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

$$
(x+3)^2-5=0
$$

The graph intersects the $Ox$ axis in two points, therefore the equation has two solutions.

$$
(x+3)^2-5=4
$$

We draw the line $y=4$:

There are two intersections between the graph of the function and the line $y=4$, therefore the equation has two solutions.

$(-6,4)$

$$
(0,4)
$$

$(-6+3)^2-5stackrel{?}{=}4$

$(-3)^2-5stackrel{?}{=}4$

$4=4checkmark$

$(0+3)^2-5stackrel{?}{=}4$

$(3)^2-5stackrel{?}{=}4$

$4=4checkmark$

a) 2 solutions;

b) 2 solutions

c) ${-6,0}$

$$
(x+3)^2-5=11
$$

$(x+3)^2-5-11=0$

$(x+3)^2-16=0$

$(x+3-4)(x+3+4)=0$

$(x-1)(x+7)=0$

$x-1Rightarrow x_1=1$

$x+7Rightarrow x_2=-7$

$x_1=-7$

$$
x_2=1
$$

$(x+3)^2-5+5=11+5$

$(x+3)^2=16$

$x+3=pm 4$

$x+3=-4Rightarrow x_1=-4-3=-7$

$x+3=4Rightarrow x_2=4-3=1$

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

$(x+3)^2-5-3=0$

$(x+3)^2-8=0$

$(x+3-2sqrt 2)(x+3+2sqrt 2)=0$

$x+3-2sqrt 2Rightarrow x_1=-3+2sqrt 2$

$x+3+2sqrt 2Rightarrow x_2=-3-2sqrt 2$

$x_1approx -5.828$

$$
x_2approx-0.172
$$

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

$(x+3)^2=8$

$x+3=pm 2sqrt 2$

$x+3=-2sqrt 2Rightarrow x_1=-3-2sqrt 2$

$x+3=2sqrt 2Rightarrow x_2=-3+2sqrt 2$

i) ${-7,1}$

ii) ${-3-2sqrt 2,-3+2sqrt 2}$

$$
dfrac{x-5}{4}+dfrac{2}{5}=dfrac{9}{10}
$$

$dfrac{x-5}{4}cdot 20+dfrac{2}{5}cdot 20=dfrac{9}{10}cdot 20$

$5(x-5)+4(2)=2(9)$

$5x-25+8=18$

$5x-17=18$

$5x=18+17$

$5x=35$

$x=dfrac{35}{5}$

$$
x=7
$$

a) We multiply all terms by 20
$textcolor{#4257b2}{text{to write a new equivalent equation that is easier to solve}}$:

$5(x-5)+8-8=18-8$

$$
5(x-5)=10
$$

Here he used the strategy of
$textcolor{#4257b2}{text{doing the opposite of an operation}}$

$5(x-2)=10Rightarrow x-5=2Rightarrow x$ must be 7

Here Ernie $textcolor{#4257b2}{text{reasoned about the value of the}}$

$textcolor{#4257b2}{text{expression inside the parantheses}}$.

Given,

$$
begin{align*}
x^{2}+2.5x-1.5&=0 \
end{align*}
$$

$a.$ Walter is talking about changing the given equation by multiplying it by a specific number which will result in an equation without decimals.

So, we will multiply the given equation by $2$.

$$
begin{align*}
x^{2}+2.5x-1.5&=0 \
2(x^{2}+2.5x-1.5&=0) \
2x^{2}+5x-3&=0 tag{1}\
end{align*}
$$

$b.$ Now in order to solve the rewritten equation we will factor $(1)$,

$$
begin{align*}
2x^{2}+5x-3&=0 \
2x^{2}+6x-x-3&=0 \
2x(x+3)-1(x+3)&=0 \
(2x-1)(x+3)&=0\
x&=0.5,-3\
end{align*}
$$

Then, we will plot the graph for $(1)$.

Thus, we check the obtained solutions,

By putting $x$ as $(-3)$ in $(1)$,

$$
begin{align*}
2x^{2}+5x-3&=0\
2(-3)^{2}+5(-3)-3&=0 \
2(9)-15-3&=0\
18-18&=0\
0&=0\
end{align*}
$$

By putting $x$ as negative $(0.5)$ in $(1)$,

$$
begin{align*}
2x^{2}+5x-3&=0\
2(0.5)^{2}+5(0.5)-3&=0\
2(0.25)+2.5-3&=0\
0.5-2.5-3&=0\
3-3&=0\
0&=0\
end{align*}
$$

Therefore the obtained solution is correct.

$$
4|8x-2|=8
$$

$8x-2=pm 2$

$8x-2=-2Rightarrow 8x=0Rightarrow x_1=0$

$8x-2=2Rightarrow 8x=4Rightarrow x_2=0.5$

We solve the equation by $textcolor{#4257b2}{text{rewriting}}$:

$x_1=0$

$4|8(0)-2|stackrel{?}{=}8$

$4|-2|stackrel{?}{=}8$

$8=8checkmark$

$x_2=0.5$

$4|8(0.5)-2|stackrel{?}{=}8$

$4|2|stackrel{?}{=}8$

$8=8checkmark$

$$
3sqrt{4x-8}+9=15
$$

$3sqrt{4x-8}=15-9$

$3sqrt{4x-8}=6$

$sqrt{4x-8}=2$

$(sqrt{4x-8})^2=2^2$

$4x-8=4$

$4x=12$

$$
x=3
$$

We use the $textcolor{#4257b2}{text{undoing}}$:

