a) The first and second differences of $y$ is shown in the table below.

Since the second difference is constant, then the function defined by the given table is quadratic.

b) The first and second differences of $y$ is shown in the table below.

Since the second difference is constant, then the function defined by the given table is quadratic.

a) To graph the quadratic function of the form $f(x)=a(x-h)^2+k$

1) Plot the vertex $(h,k)$ and evaluate the function for $x$ values close to $x=h$,
for example $x=h-1$ , $x=h+1$

b) For quadratic function of the form $f(x)=a(x-m)(x-n)$, the $x$-intercepts are $(m,0)$ and $(n,0)$ while you can obtain the vertex as $h=dfrac{m+n}{2}$, $k=f(h)$

In this case, the $x$-intercepts are $(-4,0)$ and $(6,0)$

$h=dfrac{4-6}{2}=-1$, $k=f(-1)=2(-1+4)(-1-6)=-50$

a) To graph the quadratic function of the form $f(x)=a(x-h)^2+k$, plot the vertex $(h,k)$ and evaluate the function for $x$ values close to $x=h$,
for example $x=h-1$ , $x=h+1$

b) For quadratic function of the form $f(x)=a(x-m)(x-n)$, the $x$-intercepts are $(m,0)$ and $(n,0)$ while you can obtain the vertex as $h=dfrac{m+n}{2}$, $k=f(h)$

a) The quadratic function given by $f(x)=a(x-h)^2+k
,$ has vertex at $(h,k)$, axis of symmetry at $x=h$. Hence, the given quadratic function, $f(x)=-3(x-2)^2+5
,$ has the following characteristics:

$$
begin{align*}
text{Vertex: }&
(2,5)
\text{Axis of Symmetry: }&
x=2
.end{align*}
$$

Since $a<0$ ($a=-3$), then the given quadratic function is a parabola that opens down with a maximum value of $k,$ equal to $5
.$ Hence,

$$
begin{align*}
text{Domain: }&
(-infty,infty)
\text{Range: }&
(-infty,-3]
.end{align*}
$$

b) In the form, $f(x)=a(x-h)^2+k
,$ the given function, $f(x)=2(x+4)(x-6)
,$ is equivalent to

$$
begin{align*}
f(x)&=2(x^2-2x-24)
\
f(x)&=2x^2-4x-48
\
f(x)&=(2x^2-4x)-48
\
f(x)&=2(x^2-2x)-48
\
f(x)&=2(x^2-2x+1)-48-2(1)
\
f(x)&=2(x-1)^2-48-2
\
f(x)&=2(x-1)^2-50
.end{align*}
$$

The quadratic function given by $f(x)=a(x-h)^2+k
,$ has vertex at $(h,k)$, axis of symmetry at $x=h$. Hence, the quadratic function above has the following characteristics:

$$
begin{align*}
text{Vertex: }&
(1,-50)
\text{Axis of Symmetry: }&
x=1
.end{align*}
$$

Since $a>0$ ($a=2$), then the given quadratic function is a parabola that opens up with a minimum value of $k,$ equal to $-50
.$ Hence,

$$
begin{align*}
text{Domain: }&
(-infty,infty)
\text{Range: }&
[-50,infty)
.end{align*}
$$

$$
begin{align*}
text{a) Vertex: }&
(2,5)
\text{Axis of Symmetry: }&
x=2
\text{Domain: }&
(-infty,infty)
\text{Range: }&
(-infty,-3]
.end{align*}
$$

$$
begin{align*}
text{b) Vertex: }&
(1,-50)
\text{Axis of Symmetry: }&
x=1
\text{Domain: }&
(-infty,infty)
\text{Range: }&
[-50,infty)
.end{align*}
$$

