textbf{(a)} Below is sequence of machines:
$$g(x) rightarrow k(x) rightarrow h(x) rightarrow f(x)$$
Hence we obtain the function $f circ h circ k circ g(x)$.\\
Putting $x=6$ in $g(x)$:
begin{align*}
g(6)&=-(6-2)^2\
&=-(4)^2\
&=-16
intertext{Putting $x=g(6)=-16$ in $k(x)$:}
k(-16)&=-dfrac{(-16)}{2}-1\
&=8-1\
&=7
intertext{Putting $x=k(-16)=7$ in $h(x)$:}
h(6)&=2^7-7\
&=128-7\
&=121
intertext{Putting $x=h(7)=121$ in $f(x)$:}
f(121)&=sqrt{121}\
&=11
end{align*}\

textbf{(b)} Below is sequence of machines:
$$f(x) rightarrow g(x) rightarrow k(x) rightarrow h(x)$$
Hence we obtain the function $f circ h circ k circ g(x)$.\\
Putting $x=64$ in $f(x)$:
begin{align*}
f(64)&=sqrt{64}\
&=8
intertext{Putting $x=f(64)=8$ in $g(x)$:}
g(8)&=-(8-2)^2\
&=-(6)^2\
&=-36
intertext{Putting $x=g(8)=-36$ in $k(x)$:}
k(-36)&=-dfrac{(-36)}{2}-1\
&=18-1\
&=17
intertext{Putting $x=k(-36)=17$ in $h(x)$:}
h(17)&=2^{17}-7\
&=131072-7\
&=131065
end{align*}\

(a) $g(x) rightarrow k(x) rightarrow h(x) rightarrow f(x)$ (b) $f(x) rightarrow g(x) rightarrow k(x) rightarrow h(x)$

Let us consider the given function
$$
color{#4257b2} f(x)=frac{1}{x-2}.
$$

a.

We need to calculate the value of $f(4)$. Therefore, we substitute $x=4$ in the given function and calculate the output. This gives

$$
f(4)=frac{1}{4-2} = frac{1}{2}.
$$

Therefore
$$
color{#c34632} f(4)=frac{1}{2}.
$$

b.

We need to find the value of $x$ for which $f(x)=1$. Therefore, we substitute $f(x)=1$ in the equation of the given function and calculate $x$. This gives

$$
1=frac{1}{x-2} implies x-2=1 implies x=3
$$

Therefore the input that gives an output of $1$ is
$$
color{#c34632} x=3.
$$

$$
text{a.} hspace{3mm} f(4)=frac{1}{2} hspace{20mm} text{b.} hspace{3mm} x=3
$$

The inner working of the machine is:

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

a-

If 3 is dropped in the machine:

$f(3)=3^2+2(3)+1=9+6+1=16$

The output is 16

b-

If -4 is dropped in the machine:

$f(-4)=(-4)^2+2(-4)+1=16-8+1=9$

The output is 9

a-

If -22.872 is dropped in the machine:

$f(-22.872)=(-22.872)^2+2(-22.872)+1=523.1284-45.744+1=478.384$

The output is 478.384

$g(x)=sqrt {x-5}$          (Given function machine $g$)

$h(x)=x^2-6$          (Given function machine $h$)

$x=6$          (Input value)

a-

Assuming the first function machine is $g(x)$ and the second is $h(x)$

$g(6)=sqrt {6-5}=sqrt {1}=1$

$h(1)=1^2-6=1-6=-5 neq 5$

Then this is not the correct order.

Assuming the first function machine is $h(x)$ and the second is $g(x)$

$h(6)=6^2-6=36-6=30$

$g(30)=sqrt {30-5}=sqrt {25}=5$

Then, Angelica must put the function machine $h(x)$ in the first and $g(x)$ in the last.

b-

Yes it is possible to get a final output of $-5$. by putting the function machine $g(x)$ in the first and the function machine $h(x)$ in the last.

$g(6)=sqrt {6-5}=sqrt {1}=1$

$h(1)=1^2-6=1-6=-5$

a-          Angelica must put the function machine $h(x)$ in the first and $g(x)$ in the last to get a final output of 5 from the input of 6.

b-          It is possible for Angelica to get a final output of $-5$. by putting the function machine $g(x)$ in the first and the function machine $h(x)$ in the last.

a-          The graph of the function $y=-2x+7$

b-          The graph of the function $y=dfrac {3}{5}x+1$

c-          The graph of the function $3x+2y=6$

d-          The graph of the function $y=x^2$

a-          The graph of the function $y=-2x+7$

b-          The graph of the function $y=dfrac {3}{5}x+1$

c-          The graph of the function $3x+2y=6$

d-          The graph of the function $y=x^2$

b-          What makes the equation different is that the power of the variable $x$ is 2. While in the other equations the power of the variable $x$ is 1.

a-          The graph of the part (d) is a curved line opens upwards but the other three graphs are straight lines.

b-          The power of the variable $x$ is 2. While in the other equations the power of the variable $x$ is 1.

c-          The graph of part (d) is called Parabola.

a-

$4(x-1)-2(3x+5)=-3x-1$          (Write the equation)

$4x-4-6x-10=-3x-1$          (Distributive property)

$4x-6x+3x=4+10-1$          (Grouping similar terms)

$x=13$          (Solve for $x$)

b-

$3x-5=2.5x+3-(x-4)$          (Write the equation)

$3x-5=2.5x+3-x+4$          (Distributive property)

$3x-2.5x+x=5+3+4$          (Grouping similar terms)

$1.5x=12$          (Simplify)

$x=8$          (Solve for $x$)

a-          $x=13$

b-          $x=8$

a.

Let us multiply the given expression by multiplying each monomial form the first bracket with each monomial from the second bracket

$$
begin{align*}
(5m-1)(m+2) &= 5m cdot (m+2) -1 cdot (m+2) \
&= 5m cdot m + 5m cdot 2 -1 cdot m -1 cdot 2 \
&= 5m^2 +10m -m -2 \
&= 5m^2 +9m -2.
end{align*}
$$

Therefore
$$
color{#c34632} (5m-1)(m+2)=5m^2 +9m -2.
$$

b.

Let us multiply the given expression by multiplying each monomial form the first bracket with each monomial from the second bracket

$$
begin{align*}
(6-x)(2+x) &= 6 cdot (2+x) -x cdot (2+x) \
&= 6 cdot 2 + 6 cdot x -x cdot 2 -x cdot x \
&= 12 +6x -2x -x^2 \
&= -x^2+4x+12.
end{align*}
$$

Therefore
$$
color{#c34632} (6-x)(2+x)=-x^2+4x+12.
$$

c.

Let us multiply the given expression

$$
begin{align*}
(5x-y)^2 &=(5x-y)(5x-y) \
&= 5x cdot (5x-y) -y cdot (5x-y) \
&= 5x cdot 5x + 5x cdot (-y) -y cdot (-5x) -y cdot (-y) \
&= 25x^2 -5xy -5xy +y^2 \
&= 25x^2 -10xy +y^2.
end{align*}
$$

Therefore
$$
color{#c34632} (5x-y)^2= 25x^2 -10xy +y^2.
$$

d.

Let us multiply the given expression by multiplying the first monomial with each monomial from the bracket

$$
begin{align*}
3x(2x-5y+4) &= 3x cdot 2x + 3x cdot (-5y) +3x cdot 4 \
&= 6x^2 -15xy +12x.
end{align*}
$$

Therefore
$$
color{#c34632} 3x(2x-5y+4)= 6x^2 -15xy +12x.
$$

$$
begin{align*}
&text{a.} hspace{3mm} (5m-1)(m+2) =5m^2 +9m -2 \
&text{b.} hspace{3mm} (6-x)(2+x) =-x^2+4x+12 \
&text{c.} hspace{3mm} (5x-y)^2 = 25x^2 -10xy +y^2 \
&text{d.} hspace{3mm} 3x(2x-5y+4) = 6x^2 -15xy +12x \
end{align*}
$$

a.

Let us sketch a function that has a domain of all real numbers between and including $-3$ and $10$ and a range of all real numbers between and including $-4$ and $6$. Therefore, we can write:

$$
-3 leq x leq 10 hspace{6mm} text{and} hspace{6mm} -4 leq y leq 6.
$$

We draw the lines

$$
x=-3, hspace{6mm} x=10, hspace{6mm} y=-4 hspace{6mm} text{and} hspace{6mm} y=6.
$$

as seen on graph below. Any function that is inside that rectangle including the boundries is our wanted function.

b.

Let us sketch a function that has a domain of all real numbers and a range of values $2$,$4$,$5$ and $8$.. Therefore, we can write:

$$
-infty < x < infty hspace{6mm} text{and} hspace{6mm} y in { 2,4,5,8 }.
$$

There is a big number of possible solutions. Let us give two examples.

My team’s function is the radical function$colon$          $y=2 cdot sqrt {9-x}-4$

The below is the graph of the function.

The key points of the graph are$colon$

The starting point of the graph$colon$          $(9,-4)$

The x-intercept$colon$          $(5,0)$

The y-intercept$colon$          $(0,2)$

There are so many integer inputs that give integer values as output. such as$colon$

$(9,-4), (8,-2), (5,0), (0,2)$ and $(-7,4)$

The values of $x$ that do not make sense are the values that are out of the domain.

The values of $y$ that do not make sense are the values that are out of the range.

The domain is$colon$          $x leq 9$

The range is$colon$          $y geq -4$

It is clear from showing the key point accurately that the graphing calculator shows an accurate graph.

The key points of the graph are$colon$          $(9,-4)$,          $(5,0)$ and $(0,2)$

Integer inputs that give integer values as output. such as$colon$          $(9,-4), (8,-2), (5,0), (0,2)$ and $(-7,4)$

The values of $x$ that make sense are the values of the domain$colon$          $x leq 9$

The values of $y$ that make sense are the values of the range$colon$          $y geq -4$

It is clear from showing the key point accurately that the graphing calculator shows an accurate graph.

If the x-intercepts are two points $(-4, 0)$ and $(2, 0)$ then function is a quadratic function.

Since the range of the function is:          $y geq -8$, then the parabola of the equation opens upwards and the vertex is a minimum

x-coordinate of the vertex $x_v=dfrac {x_1+x_2}{2}$ where $x_1$ and $x_2$ are the x-coordinates.
$x_v=dfrac {-4+2}{2}=-1$

The vertex is the point $(-1, -8)$

Substituting for the three points x-intercepts $(-4, 0)$ and $(2, 0)$ and the vertex $(-1, -8)$ in the general form of the quadratic equation:          $y=ax^2+bx+c$
$0=a(-4)^2+b(-4)+c$

$0=16a-4b+c$          (1)

$0=a(2)^2+b(2)+c$

$0=4a+2b+c$          (2)

$-8=a(-1)^2+b(-1)+c$

$-8=a-b+c$          (3)

Subtracting (2) from (1)

$0=16a-4b+c$          (1)

–

$0=4a+2b+c$          (2)

——————————-

$0=12a-6b$          (4)

Subtracting (3) from (2)

$0=4a+2b+c$          (2)

–

$-8=a-b+c$          (3)

——————————-

$8=3a+3b$          (5)

Solving the system of equation (4) and (5)

$0=12a-6b$          (4)

$8=3a+3b$          (5)

Multiply (5) by 2 and add to (4)

$0=12a-6b$          (4)
$16=6a+6b$          (5)
$18a=16$

$$
a=dfrac {8}{9}
$$

Substituting in (4)

$0=12 times dfrac {8}{9}-6b$          (4)
$6b=4 dfrac {8}{3}$

$6b=dfrac {32}{3}$

$b=dfrac {32}{18}$

$$
b=dfrac {16}{9}
$$

Substituting for $a$ and $b$ in (1)

$0=16(dfrac {8}{9})-4(dfrac {16}{9})+c$          (1)

$c=-dfrac {4 times 16}{9}$

$$
c=-dfrac {64}{9}
$$

The equation is:

$$
y=dfrac {8}{9}x^2+dfrac {16}{9}x-dfrac {64}{9}
$$

It is only one solution.

$$
y=dfrac {8}{9}x^2+dfrac {16}{9}x-dfrac {64}{9}
$$

The equation is:          $y=dfrac {8}{9}x^2+dfrac {16}{9}x-dfrac {64}{9}$

It is only one solution.