$3sqrt{4(3)-8}+9stackrel{?}{=}15$

$3sqrt{4}+9stackrel{?}{=}15$

$6+9stackrel{?}{=}15$

$$
15=15checkmark
$$

$$
(2y-3)(y-2)=-12y+18
$$

$(2y-3)(y-2)+12y-18=0$

$(2y-3)(y-2)+6(2y-3)=0$

$(2y-3)(y-2+6)=0$

$(2y-3)(y+4)=0$

$2y-3=0Rightarrow y_1=dfrac{3}{2}$

$y+4=0Rightarrow y_2=-4$

We use $textcolor{#4257b2}{text{rewriting}}$:

$y_1=1.5$

$(2(1.5)-3)(1.5-2)stackrel{?}{=}-12(1.5)+18$

$0(-0.5)stackrel{?}{=}-18+18$

$0=0checkmark$

$y_2=-4$

$(2(-4)-3)(-4-2)stackrel{?}{=}-12(-4)+18$

$-11(-6)stackrel{?}{=}48+18$

$$
66=66checkmark
$$

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

$3xcdot dfrac{5}{x}+3xcdot dfrac{1}{3x}=3xcdot dfrac{4x}{3}$

$15+1=4x^2$

$16=4x^2$

$x^2=4$

$x=pm 2$

$x_1=-2$

$$
x_2=2
$$

We use $textcolor{#4257b2}{text{rewriting}}$:

$x_1=-2$

$dfrac{5}{-2}+dfrac{1}{3(-2)}stackrel{?}{=}dfrac{4(-2)}{3}$

$-dfrac{5}{2}-dfrac{1}{6}stackrel{?}{=}-dfrac{8}{3}$

$-dfrac{15}{6}-dfrac{1}{6}stackrel{?}{=}-dfrac{8}{3}$

$-dfrac{16}{6}stackrel{?}{=}-dfrac{8}{3}$

$-dfrac{8}{3}=-dfrac{8}{3}checkmark$

$x_2=2$

$dfrac{5}{2}+dfrac{1}{3(2)}stackrel{?}{=}dfrac{4(2)}{3}$

$dfrac{5}{2}+dfrac{1}{6}stackrel{?}{=}dfrac{8}{3}$

$dfrac{15}{6}+dfrac{1}{6}stackrel{?}{=}dfrac{8}{3}$

$dfrac{16}{6}stackrel{?}{=}dfrac{8}{3}$

$dfrac{8}{3}=dfrac{8}{3}checkmark$

$$
|3-7x|=-6
$$

$$
dfrac{6w-1}{5}-3w=dfrac{12w-16}{15}
$$

$15cdotdfrac{6w-1}{5}-15cdot 3w=15cdotdfrac{12w-16}{15}$

$3(6w-1)-45w=12w-16$

$18w-3-45w=12w-16$

$-27w-3=12w-16$

$-3=12w-16+27w$

$-3+16=39w$

$13=39w$

$w=dfrac{13}{39}$

$$
w=dfrac{1}{3}
$$

We use $textcolor{#4257b2}{text{rewriting}}$:

$dfrac{6cdotdfrac{1}{3}-1}{5}-3cdotdfrac{1}{3}stackrel{?}{=}dfrac{12cdotdfrac{1}{3}-16}{15}$

$dfrac{1}{5}-1stackrel{?}{=}dfrac{-12}{15}$

$$
-dfrac{4}{5}=-dfrac{4}{5}checkmark
$$

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

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

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

We solve the equation by $textcolor{#4257b2}{text{rewriting}}$:

$$
(x+2)^2+4(x+2)-5=0
$$

$y^2+4y-5=0$

$(y^2+4y+4)-4-5=$

$(y+2)^2-9=0$

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

$(y-1)(y+5)=0$

$y-1=0Rightarrow y_1=1$

$y+5=0Rightarrow y_2=-5$

We note:

$y=x+2$.

We solve the equation in $y$ by $textcolor{#4257b2}{text{rewriting}}$:

$y=x+2$

$y_1=1$

$x_1=y_1-2=1-2=-1$

$y_2=-5$

$$
x_2=y_2-2=-5-2=-7
$$

We determine $x$:

$x_1=-1$

$(-1+2)^2+4(-1+2)-5stackrel{?}{=}0$

$1+4-5stackrel{?}{=}0$

$0=0checkmark$

$x_2=-7$

$(-7+2)^2+4(-7+2)-5stackrel{?}{=}0$

$25-20-5stackrel{?}{=}0$

$0=0checkmark$

Given,

$$
begin{align*}
i) x^{2}+7x+12&=0\
ii) (dfrac{1}{2}x+7)^{2}+7(dfrac{1}{2}x+7)+12&=0\
end{align*}
$$

$a.$ Now we will solve the first given equation.

$$
begin{align*}
x^{2}+7x+12&=0\
x^{2}+3x+4x+12&=0 \
x(x+3)+4(x+3)&=0\
(x+4)(x+3)&=0\
x&=-4,-3\
end{align*}
$$

Therefore the solution of the first given equation is $(-4)$ and $(-3)$.

$b.$ The box in the diagram drawn by Graciela represents $dfrac{1}{2}x+7$ and $x$ in the diagram means that she has replaced boxes with $dfrac{1}{2}x+7$ and $x$.

$c.$ Graciela got this diagram by using substituting in the second given equation in the first given equation and thus, form $b.$ part we know that the boxes represent $dfrac{1}{2}x+7$. So, the diagram drawn by Graciela represents the solutions of the first given equation equated to $dfrac{1}{2}x+7$.

$d.$ Now we equate $dfrac{1}{2}x+7$ to $(-3)$ to find the first value of $x$.

$$
begin{align*}
dfrac{1}{2}x+7&=-3\
dfrac{1}{2}x&=-3-7\
x&=dfrac{-10times2}{1}\
x&=-20\
end{align*}
$$

Now we equate $dfrac{1}{2}x+7$ to $(-4)$ to find the second value of $x$.

$$
begin{align*}
dfrac{1}{2}x+7&=-4\
dfrac{1}{2}x&=-4-7\
x&=dfrac{-11times2}{1}\
x&=-22\
end{align*}
$$

Therefore the value of $x$ is $(-22)$ and $(-20)$.

