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.

$f(x)$ : Domain=${xin R| -inftyleq xleq +infty}$ , Range=$left{ yin R| 0leq y<+infty right}$

$g(x)$ : Domain=${xin R | -inftyleq xleq +infty}$ , Range=$left{ yin R | 0leq y<+infty right}$

$h(x)$ : Domain=${xin R| -inftyleq xleq +infty}$ , Range=$left{ yin R| 0leq y<+infty right}$

begin{table}[]
defarraystretch{1.6}%
begin{tabular}{|l|l|}
hline
a) & begin{tabular}[c]{@{}l@{}}$f(x):$ domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq 0}$\ $g(x):$ domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq 0}$\ $h(x):$ domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;yleq 0}$end{tabular} \ hline
b) & begin{tabular}[c]{@{}l@{}}$f(x):$ domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq 0}$\ $g(x):$ domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq 0}$\ $h(x):$ domain = ${ xinbold{R}}$ , range = ${ yinbold{R};|;ygeq0}$end{tabular} \ hline
end{tabular}
end{table}

When combining transformations, it is very important to consider the order of the transformations. Translations must be done last.

When combining vertical transformations written in the form $af(x)+k$, first vertically stretch by $a$ and then vertically shift by $k$.

We know that $(1,4)$ is on the graph of $y=f(x)$, thus, $4=f(1)$

We shall examine what will happen to this point upon performing the transformations:

(1) reflecting in the $y$-axis $implies$ change $x$ to $-x$

$y=f(-x)implies (-1,4)$

(2) horizontal compression by $dfrac{1}{4}$ $implies$ multiply $x$-values by $frac{1}{4}$

$y=f(-4x)implies left(-dfrac{1}{4},4right)$

(3) vertical stretching by 3 $implies$ multiply $y$-values by 3

$y=3f(-4x) implies left( -dfrac{1}{4}, 12right)$

(4) horizontal translation 1 unit to the left $implies$ subtract $x$-values by 1

$y=3f[-4(x+1)] implies left(-dfrac{5}{4}, 12right)$

(5) vertical translation 2 units down $implies$ subtract $y$-values by $2$

$y=3f[-4(x+1)]-2implies left(-dfrac{5}{4}, 10right)$

$left(-dfrac{5}{4}, 10right)$

a) which transforamations we use to get function $y=-2f[dfrac{1}{3}x+4]-1$?

The graph of $y=f(x)$ is reflected across the x-axiss $longrightarrow$ $y=-f(x)$

then, stretched vertically by the factor 2 $longrightarrow$ $y=-2f(x)$

then, stretched horizontaly by the factor 3 $longrightarrow$ $y=-2f(dfrac{1}{3}x)$

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

then translated 1 unit down $longrightarrow$ $y=-2f(dfrac{1}{3}x+4)-1$

Use these steps, we get function $textcolor{#4257b2}{y=-2f(dfrac{1}{3}x+4)-1}$

b) In part a) we have a function $y=-2f(dfrac{1}{3}x+4)-1$, if $f(x)=x^{2}$ then, we determine a function as a follows:

$$
begin{align*}\
y&=-2f(dfrac{1}{3}x+4)-1\
&=-2(dfrac{1}{3}x+4)^{2}-1\
end{align*}
$$

Graph of this function is:

a) reflection in $x$-axis, vertical stretching by factor 2, horizontal stretching by factor 3, translation 4 units to the left and 1 unit down

b) The graph has been plotted in the answers.

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

then, reflected across the y -axiss $longrightarrow$ $y=sqrt{-x2}$

then translated 3 units right $longrightarrow$ $y=sqrt{-2(x-3)}$

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

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

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.

Domain$={xin R | xleq3}$

Range$=left{yin R | ygeq-2 right}$

b) The graph of $y=dfrac{1}{x}$ is stretched vertically by the factor 3 $longrightarrow$ $y=dfrac{3}{x}$

then, reflected across the x-axiss $longrightarrow$ $y=dfrac{-3}{x}$

then translated 4 units left $longrightarrow$ $y=dfrac{-3}{x+4}$

then translated 1 unit up $longrightarrow$ $y=dfrac{-3}{x+4}+1$

Use these steps, we get a function $textcolor{#4257b2}{y=dfrac{-3}{x+4}+1}$

Use graph, we can determine domain and range of this function:

Domain$={xin R | xne -4}$

Range$={yin R | yne 1}$

begin{table}[]
defarraystretch{1.9}%
begin{tabular}{|l|l|}
hline
a) & begin{tabular}[c]{@{}l@{}}$y=-sqrt{2(x-3)}-2$\ domain = ${ xinbold{R};|;xgeq 3}$\ range = ${ yinbold{R};|;yleq -2}$end{tabular} \ hline
b) & begin{tabular}[c]{@{}l@{}}$y=dfrac{-3}{x+4}+1$\ domain = ${ xinbold{R};|;xneq 4}$\ range = ${ yinbold{R};|;yneq 1}$end{tabular} \ hline
end{tabular}
end{table}

The graphical , x-intercepts are where the graph crosses the x-axis.

