begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression\ by a factor of $1/k$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

The graph of $y=2f(-2x)+3$ has the following transformations from $f(x)$

(1) Reflecting the function in the $y$-axis

(2) Horizontal compression by a factor of $dfrac{1}{2}$

(3) Vertical stretching by a factor of 2

(4) Vertical translation 3 units upward

Therefor,e, the one that is NOT applied is reflection in $x$-axis (option b).

$y=2x+30$

$y-30=2x$

$$
x=frac{y-30}{2}
$$

Solve for $x$ in the original equation

$$
y=dfrac{x-30}{2}
$$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression\ by a factor of $1/k$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

Let’s examine the transformations applied to the function $f(x)=-|x+2|+3$

The range of $y=|x|$ is the set of non-negative numbers, $ygeq 0$.

A horizontal translation 2 units to the right $y=|x-2|$ will not change the range.

Reflecting the function in the $x$-axis $y=-|x-2|$ will make the range $yleq 0$.

A vertical translation 3 units up $y=-|x-2|+3$ will make the range $yleq 3$

Therefore, the answer is option (a)

We can also confirm this by graphing.

We shall evaluate the pair of functions.

i) $h(x)=(x+6)(x+3)(x-6)$

$=(x+3)(x+6)(x-6)$

$=(x+3)(x^2-36)$

$b(x)=(x+3)(x^2-36)$

$therefore$ equivalent

ii) Remember that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$

$b(t)=(3t+2)^3$

$(3t)^3+3(3t)^2(2)+3(3t)(2)^2+2^3$

$b(t)=27t^3+3(9)(2)t^2+3(3)(4)t+8$

$b(t)=27t^3+54t^2+36t+8$

$c(t)=27t^3+54t^2+36t+8$

$therefore$ equivalent

iii) Remember that $(-a)^3=-a^3$

$h(t)=(4-x)^3=[-(x-4)]^3=-(x-4)^3$

$h(t)’=(x-4)^3$

$therefore$ NOT equivalent

iv) Since we already established that $(iii)$ is NOT equivalent and option (d) says $iii$ and $iv$, then we are now sure that $(iv)$ will NOT be equivalent.

Therefore, the answer is $i$ and $ii$ which is option (b)

$bold{Concept:}$ An algebraic expression has restrictions on values that would make the denominator equal to zero.

$bold{Solution:}$ Since the restrictions $yneq -1,0,dfrac{1}{2}$

The denominators (or terms after “$div$” ) should not contain

$(y+1)$ , $y$ and $(2y-1)$ which would make denominators equal to zero.

The option containing all these expressions is option $(c)$

Factor the quadratic expressions

$dfrac{x^2-5x+6}{x^2-1}times dfrac{x^2-4x-5}{x^2-4}$

$=dfrac{(x-3)(x-2)}{(x+1)(x-1)}times dfrac{(x-5)(x+1)}{(x+2)(x-2)}$

$=dfrac{(x-3)cancel{(x-2)}}{cancel{(x+1)}(x-1)}times dfrac{(x-5)cancel{(x+1)}}{(x+2)cancel{(x-2)}}$

$=dfrac{(x-3)}{(x-1)}times dfrac{(x-5)}{(x+2)}$

$$
=dfrac{(x-3)(x-5)}{(x-1)(x+2)}
$$

Remember that

$dfrac{a}{b}+dfrac{c}{d}=dfrac{ad+bc}{bd}$

$dfrac{5x-4}{x+1}+dfrac{3x}{x-4}$

$=dfrac{(5x-6)(x-4)+(x+1)(3x)}{(x+1)(x-4)}$

$=dfrac{5x(x-4)-6(x-4)+3x(x+1)}{(x+1)(x-4)}$

$=dfrac{5x^2-20x-6x+24+3x^2+3x}{(x+1)(x-4)}$

$$
=dfrac{8x^2-23x+24}{(x+1)(x-4)}
$$

$bold{Concept:}$ For a quadratic equation in standard form $f(x)=ax^2+bx+c$, the vertex is $(h,k)$ where

$h=-dfrac{b}{2a}$

$$
k=f(h)
$$

$bold{Solution:}$ In this case, $a=3$, $b=-6$, $c=15$

$h=-dfrac{b}{2a}=-dfrac{-6}{2(3)}=1$

$k=f(1)=3(1)^2-6(1)+15=3-6+15=12$

Therefore, the vertex is $(1,12)$ and the answer is option (a)

The quadratic function in factored form is $y=a(x-r)(x-s)$ where the $x$-intercepts are

$(r,0)$ and $(s,0)$.