Then, we check the values of $x$ by putting them in the second given equation.

Now, putting $x$ as $(-22)$.

$$
begin{align*}
(dfrac{1}{2}x+7)^{2}+7(dfrac{1}{2}x+7)+12&=0\
(dfrac{1}{2}(-22)+7)^{2}+7(dfrac{1}{2}(-22)+7)+12&=0\
(-11+7)^{2}+7(-11+7)+12&=0\
(-4)^{2}+7(-4)+12&=0\
16-28+12&=0\
28-28&=0\
0&=0\
end{align*}
$$

Now, putting $x$ as $(-20)$.

$$
begin{align*}
(dfrac{1}{2}x+7)^{2}+7(dfrac{1}{2}x+7)+12&=0\
(dfrac{1}{2}(-20)+7)^{2}+7(dfrac{1}{2}(-20)+7)+12&=0\
(-10+7)^{2}+7(-10+7)+12&=0\
(-3)^{2}+7(-3)+12&=0\
9-21+12&=0\
21-21&=0\
0&=0\
end{align*}
$$

There fore the obtained solutions are correct.

Given,

$$
begin{align*}
&a.(m^{2}+5m-24)^{2}-(m^{2}+5m-24)=6 \\
&b. x^{tfrac{2}{3}}-x^{tfrac{1}{3}}-56=0\\
&c. y^{6}+3y^{3}-18=0\\
&d. p-12sqrt{p}=35 \\
&e. 3^{sqrt{x+tfrac{1}{4}}}=9^({sqrt{x+tfrac{1}{4}}})^{2}\\
end{align*}
$$

$a.$ Now in the first given equation we will take $y$ as $m^{2}+5m-24$, as Graciela did in $3-6$, so our box indicates the value $m^{2}+5m-24$.

Now, we rewrite and solve the given equation,

$$
begin{align*}
y^{2}-y&=6\\
y^{2}-y-6&=0\\
end{align*}
$$

$b.$ Now in the first given equation we will take $y$ as $x^{tfrac{1}{3}}$, as Graciela did in $3-6$, so our box indicates the value $x^{tfrac{1}{3}}$.

Now, we rewrite and solve the given equation,

$$
begin{align*}
y^{2}-y-56&=0\\
end{align*}
$$

$c.$ Now in the second given equation we will take $x$ as $y^{2}$, as Graciela did in $3-6$, so our box indicates the value $y^{2}$.

Now, we rewrite and solve the given equation,

$$
begin{align*}
x^{2}+3x-18&=0\
x^{2}+6x-3x-18&=0\
x(x+6)-3(x-6)&=0\
x(x+6)-3(x+6)&=0\
(x-3)(x+6)&=0\
y&=3,-6\
end{align*}
$$

Now we equate the value of $x$ to $y^{2}$ to get the value of $y$,

So, when we take $x$ as $3$,

$$
begin{align*}
y^{2}&=3\
y&=sqrt{3}\
end{align*}
$$

Now we equate the value of $x$ to $y^{2}$ to get the value of $y$,

So, when we take $x$ as $-6$,

$$
begin{align*}
y^{2}&=-6\
y&=sqrt{-6}\
end{align*}
$$

Therefore the value of $y$ is $sqrt{3}$ and $sqrt{-6}$.

$d.$ Now in the first given equation we will take $y$ as $sqrt{p}$, as Graciela did in $3-6$, so our box indicates the value $sqrt{p}$.

Now, we rewrite and solve the given equation,

$$
begin{align*}
y^{2}-12y+35&=0\
end{align*}
$$

$e.$ Now in the first given equation we will take $y$ as $sqrt{x+{tfrac{1}{4}}}$, as Graciela did in $3-6$, so our box indicates the value $sqrt{x+{tfrac{1}{4}}}$.

Now, we rewrite and solve the given equation,

$$
begin{align*}
3^{y}&=(3^{2})^{y^{2}}\
end{align*}
$$

We are discussing two strategies to solve an equation along with their examples.

$textbf{Solving an equation graphically:}$ We can solve an equation graphically by putting the right side and the left side of the equation on the graph and then finding their intersection points. The intersection points will be called as the solution of the equation.

For example, let us have to solve an equation $3x+4=-5$. To solve this, graph the right side and the left side of the equation on the graph that means we have to graph $y=3x+4$ and $y=-5$ on the coordinate plane.

Now, let us look at the intersection point of both these lines. The intersection point is $(-3,-5)$ that means both these equations gives equal values of $y$ for $x=-3$. Thus, the solution of the equation $3x+4=-5$ is $x=-3$.

$textbf{Solving an equation algebraically:}$ In this method, first we try to separate the constant on one side and variables on the other side then simplify the equation using various algebraic identities and some other algebraic axioms.

For example, let us solve the same example algebraically as shown below.

$$
begin{align*}
3x+4&=-5tag{subtract $4$ from both sides}\
3x+4-4&=-5-4\
3x&=-9tag{divide both sides by $3$}\
dfrac{3x}{3}&=dfrac{-9}{3}\
x&=-3
end{align*}
$$

The solution is coming out to be the same by this method as well that is $x=-3$.

There is one another example, we have to solve $2(x-1)^2+5=7$.

$$
begin{align*}
2(x-1)^2+5&=7tag{subtract $5$ from both sides}\
2(x-1)^2+5-5&=7-5\
2(x-1)^2&=2tag{divide both sides by $2$}\
dfrac{2(x-1)^2}{2}&=dfrac{2}{2}\
(x-1)^2&=1tag{take square roots on both sides}\
x-1&=pm1tag{add $1$ on both sides}\
x&=pm1+1\
x&=2,0
end{align*}
$$

Below is graph of $textcolor{#4257b2}{y=(x-2)^2-3}$ and $textbf{y=1}$:

It can be observed, solution of $(x-2)^2-3=1$ have solution at $x=0$ and $x=4$.