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

$text{If $f(x)=(x-4)(x+3)$, we need determine x-intercepts}$.

a) $y=f(x)=(x-4)(x+3)$

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

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

$$
begin{align*}
y&=0\
(x-4)&(x+3)=0\
x-4=0 &lor x+3=0\
x=4 &lor x=-3\
end{align*}
$$

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

b) $text{There is $y=-2f(x)=-2(x-4)(x+3)$}$

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

$$
begin{align*}
y&=0\
-2(x-4)&(x+3)=0\
x-4=0 &lor x+3=0\
x=4 &lor x=-3\
end{align*}
$$

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

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

c) $text{The first, we need determine new function $f(-dfrac{1}{2}x)$:}$

$y=f(-dfrac{1}{2}x)=(-dfrac{1}{2}x-4)(-dfrac{1}{2}x+3)$ ($text{replace $x$ with $-dfrac{1}{2}x$}$)

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

$$
begin{align*}
y&=0\
(-dfrac{1}{2}x-4)&(-dfrac{1}{2}x+3)=0\
-dfrac{1}{3}-4=0 &lor -dfrac{1}{2}x+3=0\
-dfrac{1}{2}x=4 &lor -dfrac{1}{2}x=-3 tag{text{multiply both sides of equation by -2}}\
x=-8 & lor x=6\
end{align*}
$$

$text{Then the x-intercepts are the points $(-8,0)$ and $(6,0)$}$

d) $text{The first, we need determine new function $f(-(x+1))$:}$

$$
begin{align*}\
y&=f(-(x+1))\
&=(-(x+1)-4)(-(x+1)+3)\ tag{text{replace x with -(x+1)}}\
&=(-x-1-4)(-x-1+3)\
&=(-x-5)(-x+2) \
end{align*}
$$

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

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

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

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

b) $(4,0)$ and $(-,0)$

c) $(-8,0)$ and $(6,0)$

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

The base function $f(x)$ has

domain = ${ xinbold{R};|;xgeq -4}$

range = ${ yinbold{R};|;y<-1}$

a) $y=2f(x)$ means vertical stretching by a factor of 2 $implies$ $y$-values are multiplied by 2

Thus, the domain is unaffected but the range will be multiplied by 2.

domain = ${ xinbold{R};|;xgeq -4}$

range = ${ yinbold{R};|;y<-2}$

b) $y=f(-x)$ means reflecting in the $y$-axis so the $x$-values will be multiplied by $-1$ which also reverses the inequality symbol, while the range is unaffected.

domain = ${ xinbold{R};|;xleq 4}$

range = ${ yinbold{R};|;y<-1}$

c) $y=3f(x+1)+4$

transformations affecting domain:

horizontal translation 1 unit to the left $implies$ subtract 1 to the $x$-values

domain = ${ xinbold{R};|;xgeq -5}$

transformations affecting range:

vertical stretching by factor 3 and translation 4 units up $implies$ multiply $y$-values by 3 then add 4

$3(-1)+4=1$

range = ${ yinbold{R};|;y< 1}$

d) $y=-2f(-x+5)+1=-2f(-(x-5))+1$

transformations affecting the domain:

reflecting the $y$-axis $implies$ multiply $x$ values by $-1$ and reverse inequality symbol

translating 5 units to the right $implies$ add 5 to $x$-values

$-4(-1)+5=9$

domain = ${ xinbold{R};|;xleq 9}$

transformations affecting the range:

reflecting in $x$-axis $implies$ multiply $y$-values by $-1$ and reverse inequality symbol

vertical stretching by factor 2 $implies$ multiply $y$-values by 2

add 1 unit to $y$-values

$(-1)(2)(-1)+1=3$

range = ${ yinbold{R};|;y>3}$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
a) & begin{tabular}[c]{@{}l@{}}domain = ${ xinbold{R};|;xgeq -4}$\ range = ${ yinbold{R};|;y<-2}$end{tabular} \ hline
b) & begin{tabular}[c]{@{}l@{}}domain = ${ xinbold{R};|;xleq 4}$\ range = ${ yinbold{R};|;y<-1}$end{tabular} \ hline
c) & begin{tabular}[c]{@{}l@{}}domain = ${ xinbold{R};|;xgeq -5}$\ range = ${ yinbold{R};|;y3}$end{tabular} \ hline
end{tabular}
end{table}