The standard form of a quadratic function is
$f(x)=ax^2+bx+c$

$$
bold{a);;} text{Remember the special product} (a-b)^2=a^2-2ab+b^2
$$

$$
begin{align*}
f(x) &=-3(x-2)^2+5\
&=-3(x^2-4x+4)+5\
&=-3x^2+12x-12+5\
f(x)&=-3x^2+12x-7end{align*}
$$

$$
bold{b);;;} text{Remember the distributive property} (a+b)(c+d)=a(c+d)+b(c+d)
$$

$$
begin{align*}
f(x) &= 2(x+4)(x-6)\
&=2[x(x-6)+4(x-6)]\
&=2(x^2-6x+4x-24)\
&=2(x^2-2x-24)\
f(x)&=2x^2-4x-48\
end{align*}
$$

a) $f(x)=-3x^2+12x-7$

b) $f(x)=2x^2-4x-48$

a) In the form, $f(x)=a(x-h)^2+k
,$ the given function, $f(x)=x^2-6x+2
,$ is equivalent to

$$
begin{align*}
f(x)&=(x^2-6x)+2
\
f(x)&=(x^2-6x+9)+2-9
\
f(x)&=(x-3)^2-7
.end{align*}
$$

Since $a>0$ ($a=1$), then the given quadratic function is a parabola that opens
up
with a minimum value of $k,$ equal to $-7
.$

Hence, the
minimum value is $-7$.

b) In the form, $f(x)=a(x-h)^2+k
,$ the given function, $f(x)=2(x-4)(x+6)
,$ is equivalent to

$$
begin{align*}
f(x)&=2(x^2+2x-24)
\
f(x)&=2x^2+4x-48
\
f(x)&=(2x^2+4x)-48
\
f(x)&=2(x^2+2x)-48
\
f(x)&=2(x^2+2x+1)-48-2(1)
\
f(x)&=2(x+1)^2-48-2
\
f(x)&=2(x+1)^2-50
.end{align*}
$$

Since $a>0$ ($a=2$), then the given quadratic function is a parabola that opens
up
with a minimum value of $k,$ equal to $-50
.$

Hence, the
minimum value is $-50$.

c) In the form, $f(x)=a(x-h)^2+k
,$ the given function, $f(x)=-2x^2+10x
,$ is equivalent to

$$
begin{align*}
f(x)&=-2x^2+10x
\
f(x)&=-2(x^2-5x)
\
f(x)&=-2left( x^2-5x+dfrac{25}{4} right)+2left( dfrac{25}{4} right)
\
f(x)&=-2left( x^2-{5}{2} right)^2+dfrac{25}{2}
.end{align*}
$$

Since $a<0$ ($a=-2$), then the given quadratic function is a parabola that opens
down
with a
maximum
value of $k,$ equal to $dfrac{25}{2}
.$

Hence, the
maximum value is $dfrac{25}{2}$.

d) In the form, $f(x)=a(x-h)^2+k
,$ the given function, $f(x)=3.2x^2+15x-7
,$ is equivalent to

$$
begin{align*}
f(x)&=(3.2x^2+15x)-7
\
f(x)&=3.2(x^2+4.6875x)-7
\
f(x)&=3.2(x^2+4.6875x+5.4931640625)-7-3.2(5.4931640625)
\
f(x)&=3.2(x+2.34375)^2-7-17.578125
\
f(x)&=3.2(x+2.34375)^2-24.578125
.end{align*}
$$

Since $a>0$ ($a=3.2$), then the given quadratic function is a parabola that opens
up
with a
minimum
value of $k,$ equal to $-24.578125
.$

Hence, the
minimum value is $-24.578125$.

a) minimum value is $-7$

b) minimum value is $-50$

c) maximum value is $dfrac{25}{2}$

d) minimum value is $-24.578125$

In the form, $f(x)=a(x-h)^2+k
,$ the given function, $P(x)=-4x^2+16x-7
,$ is equivalent to

$$
begin{align*}
P(x)&=(-4x^2+16x)-7
\
P(x)&=-4(x^2-4x)-7
\
P(x)&=-4(x^2-4x+4)-7+4(4)
\
P(x)&=-4(x-2)^2-7+16
\
P(x)&=-4(x-2)^2+9
.end{align*}
$$