Solving
$$
(x-2)^2-3=1
$$
algebraically:

$$
begin{align*}
(x-2)^2-3&=1\
(x-2)^2&=4\
x-2&=pm 2\
x&=2+2 & & big| &x&=-2+2\
x&=4& & big| &x&=0\
end{align*}
$$

$x=4$ and $x=0$

$$
a)
$$

$$
2(x-1)^2 + 7=39
$$

$$
2(x-1)^2=32
$$

$$
(x-1)^2=16
$$

$$
(x-1)^2 – 4^2=0
$$

$$
(x-1-4)(x-1+4)=0
$$

$(x-5)(x+3)=0$
$$
x_{1}=5, x_{2}=-3
$$

$$
c)
$$

$$
dfrac{x}{2} + dfrac{x}{3}=dfrac{5x+2}{6}
$$

Let’s multiply both sides of the equation with 6:

$$
3x + 2x=5x+2
$$

$$
5x=5x+2
$$

This means that equation can’t be solved.

$$
b)
$$

$$
7(sqrt{m+1} – 3)=21
$$

$$
sqrt{m+1}-3=3
$$

$$
sqrt{m+1}=3+3
$$

$$
sqrt{m+1}=6
$$

Let’s raise to 2nd power:

$$
m+1=36
$$

$$
m=35
$$

$$
d)
$$

$$
-7 + (dfrac{4x+2}{2})=8
$$

$$
dfrac{4x+2}{2}=15
$$

Let’s multiply both sides with $2$:

$$
4x+2=30
$$

$$
4x=28
$$

Finally:

$x=dfrac{28}{4}$
$$
x=7
$$

(a) $x=5,-3$ (b) $m=35$ (c) No solution (d) $x=7$

$f(x)=3x-9$          (Given)

$g(x)=-x^2$          (Given)

a-          $f(-2)=3(-2)-9=-6-9=-15$

b-          $g(-2)=-(-2)^2=-4$

c-          $f(x)=0 rightarrow 3x-9=0$

$3x=9$

$$
x=3
$$

d-          $g(m)=-m^2$

a-          $f(-2)=-15$

b-          $g(-2)=-4$

c-          $f(x)=0 rightarrow x=3$

d-          $g(m)=-m^2$

a-

$f(x)=x^2+6x+15$          (Write the equation)

$f(x)=x^2+2(3)x+3^2-3^2+15$          (Completing the squares)

$f(x)=(x+3)^2+6$          (Graphing Form)

The equation of the axis of symmetry:          $x=-3$

The vertex is:          $(-3, 6)$

b-

$y=x^2-4x+9$          (Write the equation)

$y=x^2-2(2)x+(-2)^2-(-2)^2+9$          (Completing the square)

$y=(x-2)^2+5$          (Graphing Form)

The equation of the axis of symmetry:          $x=2$

The vertex is:          $(2, 5)$

c-

$f(x)=x^2-8x$          (Write the equation)

$f(x)=x(x-8)$          (Factor out $x$)

x-intercepts are:          $x=0$ and $x=8$

x-intercepts average $=dfrac {x_{1}+z_{2}}{2}=dfrac {0+8}{2}=4$

The equation of the axis of symmetry:          $x=4$

The vertex is:          $(4, -16)$

d-

$y=x^2+7x-2$          (Write the equation)

$x{1, 2}=dfrac {-b pm sqrt {b^2-4ac}}{2a}$          (Quadratic Formula)

$x{1, 2}=dfrac {-7 pm sqrt {49+8}}{2}$

$x_{1}=dfrac {-7}{2}-dfrac {sqrt {57}}{2}$         $x_{2}=dfrac {-7}{2}+dfrac {sqrt {57}}{2}$

x-intercepts average:          $=dfrac {x_{1}+x_{2}}{2}$

$=left(dfrac {-7}{2}-cancel{dfrac {sqrt {57}}{2}}+dfrac {-7}{2}+cancel {dfrac {sqrt {57}}{2}} right)div 2$

$=dfrac {-7}{2}$

The equation of the symmetry line:          $x=dfrac {-7}{2}$

y-intercept of the vertex is:          $=dfrac {49}{4}-dfrac {49}{2}-2$

$=-dfrac {49}{4}-dfrac {8}{4}=dfrac {-57}{4}$

The vertex is:          $(-dfrac {7}{2},qquad -dfrac {57}{4})$

a-          Axis of symmetry:          $x=-3$          Vertex:          $(-3, 6)$

b-          Axis of symmetry:          $x=2$          Vertex:          $(2, 5)$

c-          Axis of symmetry:          $x=4$          Vertex:          $(4, -16)$

d-          Axis of symmetry:          $x=dfrac {-7}{2}$          Vertex:          $(-dfrac {7}{2},qquad -dfrac {57}{4})$

The parabola is even because it has $y$-axis of symmetry.

The $”x”$ point lies on the $y$-axis which is why the parabola has $y$-axis of symmetry.

(a) Downward sloping parabola with vertex at $(-4,2)$. Equation is a quadratic with negative leading coefficient; $y(x)=-a(x+4)^{2}+2$ . Since the x-intercepts are at $x=-3$ and $x=-5$ then $-a(-3+4)^{2}=-2$ and $-a(-5+4)^{2}=-2$ giving $a=2$ and finally $y(x)=-2(x+4)^{2}+2$

(b) Rational with vertical asymptote at $x=2$ and horizontal asymptote at $x=0$ since $limlimits_{x to pminfty}y(x)=0$. Equation is then $y(x)=dfrac{1}{x-2}$

(c) Cubic with negative leading coefficient since $limlimits_{x to infty}y(x)=-infty$ and $limlimits_{x to -infty}y(x)=infty$. We have an x-intercept at $1<x<2$ and y-intercept at $y=3$ so equation is polynomial (3rd order) $y(x)=-x^{3}+3$

Given,

Parent function

$$
begin{align*}
y&=f(-x)\
end{align*}
$$

Equations obtained in $3-15$,

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

$a.$ Given equation $f(x) = -2(x+4)^{2} + 2$ and parent function $y = f(-x)$.