Thus, the feature that is most readily available are the $x$-intercepts which is option (b).

$bold{Concept:}$ The maximum height of a parabola opening down with vertex$(h,k)$ is k.

From the standard form $y=ax^2+bx+c$, the vertex can be obtained as

$h=-dfrac{b}{2a}$

$$
k=f(h)
$$

$bold{Solution:}$ Here, we are given with the standard form

$h(t)=0.8+29.4t-4.9t^2$

$a=-4.9$ , $b=29.4$ , $c=0.8$

Calculate the coordinates of the vertex

$h=-dfrac{b}{2a}=-dfrac{29.4}{2(-4.9)}=3$

$k=f(3)=0.8+29.4(3)-4.9(3^2)=44.9$ m

Therefore, the maximum height is $44.9$ m

The answer is option (c)

$bold{Concept:}$ Profit function $P(x)$ is the difference between the revenue function $R(x)$ and the cost function $C(x)$.

The revenue function is the product of the price per item $P(x)$ and the number of items sold $x$.

$bold{Solution:}$ Let $x$ be the number of occupied seats (no. of passengers)

cost function: $C(x)=225+30x$

price function: $P(x)=60+5(22-x)$

revenue function: $R(x)=xcdot P(x)=x[60+5(22-x)]$

profit function: $P(x)=R(x)-C(x)=x[60+5(22-x)]-(225+30x)$

Simplify the profit function.

$P(x)=60x+5x(22-x)-225-30x$

$P(x)=60x+110x-5x^2-225-30x$

$P(x)=-5x^2+80x-1155$

The value of $x$ that maximizes the $P(x)$ is at the vertex $(h,k)$ where

$h=-dfrac{b}{2a}$

$a=-5$, $b=80$ , $c=335$

$h=-dfrac{80}{2(-5)}=8$

Thus, the bus should run with $8$ occupied seats.

a) This is in standard form so we can easily calculate $b^2-4ac$

$b^2-4ac=(-6)^2-4(-1)(-9)=0implies$ 1 zero

b) This is in vertex form, $y=a(x-h)^2+k$, if $a$ and $k$ has the same sign, there will be no solutions. If it has different sign, then there are two solutions. If $k=0implies$ 1 solution

In this case, $a=4$ and $k=0.5$ which are both positive, thus, no zeros.

(c) This is vertex form with $k=0$. Thus the vertex is at the $x$-axis and only 1 zero.

(d) This is in vertex form with $a=-2$ and $k=4.1$. Since they have different sign, this would result in 2 zeros.

Therefore, the answer is option (d)

$bold{Concept:}$ If a quadratic function touches the $x$-axis at one point, the vertex is at $x$-axis and $b^2-4ac=0$.

$bold{Solution:}$ $f(x)=x^2-kx+k+8$

$a=1$, $b=-k$ , $c=k+8$

$b^2-4ac=(-k)^2-4(1)(k+8)=0$

$k^2-4k-32=0$

$(k-8)(k+4)=0$

$k=8$ or $k=-4$

Thus, the answer is option (b)

$bold{Concept:}$ To find the inverse, solve for $x$ in terms of $y$ then swap the variables. The domain of $f(x)$ will be the range of $f^{-1}(x)$ while the range of $f(x)$ will be the domain of $f^{-1}(x)$

$bold{Solution:}$

$y=2(x-3)^2+5$

$y-5=2(x-3)^2$

$(x-3)^2=dfrac{y-5}{2}$

$x-3=pmsqrt{dfrac{y-5}{2}}$

$x=3pmsqrt{dfrac{y-5}{2}}$

Swap the variables

$y=3pmsqrt{dfrac{x-5}{2}}$

The range of $f^{-1}$ must be the domain of $f(x)$ which is $xgeq 3$. Thus the range of $f^{-1}$ must be $ygeq 3$ so we shall choose only one branch.

$f^{-1}=3+sqrt{dfrac{x-5}{2}}$

Now, the domain of $f^{-1}$ is the range of $f(x)$. The function $f(x)$ opens up with vertex$(3,5)$ so its range is $ygeq 5$. Therefore, the domain of $f^{-1}$ is $xgeq 5$

The answer is

$y=3+sqrt{dfrac{x-5}{2}}$ , $xgeq 5$

which is option (a)

$bold{Concept:}$ All functions are relations but not all relations are functions. A function must not have more than 1 value of $y$ for a certain value of $x$. Graphically, it should touch any vertical line not more than once.