The initial deposit is 10 cents and the daily increase is 2 cents.\
The below is the table of the values.\
begin{center}
begin{tabular}{|l|l|}
hline
x & y \ hline
0 & 10 \ hline
1 & 12 \ hline
2 & 14 \ hline
3 & 16 \ hline
4 & 18 \ hline
5 & 20 \ hline
end{tabular}
end{center}
It can be represented by the function$colon$\
$y=2x+10$\
($x$ represents the days. qquad $y$ represents the balance after $x$ days)\
\
The below is the graph of the function.\

The function $y=2x+10$ represents the relation between the days $x$ and the balance $y$

$f(x)=dfrac {1}{x}$          (Given function)

a-          $f(dfrac {1}{2})=1 div dfrac {1}{2}=2$

b-          $f(dfrac {1}{10})=1 div dfrac {1}{10}=10$

c-          $f(0.01)=1 div dfrac {1}{100}=100$

d-          $f(0.007)=1 div dfrac {7}{1000}=dfrac {1000}{7}$

a-          $f(dfrac {1}{2})=2$

b-          $f(dfrac {1}{10})=10$

c-          $f(0.01)=100$

d-          $f(0.007)=dfrac {1000}{7}$

a-

When the input of the machine is 5, The output is 14

When the input of the machine is -1, The output is -4

By using the regression function of the graphing calculator, we find the function that represents the machine work is:

$$
y=3x-1
$$

b-

The equation of Terri’s machine is:

$$
y=3x-1
$$

a-          For the input of 5, The output is 14          For the input of -1, The output is -4

b-          The equation of Terri’s machine is:          $y=3x-1$

For solving the quadratic equation, we use the quadratic formula:

$x_{1,2}=dfrac {-b pm sqrt {b^2-4ac}}{2a}$

a-

$x^2-8x+15$          (Given)

$a=1$

$b=-8$

$c=15$

Solving by the quadratic formula

$x_{1,2}=dfrac {-(-8) pm sqrt {(-8)^2-4(1)(15)}}{2(1)}$

$x_{1,2}=dfrac {8 pm sqrt {64-60}}{2}$

$x_{1,2}=dfrac {8 pm sqrt {4}}{2}$

$x_{1,2}=dfrac {8 pm 2}{2}$

$x_{1}=dfrac {10}{2}=5$ and $x_{2}=dfrac {6}{2}=3$

b-

$2x^2-5x-6$          (Given)

$a=2$

$b=-5$

$c=-6$

Solving by the quadratic formula

$x_{1,2}=dfrac {-(-5) pm sqrt {(-5)^2-4(2)(-6)}}{2(2)}$

$x_{1,2}=dfrac {5 pm sqrt {25+48}}{4}$

$x_{1,2}=dfrac {5 pm sqrt {73}}{2}$

$x_{1} approx 3.39$ and $x_{2} approx -0.89$

a-          The solutions of the equation $x^2-8x+15$ are $x_{1}=5$ and $x_{2}=3$

b-          The solutions of the equation $2x^2-5x-6$ are $x_{1} approx 3.39$ and $x_{2} approx -0.89$

a-

When graphing an equation such as $y=3x-5$, The variable $y$ depends on the variable $x$.

Variable $x$ is independent. and variable $y$ is dependent.

a-          Variable $x$ is independent. and variable $y$ is dependent.

b-          Temperature is the dependent variable. and the time of the day is the independent variable.

a.

Let us factor the expression completely. We can rewrite the middle monomial in a form of a sum or difference of two monomials. Hence, our expression now has 4 monomials which we could pair ad factorize. We have

$$
begin{align*}
x^2-x-72 &= x^2+8x-9x-72 \ &=x cdot (x+8) -9 cdot (x+8) \ &= (x+8)(x-9).
end{align*}
$$

Therefore,
$$
color{#c34632} x^2-x-72=(x+8)(x-9).
$$

b.

Let us factor the expression completely. We can take the common factors in front of the bracket. We have

$$
begin{align*}
6x^2+48x &= 6 cdot x cdot x + 6 cdot 8 cdot x \ &=6x cdot (x+8) \ &= 6x(x+8).
end{align*}
$$

Therefore,
$$
color{#c34632} 6x^2+48x=6x(x+8).
$$

c.

Let us factor the expression completely. We can rewrite the middle monomial in a form of a sum or difference of two monomials. Hence, our expression now has 4 monomials which we could pair ad factorize. We have

$$
begin{align*}
x^2-8x+16 &= x^2-4x-4x+16 \ &=x cdot (x-4) -4 cdot (x-4) \ &= (x-4)(x-4) \ &=(x-4)^2.
end{align*}
$$

Also we could have used the identity

$$
color{#4257b2} a^2-2ab+b^2=(a-b)^2.
$$
We now have

$$
begin{align*}
x^2-8x+16 &= x^2-2 cdot x cdot 4 +4^2 \ &= (x-4)^2.
end{align*}
$$

Therefore,
$$
color{#c34632} x^2-8x+16=(x-4)^2.
$$

d.

Let us factor the expression completely. We can use the difference of two squares identity

$$
color{#4257b2} a^2-b^2=(a-b)(a+b).
$$
We now have

$$
begin{align*}
x^2-49 &= x^2-7^2 \ &= (x-7)(x+7).
end{align*}
$$

Therefore,
$$
color{#c34632} x^2-49 =(x-7)(x+7).
$$

$$
begin{align*}
&text{a.} hspace{3mm} x^2-x-72=(x+8)(x-9) \
&text{b.} hspace{3mm} 6x^2+48x =6x(x+8) \
&text{c.} hspace{3mm} x^2-8x+16 = (x-4)^2 \
&text{d.} hspace{3mm} x^2-49 = (x-7)(x+7) \
end{align*}
$$

Let us consider the given function $$color{blue} f(x)=sqrt{x} -2.$$
We can now make the table of the function.
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c| }
hline
x & 0 & 1 & 4 & 9 & 16 & 25 \
hline
y & -2 & -1 & 0 & 1 & 2 & 3 \
hline
end{tabular}
end{center}
And now we can draw the graph. \

Now we can fully describe the graph.

The domain of the function are all nonnegative real numbers, we can write $x geq 0$.

The range of the function are all real numbers between $-2$ and $infty$ including $-2$, or we can write $y geq -2$.

The $x$-intercept is the point $x=4$.

The $y$-intercept is the point $y=-2$.

The function is not symmetric and does not have asymptotes.

The function is increasing for $x in [0, infty).$

The function has the minimum value of $-2$ for $x=0$.

The function is continious.

a-

$(x+13)(x-7)=0$          (Write the equation)

$(x+13)=0$ or $(x-7)=0$          (Zero product propert)

$x=-13$ or $x=7$          (Solve for $x$)

b-

$(2x+3)(3x-7)=0$          (Write the equation)

$(2x+3)=0$ or $(3x-7)=0$          (Zero product property)

$x=-dfrac {3}{2}$ or $x=dfrac {7}{3}$          (Solve fore $x$)

c-

$x(x-3)=0$          (Write the equation)

$x=0$ or $(x-3)=0$          (Zero product property)

$x=0$ or $x=3$          (Solve fore $x$)

d-

$x^2-5x=0$          (Write the equation)

$x(x-5)=0$          (Factor out $x$)

$x=0$ or $(x-5)=0$          (Zero product property)

$x=0$ or $x=5$          (Solve fore $x$)

e-

$x^2-2x-35=0$          (Write the equation)

$(x-7)(x+5)=0$          (Factor the equation)

$x-7=0$ or $(x+5)=0$          (Zero product property)

$x=7$ or $x=-5$          (Solve fore $x$)

f-

$3x^2+14x-5=0$          (Write the equation)

$(3x-1)(x+5)=0$          (Factor the equation)

$3x-1=0$ or $(x+5)=0$          (Zero product property)

$x=dfrac {1}{3}$ or $x=-5$          (Solve fore $x$)

a-          $(x+13)(x-7)=0$ The solution is $x=-13$ or $x=7$

b-          $(2x+3)(3x-7)=0$ The solution is $x=-dfrac {3}{2}$ or $x=dfrac {7}{3}$

c-          $x(x-3)=0$ The solution is $x=0$ or $x=3$

d-          $x^2-5x=0$ The solution is $x=0$ or $x=5$

e-          $x^2-2x-35=0$ The solution is $x=7$ or $x=-5$

f-          $3x^2+14x-5=0$ The solution is $x=dfrac {1}{3}$ or $x=-5$

$f(x)=dfrac {1}{x-2}$          (Given)

a-          $f(2.5)=dfrac {1}{2.5-2}=dfrac {1}{0.5}=2$

b-          $f(1.75)=dfrac {1}{1.75-2}=dfrac {1}{0.25}=4$

c-          $f(2)=dfrac {1}{2-2}=dfrac {1}{0}$          (Division by 0 is undefined)

a-          $f(2.5)=2$

b-          $f(1.75)=4$

c-          $f(2)=dfrac {1}{0}$          (Division by 0 is undefined)

$f(x)=-dfrac {2}{3}x+3$          Given

$g(x)=2x^2-5$          Given

a-

$f(3)=-dfrac {2}{3}(3)+3=1$

b-

$f(x)=-5$

$-5=-dfrac {2}{3}x+3$          (Substitute -5 for $f(x)$)

$dfrac {2}{3}x=8$          (Add $dfrac {2}{3}x$ and 5 to each side)

$x=8 times dfrac {3}{2}$          (Multiply each side by $dfrac {3}{2}$)

$$
x=12
$$

c-

$g(-3)=2(-3)^2-5$

$g(-3)=18-5=13$

d-

$g(x)=-7$

$-7=2x^2-5$

$2x^2=-2$

$x^2=-1$

$x=sqrt {-1}$

Since the radicand is less than 0. Then there is no real solution

e-

$g(x)=8$

$2x^2-5=8$

$x^2=dfrac {13}{2}$

$x=sqrt {dfrac {13}{2}}$

f-

$g(x)=9$

$2x^2-5=9$

$x^2=dfrac {14}{2}$

$x=sqrt {7}$

a-          $f(3)=-dfrac {2}{3}(3)+3=1$

b-          For $f(x)=-5$,         $x=12$

c-          $g(-3)=2(-3)^2-5=13$

d-          For $g(x)=-7$,          $x=sqrt {-1}$          There is no real solution

e-          For $g(x)=8$,          $x=sqrt {dfrac {13}{2}}$

f-          For $g(x)=9$,          $x=sqrt {7}$

Assuming the function machine is $f(x)$

$f(-2)=-8$          (Given)

$f(-1)=-1$          (Given)

$f(0)=0$          (Given)

$f(3)=27$          (Given)

The machine function is:

$$
f(x)=x^3
$$

The machine function is:          $f(x)=x^3$

$textbf{(a)}$ Given graph represents relation between cost of gas(in dollars) and volume of gas(in gallons).

It can be observed that both quantities are linearly related.

In the given volume of gas is an independent variable whereas cost of gas is a dependent variable

$textbf{(b)}$ Given graph represents relation between height of a person (in inches) and age of the person .

It can be observed that both quantities are non-linearly related.