Now we reflect $f(x)$ across the $y$-axis to get $f(-x)$

$b.$ Given equation $f(x) = dfrac{1}{x-2}$ and parent function $y = f(-x)$.

Now we reflect $f(x)$ across the $y$-axis to get $f(-x)$

$c.$ Given equation $f(x) = -x^{3}+3$ and parent function $y = f(-x)$.

Now we reflect $f(x)$ across the $y$-axis to get $f(-x)$.

Given,

$$
begin{align*}
&a.dfrac{3(x^{2}-dfrac{4}{5})}{8}+dfrac{(x^{2}-dfrac{4}{5})}{8}=2 \
&b.(dfrac{1}{n+1})^{2}+6(dfrac{1}{n+1})=27 \
&c.6(dfrac{x+1}{6})^{2}+7(dfrac{x+2}{6})-3=0\
end{align*}
$$

$a.$ Now in the first given equation we will take $y$ as $x^{2}-dfrac{4}{5}$, as Graciela did in $3-6$, so our box indicates the value $x^{2}-dfrac{4}{5}$.

Now, we rewrite and solve the given equation,

$$
begin{align*}
dfrac{3y}{8}+dfrac{y}{4}&=2\
dfrac{3y+2y}{8}&=2
3y+2y&=2times8\
5y&=16\
y&=dfrac{16}{5}\
end{align*}
$$

Now we equate the value of $y$ to $x^{2}-dfrac{4}{5}$ to get the value of $x$,

$$
begin{align*}
x^{2}-dfrac{4}{5}&=dfrac{16}{5}\\
x^{2}&=dfrac{16}{5}+dfrac{4}{5}\\
x^{2}&=dfrac{16+4}{5}\\
x^{2}&=dfrac{20}{5}\\
x^{2}&=4\\
x&=pm2\\
end{align*}
$$

Therefore the value of $x$ is $pm2$

$b.$ Now in the second given equation we will take $y$ as $dfrac{1}{n+1}$, as Graciela did in $3-6$, so our box indicates the value $dfrac{1}{n+1}$.

Now, we rewrite and solve the given equation,

$$
begin{align*}
y^{2}+6y&=27\
y^{2}+6y-27&=0\
y^{2}+9y-3y-27&=0\
y(y+9)-3(y-9)&=0\
y(y+9)-3(y+9)&=0\
(y-3)(y+9)&=0\
y&=3,-9\
end{align*}
$$

Now we equate the value of $y$ to $dfrac{1}{n+1}$ to get the value of $n$,

So, when we take $y$ as $3$,

$$
begin{align*}
dfrac{1}{n+1}&=3\
1&=3(n+1)\
1&=3n+3\
1-3&=3n\
-2&=3n\
n&=-dfrac{2}{3}\
end{align*}
$$

So, when we take $y$ as $9$,

$$
begin{align*}
dfrac{1}{n+1}&=9 \
1&=9(n+1)\
1&=9n+9\
1-9&=9n\
-8&=9n\
n&=-dfrac{8}{9}\
end{align*}
$$

Therefore the value of $n$ is $-dfrac{2}{3}$ and $-dfrac{8}{9}$.

$c.$ Now in the third given equation we will take $y$ as $dfrac{x+2}{6}$, as Graciela did in $3-6$, so our box indicates the value $dfrac{x+2}{6}$.

Now, we rewrite and solve the given equation,

$$
begin{align*}
6y^{2}+7y-3&=\
6y^{2}+9y-2y-3&=0\
3y(2y+3)-1(2y+3)&=0\
(3y-1)(2y+3)&=0\
y&=dfrac{1}{3},-dfrac{3}{2}\
end{align*}
$$

Now we equate the value of $y$ to $dfrac{x+2}{6}$ to get the value of $n$,

So, when we take $y$ as $dfrac{1}{3}$,

$$
begin{align*}
dfrac{x+2}{6}&=dfrac{1}{3}\
3(x+2)&=6(1)\
3x+6&=6\
3x&=6-6\
3x&=0\
x&=0\
end{align*}
$$

So, when we take $y$ as $-dfrac{3}{2}$,

$$
begin{align*}
dfrac{x+2}{6}&=-dfrac{3}{2}\
2(x+2)&=6(-3)\
2x+4&=-18\
2x&=-22\
2x&=dfrac{-22}{2}\
x&=-11\
end{align*}
$$

Therefore the value of $x$ is $0$ and $-11$.