$bold{Solution:}$ Notice that options (a) , (b) and (c) all contains $y^2$ which implies for some value of $x$, $y$ can take either positive or negative value. This is not a characteristic of a function. Thus, the answer is option (c) which is a quadratic function.

$-1^2-5(-1)+3$

$=1+5+3$

$$
=9
$$

Let $x=-1$ and substitute back into the original equation

a) This is false because it is supposed to be vertical line test that can be used to distinguish between relation and function.

b) This is true. Input values are usually denoted as $x$.

c) $y=3x+5$ is a linear function, a type of function.

d) For functions containing ordered pairs $(x,y)$, each value of $x$ should have a unique value of $y$, which is true for the given set. Thus, this is true. Note that it is possible that many values of $x$ can have the same value of $y$ but it is not possible the any value of $x$ can have two or more values of $y$

Therefore, the one that is NOT true is option (a)

The denominator cannot have a value of zero. In this case, $f(x)=dfrac{3}{x}$, $x$ can take any real number except zero.

This is represented by option $(d)$

$y=5x-7$

$y+7=5x$

$$
x=frac{y+7}{5}
$$

$$
y=frac{x+7}{5}
$$

The inverse of a function $y=g(x)$ can be obtained by solving for $x$ in terms of $y$ then swap the variables.

First convert the standard form to vertex form by completing the square

$y=x^2-5x-6$

$y=left(x^2-5x+dfrac{25}{4}right)-6-dfrac{25}{4}$

$y=left(x-dfrac{5}{2}right)^2-dfrac{49}{4}$

Now, solve for $x$ in terms of $y$

$left(x-dfrac{5}{2}right)^2=y+dfrac{49}{4}$

$x-dfrac{5}{2}=pmsqrt{y+dfrac{49}{4}}$

$x=dfrac{5}{2}pmsqrt{y+dfrac{49}{4}}$

Swap the variables
$y=dfrac{5}{2}pmsqrt{x+dfrac{49}{4}}$

There must be typographical error in the errors but the closes one is option (c)

a) This is true. The domain and $f$ is the range of $f^{-1}$ and the range of $f$ is the domain of $f^{-1}$

b) This is true. The inverse $f^{-1}(x)$ is a reflection of $f(x)$ in the line $y=x$

c) This is true as mentioned in part(a)

d) This is FALSE. The inverse can be found by either interchanging $x$ and $y$ and solve for $y$ or solve for $x$ first then interchange $x$ and $y$.

The answer is option (d)

The domain of the inverse function $f^{-1}(x)$ is the range of the $f(x)$

In this case, $f(x)=3(x+2)^2-5$ which is a parabola that opens up with vertex at $(-2,-5)$. Thus, the range is $f(x)$ is ${ yinbold{R};|;ygeq -5}$

Thus, the domain of the $f^{-1}(x)$ must be $xgeq -5$

The answer is option (a)

We are given with $y=f(x)=sqrt{x-1}$

To find the inverse, swap the variables and solve for $y$

$x=sqrt{y-1}$

$x^2=y-1$

$y=x^2+1$

Now, the domain of $f^{-1}(x)$ is the range of $f(x)$

The range of $f(x)$ is ${ yinbold{R};|;ygeq 0}$

Therefore, the inverse function must be

$f^{-1}(x)=x^2+1$ , $xgeq 0$

There must be a typo error in the choices but the closest one is option (c)

We can confirm further our answer by graphing. $f^{-1}(x)$ should be a reflection of $f(x)$ along the line $y=x$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression\ by a factor of $1/k$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

By inspection

$y=af(x-p)+q$

Given that $a<0$, $p<0$, $q<0$, we can rewrite this as

$y=-|a|f(x+|p|)-|q|$

From this we can see the following transformations

(1) reflection in the $x$-axis

(2) vertical stretch by a factor $|a|$

(3) horizontal translation $|p|$ units to the left

(4) vertical translation $|q|$ units down

The answer is option (c)

The answer at the back of your book is option (b) which is a typographical error.

$$
h=dfrac{12}{2(-2)}=dfrac{12}{4}=-3
$$

Vertex form of a quadratic is:

$y = a(x -h)^2 + k$ where $(h,k)$ is the vertex. We can find h by determining the axis of symmetry using:

$$
h=dfrac{-b}{2a}
$$

From the solution above we know now that we have the following information for the vertex form:

$y=-2(x-(-3))^2+k$

$y=-2(x+3)^2+k$

The only option that has a +3 is C

$bold{Concept:}$ The quadratic function in factored form $f(x)=a(x-r)(x-s)$ has a vertex $(h,k)$ where

$h=dfrac{r+s}{2}$

$$
k=f(h)
$$

$bold{Solution:}$ In this case, $r=-2$ and $s=3$

We can obtain the vertex as

$h=dfrac{-2+3}{2}=dfrac{1}{2}$

$k=f(0.5)=(0.5+2)(0.5-3)=-dfrac{25}{4}$

Therefore, the vertex is $left(dfrac{1}{2},-dfrac{25}{4}right)$

The answer is option (d)

$bold{Concept:}$ The break even point of a profit function $P(x)$ are the values of $x$ where $P(x)=0$

$bold{Solution:}$ We shall solve the quadratic function by factoring

$P(x)=-4x^2+28x-40=0$

$=-4(x^2-7x+10)=0$

$=-4(x-5)(x-2)$

$x=5$ or 2

Since $x$ is in units of thousands, the number of items for break-even is

$2000$ or $5000$

The answer is option (a)

a) This is true. The standard form is $y=ax^2+bx+c$ and the $y$-intercept is $(0,c)$

b) This is NOT true. The vertex form shows the maximum profit. It is the factored form that shows the break-even points.

c) This is true. The factored form is $y=a(x-r)(x-s)$ and the $x$-intercepts are $(r,0)$ and $(s,0)$

d) This is true. The vertex form is $y=a(x-h)^2+k$ and the vertex is $(h,k)$

Therefore, the answer is option (b)

The discriminant $D$ is the parameter that determines the number of zeros for quadratic functions.

$D=b^2-4ac$

In this case, $a=7$ , $b=12$, $c=6$

$D=12^2-4(7)(6)=-24<0$

Since $D=-24<0$, there will be no roots $(n=0)$.

Thus, the answer is option (d)

Remember that $dfrac{a}{b}+dfrac{c}{d}=dfrac{ad+bc}{bd}$

$dfrac{7}{ab}-dfrac{2}{b}+dfrac{1}{3a^2}$

$=dfrac{7(b)(3a^2)-2(ab)(3a^2)+(ab)(b)}{(ab)(b)(3a^2)}$

Factor out $(ab)$ in the numerator

$=dfrac{(ab)(7)(3a)-(ab)(2)(3a^2)+(ab)b}{(ab)(b)(3a^2)}$

$=dfrac{(ab)(21a+6a^2+b)}{(ab)(b)(3a^2)}$

$=dfrac{cancel{(ab)}(21a+6a^2+b)}{cancel{(ab)}(b)(3a^2)}$

$=dfrac{21a-6a^2+b}{3a^2b}$

Denominators can’t be zero, so restriction is $a,bneq 0$

The answer is option (b)

$$
dfrac{x^2-4}{x+3}timesdfrac{x^2-9}{2x+4}
$$

$$
dfrac{(x+2)(x-2)}{x+3}timesdfrac{(x+3)(x-3)}{2(x+2)}
$$

$dfrac{x-2}{1}timesdfrac{x-3}{2}$

$$
dfrac{(x-2)(x-3)}{2}
$$

a) This is a quadratic function so we know that its domain is ${ xinbold{R}}$. We also learned that for a quadratic function $y=a(x-h)^2+k$ with vertex $(h,k)$, the range is

${ yinbold{R};|; ygeq k}$ if $a>0$

${ yinbold{R};|;yleq k}$ if $a0$ (opens upward), the range is

range = ${ yinbold{R};|;ygeq 2}$

With the known vertex, we can write the quadratic function in vertex form

$y=a(x-h)^2+k$

$y=3(x-4)^2+2$

From the vertex form, we can obtain the transformations from the base function $y=x^2$

(1) vertical stretching by factor 3 $implies y=3x^2$

(2) translation 4 units to the right and 2 units up $implies y=3(x-4)^2+2$

b) $g(x)=5-2sqrt{3x+6}$

This is the lower half of a parabola that opens to the right. The domain is restricted to values of $x$ that will make the expression inside the radical sign greater than or equal to zero.