In the given age of person is an independent variable whereas height of person is a dependent variable

$textbf{(c)}$ Given graph represents relation between levels of ozone concentration and number of years.

It can be observed that both quantities are exponentially(decaying) related.

In the given number of years is an independent variable whereas levels of ozone concentration is a dependent variable

$textbf{(d)}$ Given graph represents relation between number of classrooms and number of students.

It can be observed that there is a non continuous relation between both quantities .

In the given number of students is an independent variable whereas number of classrooms is a dependent variable

$textbf{(e)}$ All the non negative integers can be possible inputs of the graph in part (d).

Similarly, all the non negative integers can be possible outputs of the graph in part (d).

As both number of classrooms and number of students can not be a negative quantity or a fraction

a.

Let us solve the following equation

$$
begin{align*}
3.9x-2.1 &= 11.2x+51.7 \
3.9x-11.2x &= 51.7+2.1 \
-7.3x &=53.8 \
x&= -frac{53.8}{7.3} = -7.3698.
end{align*}
$$

Therefore, the solution is
$$
color{#c34632} x=-7.3698.
$$

b.

Let us solve the following equation

$$
begin{align*}
frac{1}{5} x-2 &= frac{13}{25}-0.7x \ \
frac{1}{5} x+0.7x &= frac{13}{25}+2 \ \
frac{2}{10} x+frac{7}{10} x &= frac{13}{25}+frac{50}{25} \ \
frac{9}{10} x &= frac{63}{25} \ \
x &= frac{63}{25} cdot frac{10}{9} = frac{7}{5} cdot frac{2}{1} = frac{14}{5} = 2.8.
end{align*}
$$

Therefore, the solution is
$$
color{#c34632} x=2.8.
$$

c.

Let us solve the following equation

$$
begin{align*}
2+frac{2}{x} = 3 &implies frac{2}{x} = 3-2 \ &implies frac{2}{x} = 1 implies x = frac{2}{1} = 2.
end{align*}
$$

Therefore, the solution is
$$
color{#c34632} x=2.
$$

d.

Let us solve the following equation

$$
begin{align*}
x+frac{3}{4} &= 4+2x \
x-2x &= 4- frac{3}{4} \
-x &= 4-0.75 \
-x &= 3.25 \
x &= -3.25.
end{align*}
$$

Therefore, the solution is
$$
color{#c34632} x=-3.25.
$$

$$
begin{align*}
&text{a.} hspace{3mm} x=-7.3698 & &text{b.} hspace{3mm} x=2.8 \
&text{c.} hspace{3mm} x=2 & &text{d.} hspace{3mm} x=-3.25 \
end{align*}
$$

Let us consider the given function $$color{blue} f(x)=frac{1}{x-7}.$$
We can now make the table of the function.
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c|c|c|c|c| }
hline
x & -5 & -3 & -1 & 0 & 1 & 4 & 6 & 6.5 & 7.5 & 8 & 12 \
hline
y & -0.083 & -0.1 & -0.125 & -0.143 & -0.166 & -0.16 & -1 & -2 & 2 & 1 & 0.2 \
hline
end{tabular}
end{center}
And now we can draw the graph. \

The table is not enough to sketch the graph fully and correctly. We have to comment that we have to be very careful when choosing the values for $x$ in the table. For some value of $x$, $x=7$, our function would not be defined. And if we didn’t choose values $6.5$ and $7.5$ we would not be able to predict the shape of our function completely. It can be helpful to determine the domain of the function and calculate the $x$-intercept and $y$-intercept first and then make the table. Evaluating the asymptotes before making the table can also be very helpful.

Now we can fully describe the graph.

The domain of the function are all real numbers except for $7$, we can write $x in (-infty,7) cup (7, infty)$.

The range of the function are all real numbers, except for $0$ so we can write $y in (-infty,0) cup (0, infty)$.

The $x$-intercept does not exist.

The $y$-intercept is the point $y=-0.143$.

The function is not symmetric.

The function has one vertical asymptote $x=7$ and one horizontal asymptote $y=0$.

The function is decreasing for $x in (-infty,7) cup (7, infty).$

The function does not have minimum or maximum points.

The function is not continuous.

a. \
Let us consider the given function $$ color{blue} f(x)=frac{1}{x-7}. $$
We can now make the table of the function.
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c|c|c|c|c| }
hline
x & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \
hline
y & -0.2 & -0.25 & -0.33 & -0.5 & -1 & / & 1 & 0.5 & 0.33 & 0.25 & 0.2 \
hline
end{tabular}
end{center}
The values for integers less that $7$ are negative, while the values for integers bigger than $7$ are the same just positive. In some way we can say the $y$-values are symmetric. \

b.

Our function does not have value for $x=7$. The answer makes sense because we do not divide by $0$, and for the values $x=7$ that happens.

f.

Now we can fully describe the graph.

The domain of the function are all real numbers except for $7$, we can write $x in (-infty,7) cup (7, infty)$.

The range of the function are all real numbers, except for $0$ so we can write $y in (-infty,0) cup (0, infty)$.

The $x$-intercept does not exist.

The $y$-intercept is the point $y=-0.143$.

The function is not symmetric.

The function has one vertical asymptote $x=7$ and one horizontal asymptote $y=0$.

The function is decreasing for $x in (-infty,7) cup (7, infty).$

The function does not have minimum or maximum points.

The function is not continuous.

a.

The graph of
$$
f(x)=frac{1}{x-h}
$$
has a vertical asymptote $x=h$ and a horizontal asymptote $y=0$.
As we graphed the function
$$
f(x)=frac{1}{x-7}
$$
we can write that our function has a vertical asymptote $x=7$ and a horizontal asymptote $y=0$.

We add these lines to our graph.

b. \
Now we choose a number between $-10$ and $10$. Let us choose $color{red} -3$. The new function is
$$color{red} f(x)=frac{1}{x-(-3)}=frac{1}{x+3}. $$
We can now make the table of the function.
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c|c|c|c|c| }
hline
x & -8 & -5 & -4 & -3.5 & -2.5 & -2 & 0 & 2 & 5 \
hline
y & -0.2 & -0.5 & -1 & -2 & 2 & 1 & 0.33 & 0.2 & 0.125 \
hline
end{tabular}
end{center}
And now we can draw the graph.\

Now we can fully describe the graph.

The domain of the function are all real numbers except for $-3$, we can write $x in (-infty,-3) cup (-3, infty)$.

The range of the function are all real numbers, except for $0$ so we can write $y in (-infty,0) cup (0, infty)$.

The $x$-intercept does not exist.

The $y$-intercept is the point $y=0.33$.

The function is not symmetric.

The function has one vertical asymptote $x=-3$ and one horizontal asymptote $y=0$.

The function is decreasing for $x in (-infty,-3) cup (-3, infty).$

The function does not have minimum or maximum points.

The function is not continuous.

c.

Analyzing the graph of each function we see that the shape of the graph is the same. It just translates left or right. Therefore, we can conclude that the function
$$
f(x)=frac{1}{x-h}
$$
has the graph of the same shape as function
$$
g(x)=frac{1}{x}
$$
just it is translated to the left by $h$ units if $h$ is negative or translated to the right by $h$ units if $h$ is positive.

We do not need to add any new attributes. All we have to do is slightly modify the existing ones.

a. The vertical asymptote $x=h$ and horizontal asymptote $y=0$ have been added to the graph.

b. The function has been graphed for chosen $h=-3$.

c. We didn’t have to add any attributes, the graphs have the same shape, just translated left or right.

$$
f(x)=dfrac{1}{x+25}
$$

$$
x=-25
$$

$$
y=0
$$

We use the graph of the function that we drew in Exercise 31:

$f_0(x)=dfrac{1}{x+2}$.

We horizontally translate the graph of $f_0$ 23 units to the left:

b) We use a graphing calculator to graph $f(x)$ (in a window [-4,4,1] by [-4,4,1]):

(a)
All the functions of this type have $y-$intercept, but don’t have $x-$intercept. The function is discontinued at $x=h$, and they are decrease on the interval $(-infty,h)$ and the interval $(h,infty)$.

All the graphs of these functions have a **vertical asymptote** (the denominator’s zero):
$$textcolor{#c34632}{x=h}$$
and a **horizontal asymptote** (because the degree of the numerator is smaller than the degree of the numerator):
$$textcolor{#c34632}{y=0}$$

The **main difference** between these graphs is the distance from $y-$ axis.

(c)
Examples of families of functions:
$$f(x)=sqrt{x}+hhspace{1cm}textrm{where parent function is } f(x)=sqrt{x}$$
$$f(x)=sin(bx+c)hspace{1cm}textrm{where parent function is } f(x)=sin x$$

a-          The domain of $g(x)$ is $(-2 leq x leq 4)$

b-          The range of $g(x)$ is $(-1 leq y leq 3)$

c-          Ricky is not correct. because the parabola of the function is a continuous line. i.e the range is all the real numbers between -1 and 3.

d-

A graph of another function that is with the same domain and range of $g(x)$

a-          The domain of $g(x)$ is $(-2 leq x leq 4)$

b-          The range of $g(x)$ is $(-1 leq y leq 3)$

c-          Ricky is not correct. because the parabola of the function is a continuous line. i.e the range is all the real numbers between -1 and 3.

Let us consider
$$
color{#4257b2} (2x-1)(x+3)=4.
$$

Unfortunately, both George and Jeffrey are wrong. We can use the Zero Product Property only when the product equals zero. Therefore we multiply and get

$$
begin{align*}
(2x-1)(x+3) &=4 \
2x cdot x + 2x cdot 3 -1 cdot x-1cdot 3 &=4 \
2x^2+6x-x-3 &=4 \
2x^2 +5x -3 -4 &=0 \
2x^2 +5x -7 &=0.
end{align*}
$$

We have gotten the quadratic equation, therefore we can use the quadratic formula. We have

$$
a=2 hspace{12mm} b=5 hspace{12mm} c=-7.
$$

Now we use the formula
$$
x = frac{-b pm sqrt{b^2-4ac}}{2a}.
$$

Hence, we have

$$
begin{align*}
x &= frac{-5 pm sqrt{5^2-4 cdot 2 cdot (-7)}}{2 cdot 2} \
x &= frac{-5 pm sqrt{25+56}}{4} \
x &= frac{-5 pm sqrt{81}}{4} \
x &= frac{-5 pm 9}{4} \
x &= frac{-5 + 9}{4} hspace{3mm} text{or} hspace{3mm} frac{-5 – 9}{4} \
x &= frac{4}{4} hspace{3mm} text{or} hspace{3mm} frac{-14}{4} \
x &= 1 hspace{3mm} text{or} hspace{3mm} -3.5
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} x=1 hspace{3mm} text{color{default}or} hspace{3mm} x=-3.5
$$

As instructed we check the solutions. First we substitute $x=1$ and get

$$
(2x-1)(x+3)= (2 cdot 1 -1)(1+3) = (2-1) cdot 4 = 1 cdot 4 = 4.
$$

Therefore, $x=1$ really is the solution. Now we check for $x=-3.5$. We substitute and get

$$
(2x-1)(x+3)= (2 cdot (-3.5) -1)(-3.5+3) = (-7-1) cdot (-0.5) = -8 cdot (-0.5) = 4.
$$

We can conclude that $x=-3.5$ is the correct solution too.

$$
x=1 hspace{3mm} text{or} hspace{3mm} x=-3.5
$$

a-

$x=3y+6$          (Given)

$3y=x-6$          (Subtract 6 from each side)

$y=dfrac {x-6}{3}$          (y-form)

b-

$x=5y-10$          (Given)

$5y=x+10$          (Add 10 to each side)

$y=dfrac {x+10}{5}$          (y-form)

c-

$x=y^2$          (Given)