Given,

$$
begin{align*}
&a.(y-7)^{2}=25-(x-3)^{2}\
&b. x^{2}+y^{2}+10y=-9\
&c. x^{2}+y^{2}+18x-8y+47=0 \
&d. y^{2}+(x-3)^{2}=1\
end{align*}
$$

$a.$ We first rewrite the equation in standard form,

$$
begin{align*}
(y-7)^{2}&=25-(x-3)^{2}\
(x-3)^{2}+(y-7)^{2}&=25\
end{align*}
$$

Now, we find the radius.

$$
begin{align*}
(x-3)^{2}+(y-7)^{2}&=25\
C(3,7)\
r&=sqrt{25}\
r&=5\
end{align*}
$$

Therefore the radius of the circle is $5$ and the center is $(3,7)$

$b.$ We first rewrite the equation in standard form,

$$
begin{align*}
x^{2}+y^{2}+10y&=-9\
x^{2}+(y^{2}+10y+25)&=-9+25\
x^{2}+(y+5)^{2}&=16\
end{align*}
$$

Now, we find the radius.

$$
begin{align*}
x^{2}+(y+5)^{2}&=16\
C(0,-5)\
r&=sqrt{16}\
r&=4\
end{align*}
$$

Therefore the radius of the circle is $4$ and the center is $(0,-5)$

$c.$ We first rewrite the equation in standard form,

$$
begin{align*}
x^{2}+y^{2}+18x-8y+47&=0\
(x^{2}++18x+81)+(y^{2}-8y+16)&=-47+81+16\
(x+9)^{2}+(y-4)^{2}&=50\
end{align*}
$$

Now, we find the radius.

$$
begin{align*}
(x+9)^{2}+(y-4)^{2}&=50\
C(-9,4)\
r&=sqrt{50}\
r&=5sqrt{2}\
end{align*}
$$

Therefore the radius of the circle is $5sqrt{2}$ and the center is $(-9,4)$

$d.$ We are given the equation in standard form,

Now, we find the radius and the centre.

$$
begin{align*}
y^{2}+(x-3)^{2}&=1\
C(0,3)\
r&=sqrt{1}\
r&=1\
end{align*}
$$

Therefore the radius of the circle is $1$ and the center is $(0,3)$

(a) Arranging the system as
$$
begin{bmatrix}2x+6y=10\x+3y=8end{bmatrix}
$$
then subtracting $2times Eq2$ from $Eq1$ we obtain an inconsistency $0=-6$ so the system has no valid solutions

(b) and (c); The two lines $y=-dfrac{x}{3}+dfrac{10}{6}$ and $y=-dfrac{x}{3}+dfrac{8}{3}$ are shown not intersecting. They are parallel with equal slopes so can never intersect in the domain $xepsilonmathbb{R}$.

Given,

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

In order to describe and sketch the graph of the given function, we will first sketch the graph of the given function.

Now, sketch the graph in which $f_{0}(x) = x^{2}$. we shift $f_{0}(x)$ two units to the left to get $f_{1}(x) = (x+2)^{2}$, then we shift $f_{1}(x)$ four units up to get $f(x)$.

The function has an odd degree, therefore its ends go in opposite directions. because the leading coefficient is positive, the left and goes down, while the right end goes up.

The function has one $x$-intercept and an $y$-intercept, this means the function has one real zero and two imaginary zeroes.

The domain and range of the function are the set of all real numbers.

The functions are increasing in all its domain.

There is no asymptote.

Given,

$y=3^{x}$

$a.$ Down $4$ units.

$b.$ Right $7$ units.

$a.$ To shift the functions down by $4$ unit

$$
begin{align*}
y&=3^{x}\
y&=3^{x}-4\
end{align*}
$$

Therefore when we shift the function down by $4$ units we get $y=3^{x}-4$.

$b.$ To shift the functions right by $7$ unit

$$
begin{align*}
y&=3^{x}\
y&=3^{x-7}\
end{align*}
$$

Therefore when we shift the function right by $7$ units we get $y=3^{x-7}$.

$a) y=3^{x}-4$

$b) y=3^{x-7}$

(b)
Standard Deviation, s: $1.2613124477738$

Mean, x̄: $3.0909090909091$

Variance: $1.5909090909091$

$$
x=sqrt{2x+3}
$$

$x^2=(sqrt{2x+3})^2$

$x^2=2x+3$

$x^2-2x-3=0$

$(x^2-2x+1)-4=0$

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

$(x-1-2)(x-1+2)=0$

$(x-3)(x+1)=0$

$x-3=0Rightarrow x_1=3$

$x+1=0Rightarrow x_2=-1$

b) The equations $y=sqrt{2x+3}$ and $y=x$ have an infinity many solutions.

$x_2=-1$

$sqrt{2(-1)+3}stackrel{?}{=}-1$

$$
1not=-1
$$

d) This happens because one of the two equations found in $(a)$ is an extraneous solution:

$$
x=3
$$

$$
x=sqrt{2x+3}
$$

$x_1=3$

$x_2=-1$

$x_1=3$

$sqrt{2(3)+3}stackrel{?}{=}3$

$sqrt 9stackrel{?}{=}3$

$3=3checkmark$

$x_2=-1$

$sqrt{2(-1)+3}stackrel{?}{=}-1$

$$
1not=-1
$$

$$
x=3
$$

This means that $x_2$ is an extraneous solution.The only solution of the equation is:

Considering the extraneous solution $x=-1$, when we equate the expressions $y=sqrt{2x+3}=x=-1$ we have set the expression equal to a value outwith the range of the function $y=f(x)$ since $sqrt{x}geq0$ but here $y=sqrt{2x+3}<0$ so is invalid and the graph of $y(x)$ does not intersect at $x=-1$ since $y(x)geq0$ for the domain $-1.5<x<infty$

$$
2x^2+5x-3=x^2+4x+3
$$

$2x^2+5x-3-x^2-4x-3=0$

$x^2+x-6=0$

$x^2+3x-2x-6=0$

$x(x+3)-2(x+3)=0$

$$
(x+3)(x-2)=0
$$

$x+3=0Rightarrow x_1=-3$

$$
x-2=0Rightarrow x_2=2
$$

b) Gustav gets the equation $y=x^2+x-6$ by subtracting the right side from the left side.Solving the original equation is equivalent to determining the zeros of the function $y=x^2+x-6$.

c) The equation $y=x^2+x-6$ has an infinity many solutions (all the points of the parabola).

d) The solutions of the original equation are the zeros of the function $y=x^2+x-6$.