With $P(x)$ and $x$ given as a thousand units, and since the maximum is given at $k$ (equal to $9
)$ which occurs at $h$ (equal to $2
),$ then

$$
begin{align*}
text{Maximum Profit: }&
$9000
\text{Number of Items to Maximize Profit: }&
2000text{ items}
.end{align*}
$$

$$
begin{align*}
text{Maximum Profit: }&
$9000
\text{Number of Items to Maximize Profit: }&
2000text{ items}
end{align*}
$$

In the form, $f(x)=a(x-h)^2+k
,$ the given function, $C(x)=0.3x^2-1.2x+2
,$ is equivalent to

$$
begin{align*}
C(x)&=(0.3x^2-1.2x)+2
\
C(x)&=0.3(x^2-4x)+2
\
C(x)&=0.3(x^2-4x+4)+2-0.3(4)
\
C(x)&=0.3(x-2)^2+2-1.2
\
C(x)&=0.3(x-2)^2+0.8
.end{align*}
$$

With $C(x)$ and $x$ given as a thousand units, and since the minimum is given at $k$ (equal to $0.8
)$ which occurs at $h$ (equal to $2
),$ then

$$
begin{align*}
text{Minimum Cost: }&
$800
\text{Number of Items to Minimize Cost: }&
2000text{ items}
.end{align*}
$$

$$
begin{align*}
text{Minimum Cost: }&
$800
\text{Number of Items to Minimize Cost: }&
2000text{ items}
end{align*}
$$

Let $x$ be the first number. Then the second number is $(6-x).$ The product, $y,$ of these numbers is given by

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

In the form, $f(x)=a(x-h)^2+k
,$ the equation above is equivalent to

$$
begin{align*}
y&=-(x^2-6x)
\
y&=-(x^2-6x+9)+9
\
y&=-(x-3)^2+9
.end{align*}
$$

The equation above has a maximum value of $k$ (equal to $9$). Hence, the largest possible product between the two numbers is $9
.$

$$
9
$$

a) Using $f(x)=y,$ the given function, $f(x)=x^2-4x+3
,$ is equivalent to

$$
begin{align*}
y&=x^2-4x+3
.end{align*}
$$

Interchanging the $x$ and $y$ variables and then solving for $y$ result to

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

Hence, the inverse is
$f^{-1}=2pmsqrt{x+1}$.

b) In the form $f(x)=a(x-h)^2+k,$ the given function, $f(x)=x^2-4x+3
,$ is equivalent to

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

Since $a>0$ ($a=1$), then the graph is a parabola that opens up with a minimum value of $k$ (equal to $-1$). Hence, $f(x)$ has the following characteristics:

$$
begin{align*}
text{Domain: }&
(-infty,infty)
\text{Range: }&
[-1,infty)
.end{align*}
$$

With $f^{-1}(x)=f^{-1}=2pmsqrt{x+1}$ then $f^{-1}(x)$ has the following characteristics:

$$
begin{align*}
text{Domain: }&
[-1,infty)
\text{Range: }&
(-infty,infty)
.end{align*}
$$

c) Using a table of values, the graph of $f(x)=x^2-4x+3
,$ (red graph) is shown below.

Reflecting the graph of $f(x)$ about the line $y=x$ (black graph), the graph of $f^{-1}(x)$ (blue graph) is shown below.

$$
begin{align*}
text{ a) } f^{-1}&=2pmsqrt{x+1}
\\
text{b) Domain of $f(x)$: }&
(-infty,infty)
\text{Range of $f(x)$: }&
[-1,infty)
\text{Domain of $f^{-1}(x)$: }&
[-1,infty)
\text{Range of $f^{-1}(x)$: }&
(-infty,infty)
\\
text{c) see graph}
end{align*}
$$

Solving for $x$ in the given function, $R(x)=-2.8(x-10)^2+15
,$ results to

$$
begin{align*}
R(x)-15&=-2.8(x-10)^2
\
dfrac{R(x)-15}{-2.8}&=(x-10)^2
\\
pmsqrt{dfrac{R(x)-15}{-2.8}}&=x-10
\\
10pmsqrt{dfrac{R(x)-15}{-2.8}}&=x
.end{align*}
$$