$y=sqrt {x}$          (y-form)

d-

$x=2y^2-4$          (Given)

$2y^2=x+4$          (Add 4 to each side)

$y^2=dfrac {x+4}{2}$          (Divide each side by 2)

$y= sqrt {dfrac {x+4}{2}}$          (y-form)

e-

$x=(y-5)^2$          (Given)

$y-5=sqrt {x}$          (Square root of each side)

$y=sqrt {x}+5$          (y-form)

a-          $x=3y+6 qquad rightarrow qquad y=dfrac {x-6}{3}$

b-          $x=5y-10 qquad rightarrow qquad y=dfrac {x+10}{5}$

c-          $x=y^2 qquad rightarrow qquad y=sqrt {x}$

d-          $x=2y^2-4 qquad rightarrow qquad y= sqrt {dfrac {x+4}{2}}$

e-          $x=(y-5)^2 qquad rightarrow qquad y=sqrt {x}+5$

$f(x)=2x-7$          (Given)

a-          $f(0)=2(0)-7=0-7=-7$

b-          $f(x)=0$

$2x-7=0$

$2x=7$

$x=dfrac {7}{2}$

c-          The answer to part (a) tells us that the y-intercept for the function $f(x)=2x-7$ is the point $(0, -7)$

While the answer to part (b) tells us that the x-intercept for the function $f(x)=2x-7$ is the point $(dfrac {7}{2}, 0)$

a-          $f(0)=2(0)-7=0-7=-7$

b-          $f(x)=0 qquad rightarrow qquad x=dfrac {7}{2}$

c-          Part (a) tells us that the y-intercept is the point $(0, -7)$

While part (b) tells us that the x-intercept is the point $left(dfrac {7}{2}, 0 right)$

Assuming the length of the piece that Jill cuts of is $x$ and the remaining piece is $y$, the equation that represents the situation is:\
$y=30-x$\
\
The below table represents the situation.\
begin{center}
begin{tabular}{|l|l|}
hline
$x$ & $y=30-x$ \ hline
5 & 25 \ hline
10 & 20 \ hline
15 & 15 \ hline
20 & 10 \ hline
25 & 5 \ hline
end{tabular}
end{center}

The equation that represents the situation is $y=30-x$

Let’s note:

$t$=the time of the day

$f(t)$=the number of people in the school’s campus as a function of the time of the day $t$

The error is multiplying only one side of the equation by $x$ and ignoring the other.

$x cdot dfrac {5}{x}=x-4$          (The error)

The correct solution is:

$x cdot dfrac {5}{x}=x cdot (x-4)$          (Multiplying each side by $x$)

$5=x^2-4x$          (Distributive property)

$x^2-4x-5=0$

$(x-5)(x+1)=0$          (Factoring the polynomial)

$x-5=0$ or $x+1=0$

$x=5$ or $x=-1$

Checking the solutions

For $x=5$

$x^2-4x-5=0$

$5^2-4(5)-5$          (Substituting 5 for $x$)

$25-20-5=0$          $checkmark$

For $x=-1$

$(-1)^2-4(-1)-5=0$

$1+4-5=0$          (Substituting -1 for $x$)

$0=0$          $checkmark$

The error is multiplying only one side of the equation by $x$ and ignoring the other.

The correct solution is:          $x=5$ or $x=-1$

Let us consider the given function $$color{blue} f(x)=frac{2}{x}.$$
We can now make the table of the function.
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c|c|c|c|c| }
hline
x & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \
hline
y & -0.5 & -0.66 & -1 & -2 & / & 2 & 1 & 0.66 & 0.5 \
hline
end{tabular}
end{center}
And now we can draw the graph.
\

Now we can fully describe the graph.

The domain of the function are all real numbers except for $0$, we can write $x in (-infty,0) cup (0, infty)$.

The range of the function are all real numbers, except for $0$ so we can write $y in (-infty,0) cup (0, infty)$.

The $x$-intercept and $y$-intercept do not exist.

The function is symmetric.

The function has one vertical asymptote $x=0$ and one horizontal asymptote $y=0$.

The function is decreasing for $x in (-infty,0) cup (0, infty).$

The function does not have minimum or maximum points.

The function is not continuous.

a-

$dfrac {6}{x}=x-1$          (Write the equation)

$6=x^2-x$          (Multiplying each side by $x$)

$x^2-x-6=0$          (Subtracting 6 from each side)

$(x-3)(x+2)=0$          (Factoring)
$x-3=0$ or $x+2=0$          (Zero product property)

$x=3$ or $x=-2$          (Solving for $x$)

Checking the solution

For $x=3$

$dfrac {6}{3}=3-1$

$2=2$          checkmark

For $x=-2$

$dfrac {6}{-2}=-2-1$

$-3=-3$          checkmark

b-

$dfrac {9}{x}=x$          (Write the equation)

$9=x^2$          (Multiplying each side by $x$)

$x=sqrt 9$

$x=3$          (Solving for $x$)

Checking the solution

For $x=3$

$dfrac {9}{3}=3$

$3=3$          checkmark

a-          $x=3$ or $x=-2$

b-          $x=3$

The machine function is:

$f(x)=x^2+2x+1$          (Given)

If the output is 1, then $f(x)=1$

$1=x^2+2x+1$          (Substitut 1 for $f(x)$)

$x^2+2x=0$          (Subtract 1 from each side)

$x(x+2)=0$          (Factor out $x$

$x=0$ or $(x+2=0)$          (Zero product property)

$x=0$ or $x=-2$          (Solve for $x$)

The number that could have dropped in the machine function are 0, -2

The value that allows us to find the y-intercept is: $x=0$

a-          $y=3x+6$          (Write the equation)

For y-intercept, Substitute $x=0$

$y=3(0)+6=6$

y-intercept is $(0, 6)$

b-          $x=5y-10$          (Write the equation)

$y=dfrac {x+10}{5}$          (Write the equation in the term of $y$)

For y-intercept, Substitute $x=0$

$y=dfrac {0+10}{5}=2$

y-intercept is $(0, 2)$

c-          $y=x^2$          (Write the equation)

For y-intercept, Substitute $x=0$

$y=(0)^2=0$

y-intercept is $(0, 0)$

d-          $y=2x^2-4$          (Write the equation)

For y-intercept, Substitute $x=0$

$y=2(0)^2-4=-4$

y-intercept is $(0, -4)$

e-          $y=(x-5)^2$          (Write the equation)

For y-intercept, Substitute $x=0$

$y=(0-5)^2=25$

y-intercept is $(0, 25)$

f-          $y=3x^3-2x^2+13$          (Write the equation)

For y-intercept, Substitute $x=0$

$y=3(0)^3-2(0)^2+13=13$

y-intercept is $(0, 13)$

The value that allows us to find the y-intercept is: $x=0$

a-          y-intercept is $(0, 6)$

b-          y-intercept is $(0, 2)$

c-          y-intercept is $(0, 0)$

d-          y-intercept is $(0, -4)$

e-          y-intercept is $(0, 25)$

e-          y-intercept is $(0, 13)$

a.

Let us solve the given equation
$$
color{#4257b2} x^2+7x-8=0.
$$

This one can be solved by factoring and applying the Zero Productivity Rule. We get

$$
begin{align*}
x^2+7x-8 &=0 \
x^2 +8x-x-8 &=0 \
x cdot (x+8) -(x+8) &=0 \
(x+8)(x-1) &=0.
end{align*}
$$

Now we apply the Zero Product Rule which yields

$$
begin{align*}
(x+8)(x-1) =0 \
x+8 = 0 hspace{3mm} text{or} hspace{3mm} x-1=0 \
x=-8 hspace{3mm} text{or} hspace{3mm}x=1.
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} x=1 hspace{3mm} text{color{default}or} hspace{3mm} x=-8.
$$

b.

Let us solve the given equation
$$
color{#4257b2} (x+2)^2=49.
$$

This one can be solved by factoring and applying the Zero Productivity Rule. We get

$$
begin{align*}
(x+2)^2 &=49 \
(x+2)^2-49 &=0 \
(x+2)^2-7^2 &=0.
end{align*}
$$

Using the difference of two squares identity gives us

$$
begin{align*}
(x+2)^2-7^2 &=0 \
(x+2-7)(x+2+7) &=0 \
(x-5)(x+9) &=0 \
end{align*}
$$

Now we apply the Zero Product Rule which yields

$$
begin{align*}
(x-5)(x+9) =0 \
x-5 = 0 hspace{3mm} text{or} hspace{3mm} x+9=0 \
x=5 hspace{3mm} text{or} hspace{3mm} x=-9.
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} x=5 hspace{3mm} text{color{default}or} hspace{3mm} x=-9
$$

c.

Let us solve the given equation
$$
color{#4257b2} 5x^2-x-7=0.
$$

This time we can use the quadratic formula
$$
x = frac{-b pm sqrt{b^2-4ac}}{2a}.
$$
We have

$$
a=5 hspace{12mm} b=-1 hspace{12mm} c=-7.
$$

Hence, we have

$$
begin{align*}
x &= frac{-(-1) pm sqrt{(-1)^2-4 cdot 5 cdot (-7)}}{2 cdot 5} \
x &= frac{1 pm sqrt{1+140}}{10} \
x &= frac{1 pm sqrt{141}}{10} \
x &= frac{1 + sqrt{141}}{10} hspace{3mm} text{or} hspace{3mm} frac{1 – sqrt{141}}{10}.
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} x=frac{1 + sqrt{141}}{10} hspace{3mm} text{color{default}or} hspace{3mm} x=frac{1 – sqrt{141}}{10}
$$

d.

Let us solve the given equation
$$
color{#4257b2} x^2+4x=-1.
$$

We rewrite it
$$
x^2+4x+1=0
$$

and now we can use the quadratic formula
$$
x = frac{-b pm sqrt{b^2-4ac}}{2a}.
$$
We have

$$
a=1 hspace{12mm} b=4 hspace{12mm} c=1.
$$

Hence, we have

$$
begin{align*}
x &= frac{-4 pm sqrt{4^2-4 cdot 1 cdot 1}}{2 cdot 1} \
x &= frac{-4 pm sqrt{16-4}}{2} \
x &= frac{-4 pm sqrt{12}}{2} \
x &= frac{-4 pm sqrt{4 cdot 3}}{2} \
x &= frac{-4 pm 2sqrt{3}}{2} \
x &= frac{2 cdot (-2 pm sqrt{3})}{2} \
x &= -2 pm sqrt{3} \
x &= -2 + sqrt{3} hspace{3mm} text{or} hspace{3mm} -2 – sqrt{3}.
end{align*}
$$

Therefore, the solutions are
$$
color{#c34632} x= -2 + sqrt{3} hspace{3mm} text{color{default}or} hspace{3mm} x= -2 – sqrt{3}.
$$

$$
begin{align*}
&text{a.} hspace{3mm} x=1 hspace{3mm} text{or} hspace{3mm} x=-8 \
&text{b.} hspace{3mm} x=5 hspace{3mm} text{or} hspace{3mm} x=-9 \
&text{c.} hspace{3mm} x=frac{1 + sqrt{141}}{10} hspace{3mm} text{or} hspace{3mm} x=frac{1 – sqrt{141}}{10} \
&text{d.} hspace{3mm} x= -2 + sqrt{3} hspace{3mm} text{color{default}or} hspace{3mm} x= -2 – sqrt{3} \
end{align*}
$$

Let us graph the line $$color{blue} y=3x+3.$$
We can now make the table of the function.
begin{center}
begin{tabular}{ |c|c|c|c| }
hline
x & -1 & 0 \
hline
y & 0 & 3 \
hline
end{tabular}
end{center}
And now we can draw the graph.\

a.

Let us sketch the line that is perpendicular to the line
$$
y=3x+3
$$
and passes through the point $(-3,2)$.

b.