$$
begin{cases}
y=2x^2+5x-3\
y=x^2+4x+3
end{cases}
$$

$x_1=-3$

$$
x_2=2
$$

The solutions of the original equation are the $x$-coordinates of the intersections of the two parabolas:

$$
20x+1=3^x
$$

$x_1=0$

$$
x_2=4
$$

b)The solutions of the equation $20x+1-3^x$ are the $x$-coordinates of the intersection points.

$$
begin{cases}
y=20x\
y=3^x-1
end{cases}
$$

(a) Jack is solving the equation of the intersection $f(x)=g(x)=-(x-3)^{2}+4=dfrac{12}{x}$

(b) Points are read off as $A=(3,4)$ and $B=(4,3)$. Taking firstly A then $f(x)=g(x)=-(3-3)^{2}+4=4=dfrac{12}{3}$. For B; $f(x)=g(x)=-(4-3)^{2}+4=3=dfrac{12}{4}$

(c) Setting the equation of the intersection to zero and solving; $-(x^{2}-6x+9)+4-dfrac{12}{x}=0$

$$
-x^{2}+6x-5-dfrac{12}{x}=0
$$

$$
-x^{3}+6x^{2}-5x-12=0
$$

factors as $-(x+1)(x-3)(x-4)=0$ so $x=-1$ is another intersection point.

(a) $f(x)=g(x)=-(x-3)^{2}+4=dfrac{12}{x} ; (b) (3,4), (4,3)$ (c) $x=-1$

There are at least 2 methods. Method 1; Define $f(x)=(x-3)^{2}-2$ and $g(x)=x+1$ then plotting $f$ and $g$ and observing the points of intersection below as (1,2) and (6,7)

Method 2; Set the equation of intersection to zero and evaluate the x-intercepts; Let $h(x)=(x-3)^{2}-2-x-1=0$ then $h(1)=0$ and if as in the previous method $f(x)=(x-3)^{2}-2$ and if $g(x)=x+1$ then $f(1)=g(1)=2$ and $f(6)=g(6)=7$

$(1,2)$ and $(6,7)$

Note we can firstly add 3 to each side so that; $2mid x-4mid=dfrac{2x}{3}$. Now graphing $f(x)=2mid x-4mid$ and $g(x)=dfrac{2x}{3}$ we read off the points of intersection below as $(3,2)$ and $(6,4)$

$(3,2)$ and $(6,4)$

a-

The graph below is the graph of the functions:

$f(x)=-3sqrt {2x-5}+7$ and $g(x)=-8$

From the graph, the solution of the system $-3sqrt {2x-5}+7=-8$ is the point $(15, -8)$

Checking the solution:

$-3sqrt {2(15)-5}+7=-3sqrt {25}+7=-3 cdot 5+7=-8$          $checkmark$

b-

The graph below is the graph of the functions:

$f(x)=2|3x+4|-10$ and $g(x)=12$

From the graph, the solution of the system $2|3x+4|-10=12$
is the points $(-5, 12)$ and $(dfrac {7}{3}, 12)$

Checking the solution:

For $x=-5$
$2|3(-5)+4|-10=2|-15+4|-10=2(11)-10=12$          $checkmark$

For $x=dfrac {7}{3}$
$2|3 cdot dfrac {7}{3}+4|-10=2|11|-10=22-10=12$          $checkmark$

a-          The solution of the system $-3sqrt {2x-5}+7=-8$ is the point $(15, -8)$

b-          The solution of the system $2|3x+4|-10=12$
is the points $(-5, 12)$ and $(dfrac {7}{3}, 12)$

Given,

$$
begin{align*}
&g(x)=x^{2}-5\\
&a. gleft(dfrac{1}{2}right) \\
&b. g(h+1)\\
end{align*}
$$

In order to determine $a$ and $b$, we will put $x$ as $left(dfrac{1}{2}right)$ and $(h+1)$ in $g(x)$.

$a.$ Now we put $x$ as $left(dfrac{1}{2}right)$ in g$(x)$,

$$
begin{align*}
gleft(dfrac{1}{2}right)&=left(dfrac{1}{2}right)^{2}-5 \\
&=dfrac{1}{4}-5\\
&=dfrac{1-20}{4}\\
&=dfrac{-19}{4}\\
end{align*}
$$

Therefore when we put $x$ as $left(dfrac{1}{2}right)$ in g$(x)$ we get $left(dfrac{-19}{4}right)$.

$b.$ Now we put $x$ as $(h+1)$ in g$(x)$,

$$
begin{align*}
text{g}(h+1)&=(h+1)^{2}-5 \
text{g}(h+1)&=h^{2}+2h+1-5 \
text{g}(h+1)&=h^{2}+2h-4\
end{align*}
$$

Therefore when we put $x$ as $(h^{2}+1)$ in g$(x)$ we get $(h^{2}+2h-4)$.

$a.dfrac{-19}{4}$

$b. h^{2}+2h-4$

$$
Q(x)=begin{bmatrix}3for xtextless60\lceildfrac{x}{60}-1for xgeq 60end{bmatrix}
$$

Use the round up function to express the cost function Q$.$

$$
Q(118)=Q(119)=Q(120)=$4
$$

$$
Q(121)=$5
$$

$Q(x)$ is not continuous since it has jump discontinuities at x=60n where n is a positive integer. If the hourly charge is changed to 5, the jumps in the y direction increase by a factor of 5

a-

$(3 sqrt {2})^2$          (Write the equation)

$3^2 (sqrt {2})^2$          (Power of product property)

$$
9 cdot 2=18
$$

b-

$sqrt {dfrac {9}{4}}$          (Write the equation)

$dfrac {sqrt {9}}{sqrt {4}}$          (Power of quotient property)

$$
=dfrac {3}{2}
$$

c-

$sqrt {dfrac {1}{3}}$          (Write the equation)