Since $x$ (number of items sold) is a positive number, then $x=10+sqrt{dfrac{R(x)-15}{-2.8}}
.$

$$
x=10+sqrt{dfrac{R(x)-15}{-2.8}}
$$

Almost all linear functions, when reflected about the line $y=x,$ will still result to a line that will satisfy the Vertical Line Test. Quadratic functions, however, whose graph is a parabola that opens up or down, when reflected about the line $y=x$ will produce a parabola that opens to the left or right. Such resulting graph will not satisfy the Vertical Line Test.

a) We can sketch the graph by substituting some values of $x$ as follows.

a) The domain is the set of all possible values of $x$. Remember that the expression under the radical sign cannot have negative sign. Therefore, the domain is the solution of

$x+3geq 0implies xgeq 3$

domain: ${ x in bold{R};|;xgeq 3}$

The range is the set of all possible values of $f(x)$.

As you can observe from the graph,

range: ${ f(x)inbold{R};|;f(x)leq 0}$

b) To find $f^{-1}$, replace $y$ with $x$ and $x$ with $y$, then solve for $y$

$y=-sqrt{x+3}$

$x=-sqrt{y+3}$

$x^2=y+3$

$y=x^2-3$

The domain of $f^{-1}(x)$ must be the range of $f(x)$, therefore

$f^{-1}(x)=x^2-3$ where $xleq 0$

a) domain: ${ x in bold{R};|;xgeq 3}$

range: ${ f(x)inbold{R};|;f(x)leq 0}$

b) $f^{-1}(x)=x^2-3$ where $xleq 0$

In the following, factor the expression under the radical such that one factor is a perfect square, then apply the rule of radical as

$$
sqrt{a^2cdot b}=asqrt{b}
$$

a) $sqrt{48}=sqrt{16cdot 3}=sqrt{4^2cdot 3}=4sqrt{3}$

b) $sqrt{68}=sqrt{4cdot 17}=sqrt{2^2cdot 17}=2sqrt{17}$

c) $sqrt{180}=sqrt{36cdot 5}=sqrt{6^2cdot 5}=6sqrt{5}$

d) $-3sqrt{75}=-3sqrt{25cdot 3}=-3sqrt{5^2cdot 3}=-3(5)sqrt{3}=-15sqrt{3}$

e) $5sqrt{98}=5cdot sqrt{49cdot 2}=5cdot sqrt{7^2cdot 2}=5cdot 7sqrt{2}=35sqrt{2}$

f) $-8sqrt{12}=-8sqrt{4cdot 3}=-8sqrt{2^2cdot 3}=-8cdot 2sqrt{3}=-16sqrt{3}$

a) $4sqrt{3}$

b) $2sqrt{17}$

c) $6sqrt{5}$

d) $-15sqrt{3}$

e) $35sqrt{2}$

f) $-16sqrt{3}$

a) Using the properties of radicals, the given expression, $sqrt{7}timessqrt{14}
,$ is equivalent to

$$
begin{align*}
&
sqrt{7(14)}
\&=
sqrt{7(7cdot2)}
\&=
sqrt{(7)^2cdot2}
\&=
7sqrt{2}
.end{align*}
$$

b) Using the properties of radicals, the given expression, $3sqrt{5}times2sqrt{15}
,$ is equivalent to

$$
begin{align*}
&
3(2)sqrt{5(15)}
\&=
6sqrt{5(5cdot3)}
\&=
6sqrt{(5)^2cdot3}
\&=
6(5)sqrt{3}
\&=
30sqrt{3}
.end{align*}
$$

c) Extracting the roots of the factors that are perfect powers of the index, the given expression, $sqrt{12}+2sqrt{48}-5sqrt{27}
,$ is equivalent to

$$
begin{align*}
&
sqrt{4cdot3}+2sqrt{16cdot3}-5sqrt{9cdot3}
\&=
sqrt{(2)^2cdot3}+2sqrt{(4)^2cdot3}-5sqrt{(3)^2cdot3}
\&=
2sqrt{3}+2(4)sqrt{3}-5(3)sqrt{3}
\&=
2sqrt{3}+8sqrt{3}-15sqrt{3}
.end{align*}
$$