Let us write the equation of the perpendicular line to the line
$$
y=3x+3.
$$
We need to find the equation

$$
color{#4257b2} y=k cdot x+n.
$$

We can find the coefficient $k$ using the fact that the two lines are perpendicular. We have

$$
k=-frac{1}{3}.
$$

In order to find $n$ we use the fact that our wanted line passes through $(-3, 2)$. We substitute $k=-1/3$, $x=-3$ and $y=2$ and get

$$
2=-frac{1}{3} cdot (-3) +n implies 2=1+n implies n=2-1=1.
$$

Therefore, the equation of the perpendicular line to the line $y=3x+3$ is

$$
color{#c34632} y=-frac{1}{3}x+1.
$$

a. We graphed the line $y=3x+3$ and the line that is perpendicular to it and passes through $(-3, 2)$.

b. The equation of the perpendicular line to $y=3x+3$ is
$$
y=-frac{1}{3}x+1.
$$

Let us consider the given functions

$$
color{#4257b2} f_1(x) = x-2 hspace{4mm} text{color{default} and} hspace{4mm} g_1(x)=2x+3.
$$

First we try to predict the shape of the new graph of the sum of the functions. Since we are summing two linear functions we predict our new function will be linear again.

$$
p(x)=f_1(x) + g_1(x) = x-2+2x+3 = 3x+1.
$$

As we predicted the function is linear.
Let us sketch the given functions along with the new function. We see the graph below.

First we try to predict the shape of the new graph of the difference of the functions. Since we are subtracting two linear functions we predict our new function will be linear again.

$$
q(x)=f_1(x) – g_1(x) = x-2-(2x+3) = x-2-2x-3 = -x-5.
$$

As we predicted the function is linear.
Let us sketch the given functions along with the new function. We see the graph below.

First we try to predict the shape of the new graph of the product of the functions. Since we are multiplying two linear functions we predict our new function will be a quadratic function.

$$
r(x)=f_1(x) cdot g_1(x) = (x-2)(2x+3) = 2x^2 +3x-4x-6 = 2x^2-x-6.
$$

As we predicted the function is quadratic. Let us sketch the given functions along with the new function. We see the graph below.

Let us divide the two give functions.

$$
t(x)=frac{f_1(x)}{ g_1(x)} = frac{x-2}{2x+3}.
$$

Let us sketch the given functions along with the new function. We see the graph below.

Let us consider the given functions

$$
color{#4257b2} f_1(x) = x-2 hspace{4mm} text{color{default} and} hspace{4mm} g_1(x)=2x+3.
$$

First we try to predict the shape of the new graph of the sum of the functions. Since we are summing two linear functions we predict our new function will be linear again.

$$
p(x)=f_1(x) + g_1(x) = x-2+2x+3 = 3x+1.
$$

As we predicted the function is linear. Let us sketch the given functions along with the new function. We see the graph below.

Now we can fully describe the graph.

The sum function is linear. The domain of the function are all real numbers. Same goes for the range, it includes all real numbers.

The $x$-intercept is $x=-1/3$, while the $y$-intercept is $y=0$.

The function is increasing for $x in (-infty, infty)$. Hence, our function does not have minimum or maximum points.

Also, we can now comment all the possible cases when summing two linear functions. One possibility is the one we have here, we can get a linear function. But if the leading coefficients of two functions we are summing are opposite numbers, the result would be a constant function. Those are the only two options.

Let us consider the given functions

$$
color{#4257b2} f_1(x) = x-2 hspace{4mm} text{color{default} and} hspace{4mm} g_1(x)=2x+3.
$$

First we try to predict the shape of the new graph of the difference of the functions. Since we are subtracting two linear functions we predict our new function will be linear again.

$$
q(x)=f_1(x) – g_1(x) = x-2-(2x+3) = x-2-2x-3=-x-5.
$$

As we predicted the function is linear. Let us sketch the given functions along with the new function. We see the graph below.

Now we can fully describe the graph.

The sum function is linear. The domain of the function are all real numbers. Same goes for the range, it includes all real numbers.

The $x$-intercept is $x=-5$, while the $y$-intercept is $y=-5$.

The function is increasing for $x in (-infty, infty)$. Hence, our function does not have minimum or maximum points.

Also, we can now comment all the possible cases when subtracting two linear functions. One possibility is the one we have here, we can get a linear function. But if the leading coefficients of two functions we are subtracting are the same numbers, the result would be a constant function. Those are the only two options.

Let us consider the given functions

$$
color{#4257b2} f_1(x) = x-2 hspace{4mm} text{color{default} and} hspace{4mm} g_1(x)=2x+3.
$$

First we try to predict the shape of the new graph of the product of the functions. Since we are multiplying two linear functions we predict our new function will be a quadratic function.

$$
r(x)=f_1(x) cdot g_1(x) = (x-2)(2x+3) = 2x^2 +3x-4x-6 = 2x^2-x-6.
$$

As we predicted the function is quadratic. Let us sketch the given functions along with the new function. We see the graph below.

Now we can fully describe the graph.

The product function is quadratic. The parabola opens upward. The domain of the function are all real numbers. The range is $[-6.125, infty)$.

The $x$-intercept is $x=-6$, while the $y$-intercept are $y=-1.5$ and $y=2$.

The function is decreasing for $x in (-infty, 0.25)$ and decreasing for $x in (0.25, infty)$. Hence, our function has a minimum point $(0.25, -6.125)$.

Also, we can now comment the product of two linear functions can be a quadratic function. But if one of the linear functions is a constant vertical function then the product will be a linear function. Therefore, the product of two linear functions is a quadratic function, with the exception when one of the functions is a constant vertical function then the product is a linear function.

Let us consider the given functions

$$
color{#4257b2} f_1(x) = x-2 hspace{4mm} text{color{default} and} hspace{4mm} g_1(x)=2x+3.
$$

Let us divide the two give functions.

$$
t(x)=frac{f_1(x)}{ g_1(x)} = frac{x-2}{2x+3}.
$$

Let us sketch the given functions along with the new function. We see the graph below.

Now we can fully describe the graph.

The quotient function has the shape of hyperbola. The domain of the function are all real numbers except for $-3/2$, therefore the domain is
$$
(-infty,-frac{3}{2}) cup (-frac{3}{2}, infty).
$$
The range are all real numbers except for $1/2$, therefore

$$
(-infty,frac{1}{2}) cup (frac{1}{2}, infty).
$$

The $x$-intercept is $x=2$, while the $y$-intercept is $y=-2/3$.

The function is increasing for $x in (-infty,-frac{3}{2}) cup (-frac{3}{2}, infty)$. Hence, our function does not have a minimum or maximum point.

Also, we can now comment the asymptotes of the quotient function. This function has one vertical and one horizontal asymptotes. The vertical asymptote is
$$
x=-frac{3}{2}
$$
and the horizontal asymptote is
$$
y=frac{1}{2}.
$$

a.

Let us consider the given functions

$$
color{#4257b2} f_1(x) = x-2 hspace{4mm} text{color{default} and} hspace{4mm} g_1(x)=2x+3.
$$

First, we evaluate the sum function. We have

$$
f_1(x) + g_1(x) = x-2+2x+3 = 3x+1.
$$

The sum function $text{color{#c34632} is}$ a polynomial function. It’s degree is $color{#c34632}1$.

Now we evaluate the difference function. We have

$$
f_1(x) – g_1(x) = x-2-(2x+3) = x-2-2x-3 = -x-5.
$$

The difference function $text{color{#c34632} is}$ a polynomial function. It’s degree is $color{#c34632} 1$.

Next, we consider the product function. We multiply and get

$$
f_1(x) cdot g_1(x) = (x-2)(2x+3) = 2x^2 +3x-4x-6 = 2x^2-x-6.
$$

The product function $text{color{#c34632} is}$ a polynomial function. It’s degree is $color{#c34632} 2$.

Finally we consider the quotient function. We have

$$
frac{f_1(x)}{ g_1(x)} = frac{x-2}{2x+3}.
$$

The quotient function $text{color{#c34632} is not}$ a polynomial function because the variable is in the denominator.

b.

Let us discuss if the results from part (a) be generalized.

The sum function of two linear functions can be a linear function or a constant form of linear function. Therefore, the function is always linear, hence it is always a polynomial function, but the degree of the polynomial is not always the same. If the leading coefficients of the functions we are summing are opposite numbers then the degree is $0$, otherwise the degree is $1$.

The difference function of two linear functions can be a linear function or a constant form of linear function. Therefore, the function is always linear, hence it is always a polynomial function, but the degree of the polynomial is not always the same. If the leading coefficients of the functions we are summing are the same numbers then the degree is $0$, otherwise the degree is $1$.

The product function of two linear functions can be a quadratic function or a linear function. Therefore, the function is always a polynomial function, but the degree of the polynomial is not always the same. If none of the functions we are multiplying is not a constant vertical function then the degree of the polynomial is $2$, otherwise the degree is $1$.

The quotient function of two linear functions can be a polynomial function only if the function in the denominator is a constant vertical function. In that case the degree of the polynomial is $1$. But if the variable is in the denominator, the quotient function is not a polynomial function.

c.

One-variable polynomials are not closed under division. As we have seen in previous exercises dividing two polynomials might not always be a polynomial. If there is a variable in the denominator the function is not a polynomial function. The examples are given in the exercise $49$ and $53$.

Let us consider the given functions

$$
color{#4257b2} p(x) = x^3-3x-1 hspace{4mm} text{color{default} and} hspace{4mm} q(x)=x-1.
$$

We graph the function
$$
color{#c34632} p(x) = x^3-3x-1
$$
as seen below. Now we can describe the graph.

The function is polynomial function. The domain of the function are all real numbers. Same goes for the range, it includes all real numbers.

The $x$-intercept are $x=-1.532$, $x=-0.347$ and $x=1.879$, while the $y$-intercept is $y=-1$.

The function is increasing for $x in (-infty, -1) cup (1,infty)$ and the function is decreasing for $x in (-1,1)$. Hence, our function has a local maximum point $(-1,1)$ and a local minimum point $(1,-3)$.

The function is not symmetric, but it is continuous.

Now we graph the function
$$
color{#c34632} q(x) = x-1
$$
as seen below. We can describe the graph.

The function is linear and polynomial function. The domain of the function are all real numbers. Same goes for the range, it includes all real numbers.

The $x$-intercept are $x=1$, while the $y$-intercept is $y=-1$.

The function is increasing for $x in (-infty, infty)$. Hence, our function does not have minimum or maximum points.

The function is not symmetrical, but it is continuous.

a.

Let us consider the given functions

$$
color{#4257b2} p(x) = x^3-3x-1 hspace{4mm} text{color{default} and} hspace{4mm} q(x)=x-1.
$$

Now we evaluate the sum function.

$$
g(x)=p(x) + q(x) = x^3-3x-1+x-1 =color{#c34632} x^3-2x-2.
$$

Let us sketch the given functions along with the sum function. We see the graph below.

Now we can describe the graph.

The function is polynomial function. The domain of the function are all real numbers. Same goes for the range, it includes all real numbers.

The $x$-intercept are $x=1.769$, while the $y$-intercept is $y=-2$.

The function is increasing for $x in (-infty, -0.8) cup (0.8,infty)$ and the function is decreasing for $x in (-0.8,0.8)$. Hence, our function has a local maximum point $(-0.8,-0.9)$ and a local minimum point $(0.8,-3.1)$.

The function is not symmetric, but it is continuous.

b.

Let us consider the given functions

$$
color{#4257b2} p(x) = x^3-3x-1 hspace{4mm} text{color{default} and} hspace{4mm} q(x)=x-1.
$$

Now we evaluate the difference function.

$$
f(x)=p(x) – q(x) = x^3-3x-1-(x-1) =x^3-3x-1-x+1 =color{#c34632} x^3-4x.
$$

Let us sketch the given functions along with the difference function. We see the graph below.