$3x+6geq 0implies 3xgeq -6 implies xgeq -2$

domain = ${ xinbold{R};|;xgeq -2}$

Since this function is decreasing, the maximum value is when the radicand is zero $(x=-2)$ which corresponds to $g(x)=5$

range = ${ yinbold{R};|;yleq 5}$

The transformations from the base function $y=sqrt{x}$ can be observed by rewriting as

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

(1) horizontal compression by factor $dfrac{1}{3}implies y=sqrt{3x}$

(2) reflecting in $x$-axis $implies y=-sqrt{3x}$

(3) vertical stretching by factor 2 $implies y=-2sqrt{3x}$

(4) translation 2 units to the left and 5 units up $implies y=-2sqrt{3(x+2)}+5$

c) $h(x)=dfrac{1}{frac{1}{3}(x-6)}-2$

This is a hyperbola with asymptotes at $x=6$ and $y=-2$

domain = ${ xinbold{R};|;xneq 6}$

range = ${ yinbold{R};|;yneq 2}$

It has undergone the following transformation from the base function $y=dfrac{1}{x}$

(1) horizontal stretching by a factor of $3implies y=dfrac{1}{frac{1}{3}x}$

(2) translation 6 units to the right and 2 units down $y=dfrac{1}{frac{1}{3}(x-6)}-2$

begin{table}[]
defarraystretch{1.4}%
begin{tabular}{l|l|l|l|l|}
cline{2-5}
& domain & range & parent function & transformations \ cline{2-5}
textbf{a)} & ${ xinbold{R}}$ & ${ yinbold{R};|;ygeq 2}$ & $y=x^2$ & begin{tabular}[c]{@{}l@{}}(1) vertical stretch by factor 3\ (2) translation 4 units right and 2 units upend{tabular} \ cline{2-5}
textbf{b)} & ${xinbold{R};|;xgeq -2}$ & ${ yinbold{R};|;yleq 5}$ & $y=sqrt{x}$ & begin{tabular}[c]{@{}l@{}}(1) horizontal compression by factor $frac{1}{3}$\ (2) reflection in the $x$-axis\ (3) vertical stretch by factor $2$\ (4) translation 2 units left and 5 units upend{tabular} \ cline{2-5}
textbf{c)} & ${ xinbold{R};|;xneq 6}$ & ${yinbold{R}:|;yneq -2}$ & $y=dfrac{1}{x}$ & begin{tabular}[c]{@{}l@{}}(1) horizontal stretch by factor 3\ (2) translation 6 units right and 2 units downend{tabular} \ cline{2-5}
end{tabular}
end{table}

Sacha walks 1.4 km/h faster than Jill.
Let $v$ be the speed of Jill, then $v+1.4$ is the speed of Sacha.
Let $t$ be the time spent by Jill. Since Sacha stopped for 20 min $left(text{equal to } frac{1}{3};text{hour}right)$ and finished the walk 2 hour ahead of Jill, her time spent on walking is $t-left(2+frac{1}{3}right)=t-frac{7}{3}$.

We also know that

$$
begin{equation*}text{distance} = text{speed} times text{time}end{equation*}
$$

Both of them must cover the same distance of 30 km.

distance covered by Jill: $vt=30 implies t=dfrac{30}{v}$

distance covered by Sacha: $(v+1.4)left(t-frac{7}{3}right)=30$

$(v+1.4)left(dfrac{30}{v}-dfrac{7}{3}right)=30$

By solving graphically, we get $v=3.6$ km/h

Jill’s speed: 3.6 km/h

Jill’s time: $dfrac{30}{3.6}=8frac{1}{3}$ hr or 8 hours and 20 minutes

Sacha’s speed: 3.6+1.4= 5 km/h

Sacha’s time: $8frac{1}{3}-frac{7}{3}=6$ h or (6 hours and 20 minutes including the time she talked with friend)

Let $x$ be the number of times the price is increased by $$50$. There are 2 students who will not join every time that price is increased by 50.

Students: $25-2x$

Selling Price = $550+50x$

Income = Students $times$ Price = $(25-2x)(550+50x)$

Cost: $5500+240x$

Profit = Income $-$ Cost = $(25-2x)(550+50x)-(5500+240x)$

a) The break even is the value of $x$ when Profit $=0$, that is, the zeros of the quadratic function. By graphical approach, we get $x=8.64$ which means $25-2(8.64)=7.72approx 8$ students

b) From the graph, the maximum occurs close to zero, hence, he could maximize the profit if he does not change the price and maintain at $$550$

a) 8 students

b) $$550$