By combining like radicals, the expression above is equivalent to

$$
begin{align*}
&
2sqrt{3}+8sqrt{3}-15sqrt{3}
\&=
(2+8-15)sqrt{3}
\&=
-5sqrt{3}
.end{align*}
$$

d) Extracting the roots of the factors that are perfect powers of the index, the given expression, $3sqrt{28}-2sqrt{50}+sqrt{63}-3sqrt{18}
,$ is equivalent to

$$
begin{align*}
&
3sqrt{4cdot7}-2sqrt{25cdot2}+sqrt{9cdot7}-3sqrt{9cdot2}
\&=
3sqrt{(2)^2cdot7}-2sqrt{(5)^2cdot2}+sqrt{(3)^2cdot7}-3sqrt{(3)^2cdot2}
\&=
3(2)sqrt{7}-2(5)sqrt{2}+3sqrt{7}-3(3)sqrt{2}
\&=
6sqrt{7}-10sqrt{2}+3sqrt{7}-9sqrt{2}
.end{align*}
$$

By combining like radicals, the expression above is equivalent to

$$
begin{align*}
&
(6sqrt{7}+3sqrt{7})+(-10sqrt{2}-9sqrt{2})
\&=
(6+3)sqrt{7}+(-10-9)sqrt{2}
\&=
9sqrt{7}-19sqrt{2}
.end{align*}
$$

e) Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $(4-sqrt{3})(5+2sqrt{3})
,$ is equivalent to

$$
begin{align*}
&
4(5)+4(2sqrt{3})-sqrt{3}(5)-sqrt{3}(2sqrt{3})
.end{align*}
$$

Using the properties of radicals, the expression above is equivalent to

$$
begin{align*}
&
20+8sqrt{3}-5sqrt{3}-2(sqrt{3})^2
\&=
20+8sqrt{3}-5sqrt{3}-2(3)
\&=
20+8sqrt{3}-5sqrt{3}-6
.end{align*}
$$

By combining like terms, the expression above is equivalent to

$$
begin{align*}
&
(20-6)+(8sqrt{3}-5sqrt{3})
\&=
14+3sqrt{3}
.end{align*}
$$

f) Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression, $(3sqrt{5}+2sqrt{10})(-2sqrt{5}+5sqrt{10})
,$ is equivalent to

$$
begin{align*}
&
3sqrt{5}(-2sqrt{5})+3sqrt{5}(5sqrt{10})+2sqrt{10}(-2sqrt{5})+2sqrt{10}(5sqrt{10})
.end{align*}
$$

Using the properties of radicals, the expression above is equivalent to

$$
begin{align*}
&
3(-2)(sqrt{5})^2+3(5)sqrt{5(10)}+2(-2)sqrt{10(5)}+2(5)(sqrt{10})^2
\&=
-6(5)+15sqrt{5(5cdot2)}-4sqrt{5cdot2(5)}+10(10)
\&=
-30+15sqrt{(5)^2cdot2}-4sqrt{(5)^2cdot2}+100
\&=
-30+15(5)sqrt{2}-4(5)sqrt{2}+100
\&=
-30+75sqrt{2}-20sqrt{2}+100
.end{align*}
$$

By combining like terms, the expression above is equivalent to

$$
begin{align*}
&
(-30+100)+(75sqrt{2}-20sqrt{2})
\&=
70+55sqrt{2}
.end{align*}
$$

a) $7sqrt{2}$

b) $30sqrt{3}$

c) $-5sqrt{3}$

d) $9sqrt{7}-19sqrt{2}$

e) $14+3sqrt{3}$

f) $70+55sqrt{2}$