Now we can describe the graph.

The function is polynomial function. The domain of the function are all real numbers. Same goes for the range, it includes all real numbers.

The $x$-intercept are $x=-2$, $x=0$ and $x=2$, while the $y$-intercept is $y=0$.

The function is increasing for $x in (-infty, -1.15) cup (1.15, infty)$ and the function is decreasing for $x in (-1.15,1.15)$. Hence, our function has a local maximum point $(-1.15, 3.08)$ and a local minimum point $(0.8, -3.08)$.

The function is symmetric and it is continuous.

c.

Let us consider the given functions

$$
color{#4257b2} p(x) = x^3-3x-1 hspace{4mm} text{color{default} and} hspace{4mm} q(x)=x-1.
$$

Now we multiply the given functions and get

$$
begin{align*}
h(x) &=p(x) cdot q(x) = (x^3-3x-1)(x-1) \ &=x^4-x^3-3x^2+3x-x+1 \ &=color{#c34632} x^4-x^3-4x^2+2x+1.
end{align*}
$$

Let us sketch the given functions along with the product function. We see the graph below.

Now we can describe the graph.

The function is polynomial function. The domain of the function are all real numbers. And the range is $[-2.04, infty)$.

The $x$-intercept are $x=-1.53$, $x=-0.35$, $x=1$ and $x=1.88$, while the $y$-intercept is $y=1$.

The function is increasing for $x in (-1.08, 0.31) cup (1.52, infty)$ and the function is decreasing for $x in (-infty, -1.08) cup (0.31, 1.52)$. Hence, our function has a local maximum point $(0.31, 1.31)$ and two local minimum points $(-1.08, -2.04)$ and $(1.52, -1.07)$.

The function is not symmetric, but it is continuous.

d.

Let us consider the given functions

$$
color{#4257b2} p(x) = x^3-3x-1 hspace{4mm} text{color{default} and} hspace{4mm} q(x)=x-1.
$$

Now we evaluate the quotient function.

$$
r(x)=frac{p(x)}{q(x)}=color{#c34632} frac{x^3-3x-1}{x-1}.
$$

Let us sketch the given functions along with the quotient function. We see the graph below.

Now we can describe the graph.

The domain of the function are all real numbers except $1$, hence the domain is $(-infty, 1) cup (1, infty)$. The range includes all real numbers.

The $x$-intercept are $x=-1.53$, $x=-0.35$ and $x=1.88$, while the $y$-intercept is $y=1$.

The function is increasing for $x in (-0.91, 1) cup (1, infty)$ and the function is decreasing for $x in (-infty,-0.91)$. Hence, our function has a local minimum point $(-0.91, -0.51)$.

The function is not symmetric and it is not continuous.

The quotient function has a vertical asymptote $x=1$.

$f(x)=3x^2-5$          (Given)

$g(x)=sqrt {x-5}+2$          (Given)

a-          $f(5)=3(5)^2-5=75-5=70$

b-          $g(5)=sqrt {5-5}+2=0+2=2$

c-          $f(4)=3(4)^2-5=48-5=43$

d-          $g(4)=sqrt {4-5}+2=sqrt {-1}+2$          (The radicand is less than 0 so there is no real solution)

e-          $f(x)+g(x)=3x^2-5+sqrt {x-5}+2=3x^2+sqrt {x-5}-3$

f-          $g(x)-f(x)=sqrt {x-5}+2-(3x^2-5)=-3x^2+sqrt {x-5}+7$

g-          The domain of $f(x)$ is $(-infty<x<infty)$

h-          The domain of $g(x)$ is $(5 leq x)$

i-          The domain of the function $g(x)$ is more restrictive than the domain of $f(x)$ because $g(x)$ is a radical function. the radicand is $(x-5)$ that can't be negative value.

a- $f(5)=70$          b- $g(5)=2$          c- $f(4)=43$

d- $g(4)=sqrt {-1}+2$          (The radicand is less than 0 so there is no real solution)

e- $f(x)+g(x)=3x^2+sqrt {x-5}-3$          f-          $g(x)-f(x)=-3x^2+sqrt {x-5}+7$

g- The domain of $f(x)$ is $(-infty<x<infty)$          h- The domain of $g(x)$ is $(5 leq x)$

i- The domain of the function $g(x)$ is more restrictive than the domain of $f(x)$ because $g(x)$ is a radical function. the radicand is $(x-5)$ that can't be negative value.

a.

The given function
$$
y=3x^2+2x^2+x = 5x^2+x
$$
is a polynomial function.

b. The given function
$$
y=(x-2)^2(x-2)^2 = (x-2)^4 = x^4-8x^3+24x^2-32x+16
$$
is a polynomial function.

c. The given function
$$
y=x^2+2^x
$$
is not a polynomial function because the variable is in the exponent.

d.

The given function
$$
y=3x-1
$$
is a polynomial function.

e.

The given function
$$
y=(x-2)^2-1 = x^2-4x+4-1=x^2-4x+3
$$
is a polynomial function.

f.

The given function
$$
y^2=(x-2)^2-1 implies y= sqrt{(x-2)^2-1}
$$
is not a polynomial function because the variable is inside the radical.

g. The given function
$$
y=frac{1}{x^2}+frac{1}{x}+frac{1}{2}
$$
is not a polynomial function because the variable is in the denominator.

h. The given function
$$
y=frac{1}{2}x+frac{1}{3}
$$
is a polynomial function.

i.

The given function
$$
y=x
$$
is a polynomial function.

j.

The given function
$$
y=-7
$$
is a polynomial function.

The initial height $=2$ feet.

The weekly growth $=3$ in $=dfrac {3}{12}=dfrac {1}{4}$ feet.

Assuming $x$ is the days passed from planting the tree. and $y$ is the total height of the tree after $x$ days.

The weeks passe $w=dfrac {x}{7}$

a-

The equation that represents this relation is:

$y=dfrac {1}{4}w+2$

$y=dfrac {1}{4} times dfrac {x}{7}+2$

$y=dfrac {x}{28}+2$

The below is the data table for the first 4 weeks.\
\
begin{center}
begin{tabular}{|c|c|}
hline
begin{tabular}[c]{@{}c@{}}Days\ $x$end{tabular} & begin{tabular}[c]{@{}c@{}}Height of the tree\ $y$ feetend{tabular} \ hline
0 & 2 \ hline
7 & 2.25 \ hline
14 & 2.5 \ hline
21 & 2.75 \ hline
28 & 3 \ hline
end{tabular}
end{center}

b-

If the tree is 6 feet tall, then $y=6$

$6=dfrac {x}{28}+2$          (Substituting 6 for $y$)

$dfrac {x}{28}=4$          (Subtracting 2 from each side)

$x=112$          (Solve for $x$)

$w=dfrac {x}{7}=dfrac {112}{7}=16$ weeks

The tree will be 6 feet tall after 112 days (16 weeks)

c-

The possible inputs of the graph is $(x geq 0)$

The possible outputs of the graph is $(y geq 2)$

a-          The equation that represents this relation is: $y=dfrac {x}{28}+2$

b-          The tree will be 6 feet tall after 112 days (16 weeks)

c-          The possible inputs: $(x geq 0)$          The possible outputs $(y geq 2)$

The error is equating $5.4x=23.7$

While the correct is $5.4x=-23.7$

$5.4x=-23.7$

$dfrac {5.4x}{5.4}=dfrac {23.7}{5.4}$          (Divide each side by 5.4)

$x=-4.39$          (Solve for $x$)

The correct solution is $x=-4.39$

Let us consider the given function $$color{blue} f(x)=4^x.$$
We can now make the table of the function.
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c|c|c|c|c| }
hline
x & -2 & -1 & 0 & 1 & 2 \
hline
y & 0.0625 & 0.25 & 1 & 4 & 16 \
hline
end{tabular}
end{center}
And now we can draw the graph.\

Now we can completely describe the graph.

The domain of the function are all real numbers, we can write $x in (-infty, infty)$.

The range of the function are all positive real numbers, so we can write $y in (0, infty)$.

The $x$-intercept does not exist.

The $y$-intercept is the point $y=1$.

The function is not symmetric.

The function has one horizontal asymptote $y=0$.

The function is increasing for $x in (-infty, infty).$

The function does not have minimum or maximum points.

The function is continuous.

a-

The domain is $x in (-1, 1, 2)$

The range is $y in(-2, 1, 2)$

b-

The domain is: $(-2 leq x<1)$

The range is: $(-2 leq y<2)$

c-

The domain is: $(-2 leq x< infty)$

The range is: $(-2 leq y< infty)$

d-

The domain is: $(-infty leq x< infty)$

The range is: $(-2 leq y< infty)$

begin{center}
renewcommand{arraystretch}{1.5}
begin{tabular}{|l|l|l|}
hline
Item & Domain & Range \
hline
$a-$ & $x in (-1, 1, 2)$ & $y in (-2, 1, 2)$ \
hline
$b-$ & $(-2 leq x < 1)$ & $(-2 leq y < 2)$ \
hline
$c-$ & $(-2 leq x < infty)$ & $(-2 leq y < infty)$ \
hline
$d-$ & $(-infty leq x < infty)$ & $(-2 leq y < infty)$ \
hline
end{tabular}
end{center}

Let us consider the given picture. We marked the angles of the given triangle with $alpha$, $beta$ and $gamma$. Now we have
$$
alpha +150^{circ} = 180^{circ} implies alpha =180^{circ}-150^{circ} implies alpha = 30^{circ}.
$$

Similarly,
$$
gamma +140^{circ} = 180^{circ} implies gamma =180^{circ}-140^{circ} implies gamma = 40^{circ}.
$$

Since angles in a triangle sum up to $180^{circ}$ we get

$$
begin{align*}
alpha + beta + gamma &= 180^{circ} \
30^{circ}+ beta +40^{circ} &=180^{circ} \
beta +70^{circ} &= 180^{circ} \
beta &= 1480^{circ}-70^{circ} implies beta = 110^{circ}.
end{align*}
$$

Finally, we have
$$
x=180^{circ}-110^{circ} = 70^{circ}.
$$

Therefore,
$$
color{#c34632} x=70^{circ}.
$$

$$
x=70^{circ}
$$

a.

Let us determine the points of intersection of the functions

$$
f(x)= 2x^2-5x+6 hspace{5mm} text{and} hspace{5mm} g(x)= -2x^2-x+30.
$$

We can do that by using many different methods. we can use graphs, tables and equations.

In order to determine the points of intersection of the two functions using graphs, we graph

$$
f(x)= 2x^2-5x+6 hspace{5mm} text{and} hspace{5mm} g(x)= -2x^2-x+30.
$$

Using the graph given below we see that the intersection points are

$$
color{#c34632} (-2, 24) hspace{5mm} text{color{default} and} hspace{5mm} (3, 9).
$$

Now, we can find the intersection point using tables. We form the table for function $$f(x)=2x^2-5x+6$$ first. We have
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c| }
hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \
hline
y & 39 & 24 & 13 & 6 & 3 & 4 & 9 & 18 \
hline
end{tabular}
end{center}
Now we form the table for function $$g(x)= -2x^2-x+30 $$ and get
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c| }
hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \
hline
y & 15 & 24 & 29 & 30 & 27 & 20 & 9 & -6 \
hline
end{tabular}
end{center}
Comparing these two tables we see that the intersection points are
$$color{red} (-2, 24) hspace{5mm} text{color{black} and} hspace{5mm} (3, 9). $$

In order to find the intersection points using equations we solve the following system

$$
begin{align*}
y &= 2x^2-5x+6 \
y &= -2x^2 -x+30.
end{align*}
$$

Since the left sides of both equations are equal, we conclude that the right sides must be equal too. Therefore

$$
2x^2-5x+6 = -2x^2 -x+30.
$$

Now we solve for $x$.

$$
begin{align*}
2x^2-5x+6+2x^2+x-30 &=0 \
4x^2-4x-24 &= 0 \
4x^2+8x-12x-24 &=0 \
4x(x+2) – 12(x+2) &=0 \
(x+2) cdot (4x-12) &=0 \
x+2 = 0 hspace{3mm} vee hspace{3mm}4x-12 &= 0 \
x = -2 hspace{3mm} vee hspace{3mm}4x &=12 \
x = -2 hspace{3mm} vee hspace{3mm} x &=3
end{align*}
$$

In order to calculate $y$ we substitute $x=-2$ and $x=3$ in one of the two equations from the system. This yields

$$
begin{align*}
y &=2cdot (-2)^2-5 cdot (-2)+6 = 2 cdot 4 +10+6 = 8+16 = 24 \
y &=2cdot 3^2-5 cdot 3+6 = 2 cdot 9 -15+6 = 18-9 = 9.
end{align*}
$$

Therefore, the intersection points are

$$
color{#c34632} (-2, 24) hspace{5mm} text{color{default} and} hspace{5mm} (3, 9).
$$

a. $(-2, 24)$ and $(3,9)$

b. Yes. It is very accurate.

Let us consider the given functions $$color{blue} f(x)=3x-5$$
and the function $g(x)$ defined with the following table
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c|c|c|c|c| }
hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \
hline
g(x) & 21 & 17 & 13 & 9 & 5 & 1 & -3 \
hline
end{tabular}
end{center}
Now we can draw the graph of the both functions and find where they intersect.\
As seen on the graph below the point of intersection of the two graphs is $$color{red} (2, 1).$$

The point of intersection is $(2, 1)$.

a.

Let us consider the given expression
$$
color{#4257b2} 27^{1/3}.
$$

We can transform the given expression and get

$$
27^{1/3} = sqrt[3]{27} = sqrt[3]{3^3} = 3.
$$

Therefore,
$$
color{#c34632} 27^{1/3}=3.
$$

b.

Let us consider the given expression
$$
color{#4257b2} (5x^7y^{-1})^{-2}.
$$

We can use the properties of exponents and transform the given expression

$$
(5x^7y^{-1})^{-2} = Big(5x^7 cdot frac{1}{y^1}Big)^{-2} = bigg( frac{5x^7}{y}bigg)^{-2} = bigg( frac{y}{5x^7}bigg)^2 = frac{y^2}{5^2 cdot (x^7)^2} = frac{y^2}{25x^{14}}.
$$

Therefore,
$$
color{#c34632} (5x^7y^{-1})^{-2}= frac{y^2}{25x^{14}}.
$$

c.

Let us consider the given expression
$$
color{#4257b2} (3x^9 cdot y^{-4})(6x^{-8}y^4).
$$

We can use the properties of exponents and transform the given expression

$$
begin{align*}
(3x^9 cdot y^{-4})(6x^{-8}y^4) &= bigg( 3x^9 cdot frac{1}{y^4}bigg)bigg(6 cdot frac{1}{x^8} cdot y^4bigg)
= frac{3x^9}{y^4} cdot frac{6y^4}{x^8} \
&= frac{18x^9y^4}{y^4x^8} = 18x^{9-8} cdot y^{4-4}=18x^1 cdot y^0 \ &= 18x cdot 1 = 18x.
end{align*}
$$

Therefore,
$$
color{#c34632} (3x^9 cdot y^{-4})(6x^{-8}y^4)= 18x.
$$

$$
begin{align*}
&text{a.} hspace{5mm} 27^{1/3} = 3 \
&text{b.} hspace{5mm} (5x^7y^{-1})^{-2}= frac{y^2}{25x^{14}} \
&text{c.} hspace{5mm} (3x^9 cdot y^{-4})(6x^{-8}y^4)= 18x \
end{align*}
$$

Let us consider the given condition
$$
color{#4257b2} 2^{x+4}=2^{3x-1}.
$$

In order for the equality to stand the exponents must be the same. Therefore,

$$
begin{align*}
2^{x+4}=2^{3x-1} implies x+4 &=3x-1 \ x-3x &= -1-4 \ -2x&=-5 \ x &= frac{-5}{-2} = frac{5}{2} = 2,5.
end{align*}
$$

We can conclude that if
$$
2^{x+4}=2^{3x-1}
$$
the value of $x$ is
$$
color{#c34632} x=2,5.
$$

If $2^{x+4} = 2^{3x-1}$ then the value of $x$ is $x=2,5$.

Assuming the points are: $A=(-5,0)$ and $B=(0,3)$

a-          the graph of the points is the below.

The distance between two points $(x1, y1)$ and $(x2, y2)$ is given by the formula:

$d=sqrt {(y2-y1)^2+(x2-x1)^2}$

The distance between $A$ and $B$:

$d=sqrt {(3-0)^2+(0-(-5))^2}=sqrt {9+25}=sqrt {34} approx 5.83$ units

b-

The slope the line is given by the formula:

$m=dfrac {y2-y1}{x2-x1}=dfrac {3-0}{0-(-5)}=dfrac {3}{5}$

a-          The distance between two points $=sqrt {34} approx 5.83$ units

b-          The slope the line: $m=dfrac {3}{5}$

Assuming $y$ is the length of the snake after $x$ weeks

The initial length of the snake is 26 cm

The weekly growing is 2 cm

a-

The equation that could represent this relation is:

$y=2x+26$

The graph below represents the equation

The next is the data table for the equation:\
\
begin{center}
renewcommand{arraystretch}{1.5}
begin{tabular}{|r|r|}
hline
Weeks $x$ & Snake length $y$ \
hline
0 & 26 \
hline
1 & 28 \
hline
2 & 30 \
hline
3 & 32 \
hline
4 & 34 \
hline
5 & 36 \
hline
6 & 38 \
hline
7 & 40 \
hline
8 & 42 \
hline
9 & 44 \
hline
10 & 46 \
hline
end{tabular}
end{center}

b-

When the snake is 100 cm length, $y=100$

Substituting in the equation:

$y=2x+26$

$100=2x+26$

$2x=74$

$x=37$

The snake will be 100 cm long after 37 weeks.

a-          The equation that represent this relation is:          $y=2x+26$

b-          The snake will be 100 cm long after 37 weeks.

The value that allows us to find the x-intercept is: $y=0$

a-          $y=3x+6$          (Write the equation)

For x-intercept, Substitute $y=0$

$0=3x+6$

$x=-dfrac {6}{3}=-2$

x-intercept is $(-2, 0)$

b-          $x=5y-10$          (Write the equation)

For x-intercept, Substitute $y=0$

$x=5(0)-10$          (Substitute $y=0$)

$x=0-10=-10$

x-intercept is $(-10, 0)$

c-          $y=x^2$          (Write the equation)

For x-intercept, Substitute $y=0$

$0=x^2$

$x=0$

x-intercept is $(0, 0)$

d-          $y=2x^2-4$          (Write the equation)

$2x^2=y+4$

$x^2=dfrac {y+4}{2}$

$$
x=sqrt {dfrac {y+4}{2}}
$$

For x-intercept, Substitute $y=0$

$x=pm sqrt {dfrac {0+4}{2}}=pm sqrt {2} approx pm 1.414$

x-intercept is $(1.414, 0)$ and $(-1.414, 0)$

e-          $y=(x-5)^2$          (Write the equation)

For x-intercept, Substitute $y=0$

$0=(x-5)^2$

$x-5=0$

$x=5$

x-intercept is $(5, 0)$

f-          $y=x^3-13$          (Write the equation)

$x^3=y+13$

$x=sqrt [3]{y+13}$

For x-intercept, Substitute $y=0$

$x=sqrt [3]{13} approx 2.3513$

x-intercept is $(2.3513, 0)$

The value that allows us to find the x-intercept is: $y=0$

a-          x-intercept is $(-2, 0)$

b-          x-intercept is $(-10, 0)$

c-          x-intercept is $(0, 0)$

d-          x-intercepts are $(1.414, 0)$ and $(-1.414, 0)$

e-          x-intercept is $(5, 0)$

e-          x-intercept is $(2.3513, 0)$

a-

$y=dfrac {3}{5}x+1$          (Write the equation)

$dfrac {3}{5}x=y-1$          (Subtract 1 from each side)

$x=dfrac {5}{3}(y-1)$          (Multiply each side by $dfrac {5}{3}$

$$
x=dfrac {5}{3}(y-1)
$$

b-

$3x+2y=6$          (Write the equation)

$3x=6-2y$          (Subtract $2y$ from each side)

$x=dfrac {6-2y}{3}$          (Divide each side by 3)

$$
x=dfrac {6-2y}{3}
$$

c-

$y=x^2$          (Write the equation)

$x=sqrt {y}$          (The square root of each side)

$$
x=sqrt {y}
$$

d-

$y=x^2-100$          (Write the equation)

$x^2=y+100$          (Add 100 to each side)

$x=sqrt {y+100}$          (The square root of each side)

$$
x=sqrt {y+100}
$$

a-          $x=dfrac {5}{3}(y-1)$

b-          $x=dfrac {6-2y}{3}$

c-          $x=sqrt {y}$

d-          $x=sqrt {y+100}$

Assuming the size of the paper sheet is $22 mathrm{ ~cm} times 16 mathrm{ ~cm}$\
And the side length of the square we cut from the corner is $x mathrm{ ~cm}$\
The following table demonstrates the different volumes of the box for the different valus of $x$\
\
begin {center}
renewcommand{arraystretch}{2}
begin{tabular}{|r|r|r|r|}
hline
Square length (Unint) & Remaining Length & Remaining Width & Box volume \
hline
0 & 22 & 16 & 352 \
hline
1 & 20 & 14 & 280 \
hline
2 & 18 & 12 & 216 \
hline
3 & 16 & 10 & 160 \
hline
4 & 14 & 8 & 112 \
hline
5 & 12 & 6 & 72 \
hline
6 & 10 & 4 & 40 \
hline
7 & 8 & 2 & 16 \
hline
8 & 6 & 0 & 0 \
hline
end{tabular}
end {center}

We collect data for this relationship by assuming the length of the square $x=1$ cm in the firs time and increase the length by 1 unit each time.

All the possible inputs for our function are:          $(0 <x<8)$

$l=22$

$w=16$

begin{center}
begin{tabular}{|| c|c|c| c||}
hline
$x$ & $22-2x$ & $16-2x$ & $V$ \ [0.5ex]
hline
1 & 22-2(1)=20 & 16-2(1)=14 & 1(20)(14)=280 \
hline
2 & 22-2(2)=18 & 16-2(2)=12 & 2(18)(12)=432 \
hline
3 & 22-2(3)=16 & 16-2(3)=10 & 3(16)(10)=480 \
hline
4 & 22-2(4)=14 & 16-2(4)=8 & 4(14)(8)=448 \
hline
5 & 22-2(5)=12 & 16-2(5)=6 & 5(12)(6)=360 \
hline
6 & 22-2(6)=10 & 16-2(6)=4 & 6(10)(4)=240 \[1ex]
hline
end{tabular}
end{center}

b) We give $x$ different values and determine the volume of the box:

c) We graph the volume as a function of the height $x$:

a) We note:

$l$=the length of the paper

$w$=the width of the paper

$x$=the height of the box

The length of the box is:

$22-2x$,

while the width of the box is:

$16-2x$.

$$
V=(l-2x)(w-2x)x
$$

$$
V=x(22-2x)(16-2x)
$$

a) In Problem 1-75 we got the equation of the volume in terms of the height $x$:

$$
begin{cases}
x>0\
22-2x>0\
16-2x>0
end{cases}
$$

$$
begin{cases}
x>0\
22>2x\
16>2x
end{cases}
$$

$$
begin{cases}
x>0\
11>x\
8>x
end{cases}
$$

$$
xin (0,8)
$$

$$
V=4x(11-x)(8-x)
$$

The function $V$ has the zeros $0, 8,11$.Between the two consecutive zeros 0 and 8, the function is positive, having a local maximum at some point between 0 and 8, therefore the range of the volume is limited between 0 and this relative maximum.

c) We can connect the points with a smooth curve as the function has meaning for any real value of $x$ between 0 and 8 (the function is continuous).

d) The graph has the domain $(0,8)$ and the range $(0,480]$.

It is continuous, increases when $xin (0,3)$ and decreases for $xin (3,8)$.

It has no asymptote.

Both ends go to 0.

$V=x(22-2x)(16-2x)$

The graph we made on paper is included in the above graph. The domain of our function is a subset of the set of real numbers, so the function we drew is a restriction of teh above function to the interval which makes sense for our model: $(0,8)$.

$f(x)=x^2-2x+6$          (Given)

$g(x)=2x+11$          (Given)

For solving the system we graph both functions on the same grid and the intersections of both functions represent the solution.

a-

From the graph, the point of intersection are:          $(-1, 9)$ and $(5, 21)$

b-

$$
begin{align*}
& f(x)+g(x)\
& =(x^2-2x+6)+(2x+11)\
& =x^2 cancel {-2x}+6 cancel {+2x}+11\
& =x^2+17\
\
end{align*}
$$

c-

$$
begin{align*}
& f(x)-g(x)\
& =(x^2-2x+6)-(2x+11)\
& =x^2-2x+6 -2x-11\
& =x^2-4x-5\
\
end{align*}
$$

a-          The point of intersection are:          $(-1, 9)$ and $(5, 21)$

b-                   $f(x)+g(x)=x^2+17$

c-                   $f(x)-g(x)=x^2-4x-5$

a.

The lengths of earthworms in centimeters are

$$
color{#4257b2} 4.0, hspace{3mm} 5.0, hspace{3mm} 5.8, hspace{3mm} 7.1, hspace{3mm} 7.8, hspace{3mm} 8.1, hspace{3mm} 8.2, hspace{3mm} 8.3, hspace{3mm} 8.4, hspace{3mm} 11.9, hspace{3mm} 12.0
$$

We can conclude right away that the minimum is $4$ and the maximum is $12$. The median is
$$
M=8.1.
$$
Since
$$
4.0, hspace{3mm} 5.0, hspace{3mm} 5.8, hspace{3mm} 7.1, hspace{3mm} 7.8
$$
we now have

$$
color{#4257b2} Q1=5.8.
$$

Similarly, since
$$
8.2, hspace{3mm} 8.3, hspace{3mm} 8.4, hspace{3mm} 11.9, hspace{3mm} 12.0
$$
we get
$$
color{#4257b2} Q3=8.4.
$$

Finally, we can calculate the Interquartile Range. We have

$$
IQR=Q3-Q1=8.4-5.8=2.6.
$$

Therefore, the Interquartile Range is
$$
color{#c34632} IQR=2.6.
$$

a. The Interquartile Range is $2.6$.

b. The boxplot of data is made.

a.

Let us consider the given expression
$$
color{#4257b2} 16^{5/4}.
$$

We can transform the given expression and get

$$
16^{5/4} = sqrt[4]{16^5} = big( sqrt[4]{16} big)^5 = big( sqrt[4]{2^4} big)^5 = 2^5 = 32.
$$

Therefore,
$$
color{#c34632} 16^{5/4}=32.
$$

b.

Let us consider the given expression
$$
color{#4257b2} (x^5y^4)^{1/2}.
$$

We can use the properties of exponents and transform the given expression

$$
begin{align*}
(x^5y^4)^{1/2} &= (x^5)^{1/2} cdot (y^4)^{1/2} = x^{5 cdot frac{1}{2}} cdot y^{4 cdot frac{1}{2}} = x^{5/2} cdot y^2 \ &= sqrt{x^5} cdot y^2 = sqrt{x^4 cdot x} cdot y^2 = x^2y^2 sqrt{x}. end{align*}
$$

Therefore,
$$
color{#c34632} (x^5y^4)^{1/2}= x^2y^2 sqrt{x}.
$$

c.

Let us consider the given expression
$$
color{#4257b2} (x^2y^{-1})(x^{-3}y)^0.
$$

We can use the properties of exponents and transform the given expression

$$
begin{align*}
(x^2y^{-1})(x^{-3}y)^0 &= bigg( x^2 cdot frac{1}{y^1}bigg) cdot 1 = frac{x^2}{y}.
end{align*}
$$

Therefore,
$$
color{#c34632} (x^2y^{-1})(x^{-3}y)^0= frac{x^2}{y}.
$$

$$
begin{align*}
&text{a.} hspace{5mm} 16^{5/4}=32 \
&text{b.} hspace{5mm} (x^5y^4)^{1/2}= x^2y^2 sqrt{x} \
&text{c.} hspace{5mm} (x^2y^{-1})(x^{-3}y)^0= frac{x^2}{y} \
end{align*}
$$

Let us consider the given function $$color{blue} f(x)=sqrt[3]{x+2}.$$
We can now make the table of the function.
begin{center}
begin{tabular}{ |c|c|c|c|c|c|c|c|c|c|c|c|c| }
hline
$x$ & -10 & -3 & -2 & -1 & 0 & 6 & 25 \
hline
$f(x)$ & -2 & -1 & 0 & 1 & $sqrt[3]{2} approx 1.26$ & 2 & 3 \
hline
end{tabular}
end{center}
And now we can draw the graph. The graph is seen below.\
\
We can see in the table and on the graph that $x$-intercept is $x=-2$, while the $y$=intercept is $y=sqrt[3]{2} approx 1.26$. \
\
The domain are all real number, hence we can write $color{red} x in (-infty, infty)$. The range is just the same, therefore $color{red} y in (-infty, infty)$. \

The function is graphed and the domain and the range are all the real numbers, hence $(-infty, infty)$.

a-

Domain is:          $x in (-2, -1, 2)$

Range is:          $y in (-1, 0, 1)$

b-

Domain is:          $(-1<x leq 1)$

Range is:          $(-1 leq y<2)$

c-

Domain is:          $(-1<x < infty)$

Range is:          $(-1<y<infty)$

d-

Domain is:          $(-infty<x < infty)$

Range is:          $(-infty<y<infty)$

begin{center}
begin{tabular}{|r|r|r|}
hline
Item & Domain & Range \
hline
a- & $x in (-2, -1, 2)$ & $y in (-1, 0, 1)$ \
hline
b- & $(-1 < x leq 1)$ & $(-1 leq y < 2)$ \
hline
c- & $(-1 < x < infty)$ & $(-1 < y < infty)$ \
hline
d- & $(-infty < x < infty)$ & $(-infty < y < infty)$ \
hline
end{tabular}
end{center}

Assuming the width of the rectangle is $w$

The length of the rectangle $l=4w$          (Given)

$l+w=22$          (Given)

$4w+w=22$          (Substituting $4w$ for $l$)

$5w=22$          (Simplify)

$w=4.4$ Units          (Solve for $w$)

$l=4w=4cdot 4.4=17.6$ units          (Solve for $l$)

The length of the rectangle $l=17.6$

The width of the rectangle $w=4.4$

Let us calculate exactly where the given lines
$$
color{#4257b2} y=2x-frac{7}{6} hspace{4mm} text{color{default} and} hspace{4mm} y=3-3x
$$
intersect. To find their intersection using algebra we solve the following system of equations

$$
begin{align*}
y &=2x-frac{7}{6} \
y &=3-3x.
end{align*}
$$

Since the left sides of both equations are equal, we can conclude that the right sides of the equations must be equal. Therefore we get

$$
2x-frac{7}{6} = 3-3x.
$$

Now we solve this equation for $x$.

$$
begin{align*}
2x-frac{7}{6} &= 3-3x \
2x+3x &= 3+ frac{7}{6} \
5x &= frac{18}{6} + frac{7}{6} \
5x &= frac{25}{6} \
x&= frac{25}{6} cdot frac{1}{5} = frac{5}{6}.
end{align*}
$$

To calculate $y$ we substitute $x=5/6$ in $y=3-3x$ and get

$$
y=3-3cdot frac{5}{6} = 3-frac{5}{2} = frac{6}{2}- frac{5}{2}= frac{1}{2}.
$$

Finally, we conclude that the point of intersection for two given functions is
$$
color{#c34632} (x, y) = (frac{5}{6}, frac{1}{2}).
$$

The point of intersection for the two given functions is
$$
(frac{5}{6}, frac{1}{2}).
$$

a-

$w^2+4w=0$          (Given)

$w(w+4)=0$          (Factor out $w$)

$w=0$ or $w=-4$

b-

$5w^2-2w=0$          (Given)

$w(5w-2)=0$          (Factor out $w$)

$w=0$ or $w=dfrac {2}{5}$

c-

$w^2=6w$          (Given)

$w^2-6w=0$

$w(w-6)=0$          (Factor out $w$)

$w=0$ or $w=6$

a-          $w=0$ or $w=-4$

b-          $w=0$ or $w=dfrac {2}{5}$

c-          $w=0$ or $w=6$

A nonprofit consumer health organization measured the fat content of five low-fat raspberry muffins to have
$$
color{#4257b2} 6, hspace{3mm} 8, hspace{3mm} 8, hspace{3mm} 9, hspace{3mm} 7
$$
grams of fat. Let us calculate the mean first. The mean value is
$$
frac{6+8+8+9+7}{5} = frac{38}{5} = color{#c34632} 7,6.
$$

The next step is to calculate the distances from the mean value. We have

$$
begin{align*}
6-7,6 &= -1.6 \
8-7,6 &= 0,4 \
9-7,6 &= 1,4 \
7-7,6 &= -0,6.
end{align*}
$$

Now we calculate the sum of squares of the distances from the mean value. We have

$$
begin{align*}
(-1,6)^2+(0,4)^2+(0,4)^2+(1,4)^2 + (-0,6)^2 &= 2,56+0,16+0,16+1,96+0,36 \ &=5,2. end{align*}
$$

Finally, we can calculate the sample standard deviation

$$
sqrt{frac{5,2}{5-1}} = sqrt{frac{5,2}{4}} = sqrt{1,3} approx color{#c34632} 1,14.
$$

Therefore, $text{color{#c34632} the mean value is $7,6$g and the sample standard deviation is $1,14$g.}$

The mean value is $7,6$g and the sample standard deviation is $1,14$g.

The graph of the function $h(x)=x^2-5$ crosses the x-axis when $h(x)=0$

$0=x^2-5$          (Substitute 0 for $h(x)$)

$x^2=5$

$x= pm sqrt {5} approx 2.236$

The graph crosses the x-axis at:          $x approx 2.236$ and $x approx -2.236$

The graph crosses the x-axis at:          $x approx 2.236$ and $x approx -2.236$

The equation of the x-axis is:

$$
y=0
$$

The equation of the y-axis is:

$$
x=0
$$

The equation of the x-axis is:          $y=0$

The equation of the y-axis is:          $x=0$

a.

Let us multiply the given expression

$$
begin{align*}
(x-1)(x+1) &= x cdot x + x cdot 1 -1 cdot x -1 cdot 1 \
&= x^2+x-x-1 \ &= x^2 -1.
end{align*}
$$

Therefore, we conclude
$$
color{#c34632} (x-1)(x+1) = x^2-1.
$$