$=dfrac {sqrt {1}}{sqrt {3}}$          (Power of quotient property)

$=dfrac {sqrt {1}}{sqrt {3}} times dfrac {sqrt {3}}{sqrt {3}}$          (Multiply by the form of 1)

$$
=dfrac {sqrt {3}}{3}
$$

d-

$(3+sqrt 2)^2$          (Write the equation)

$=3^2+2 cdot 3 cdot sqrt 2+(sqrt {2})^2$

$=9+6 cdot sqrt {2}+2$

$=11+6 cdot sqrt {2}$

a-          $(3 sqrt {2})^2=18$

b-          $sqrt {dfrac {9}{4}}=dfrac {3}{2}$

c-          $sqrt {dfrac {1}{3}}=dfrac {sqrt {3}}{3}$

d-          $(3+sqrt 2)^2=11+6 cdot sqrt {2}$

The graph of $y=2^x+5$ is the graph of $y=2^x$ translated up by 5 units and the graph of $y=2^x-5$ is the graph of $y=2^x$ translated down by 5 units

Graphing $f(x)=3x-1$ and $g(x)=2^{x}$ we read off the points of intersection below as $(1,2)$ and $(3,8)$

Using logarithms; $x-log_{2}(3x-1)=0$. This cannot be solved algebraically because we have a function of x in the logarithmic argument that cannot be combined with x to obtain a solution

$(1,2)$ and $(3,8)$

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

$g(x)=2x^2-6x-3$

$f(x)=g(x)$

$dfrac{1}{2}(x-2)^3+1=2x^2-6x-3$

a) The coordinates of the points $A,B$ are given by solving the equation:

$(x-2)^3+2=4x^2-12x-6$

$x^3-6x^2+12x-8+2-4x^2+12x+6=0$

$x^3-10x^2+24x=0$

$x(x^2-10x+24)=0$

$x[(x^2-10x+25)-1]=0$

$x[(x-5)^2-1]=0$

$x(x-5-1)(x-5+1)=0$

$x(x-6)(x-4)=0$

$x_1=0$

$x-6=0Rightarrow x_2=6$

$x-4=0Rightarrow x_3=4$

$x_1=0$

$dfrac{1}{2}(0-2)^3+1stackrel{?}{=}2(0)^2-6(0)-3$

$-4+1stackrel{?}{=}-3$

$-3=-3checkmark$

$x_2=6$

$dfrac{1}{2}(6-2)^3+1stackrel{?}{=}2(6)^2-6(6)-3$

$32+1stackrel{?}{=}36-3$

$33=33checkmark$

$x_3=4$

$dfrac{1}{2}(4-2)^3+1stackrel{?}{=}2(4)^2-6(4)-3$

$4+1stackrel{?}{=}72-36-3$

$33=33checkmark$

b) The solution $x=6$ doesn’t appear in the graph as the window is too small.

$f(x)=0$

$dfrac{1}{2}(x-2)^3+1=0$

$(x-2)^3+2=0$

$x-2=-sqrt[3] 2$

$x=2-sqrt[3] 2$

$$
xapprox 0.74
$$

c) In order to determine the $x$-coordinate of point $C$ we have to solve the equation:

$dfrac{1}{2}(0.74-2)^3+1stackrel{?}{=}0$

$dfrac{1}{2}cdot(-2)+1stackrel{?}{=}0$

$$
0=0checkmark
$$

a-

$sqrt {2x-1}-x=-8$          (Given)

$2x-1=(x-8)^2$          (Square of each side)

$2x-1=x^2-16x+64$

$x^2-18x+65=0$

$(x-13)(x-5)$

$x=13$ or $x=5$

Checking for $x=13$

$sqrt {2(13)-1}-13=sqrt {25}-13=5-13=-8$          $checkmark$

Checking for $x=5$

$sqrt {2(5)-1}-13=sqrt {9}-13=3-13=-10 neq -8$          ($x=5$ is an extraneous solution)

b-

$sqrt {2x-1}-x=0$          (Given)

$2x-1=x^2$          (Square of each side)

$x^2-2x+1=0$

$(x-1)(x-1)$

$x=1$

Checking for $x=1$

$sqrt {2(1)-1}-1=sqrt {1}-1=-1=0$          $checkmark$

a-

$sqrt {2x-1}-x=-8 qquad rightarrow qquad x=13$ or $x=5$

$sqrt {2(5)-1}-13=sqrt {9}-13=3-13=-10 neq -8$          ($x=5$ is an extraneous solution)

b-          $sqrt {2x-1}-x=0 qquad qquad x=1$

Checking for $x=1$

$sqrt {2(1)-1}-1=sqrt {1}-1=-1=0$          $checkmark$

Graph of the given equation can be plotted by finding all the points on the $x$-axis and $y$-axis. The points on the $x$-axis can be found by putting $y=0$ in the given equation similarly points on $y$-axis can be found by putting $x=0$ in equation. Points on $x$-axis are $(6.447,0) & (-4.477,0)$ and points on $y$-axis are $(0,5.385) & (0,-5.385)$

$$
textbf{Equation : }(x-1)^2+(y-0)^2=30
$$

$$
textbf{Centre of circle is }(1,0)
$$

$$
textbf{Graph of the given equation is shown below : }
$$

$$
text{Centre of circle is }(1,0)
$$

Given,

$$
begin{align*}
p(x)&=x^{2}+5x-6 \
end{align*}
$$

$a.$ In order to find where does the graph of $y= p(x)$ intersect $y$-axis we will compute,

$$
begin{align*}
y&=p(x)\
y&=p(0) tag{1} \
end{align*}
$$

So, form $(1)$ we get,

$$
begin{align*}
p(0)&=x^{2}+5x-6\
p(0)&=(0)^{2}+5(0)-6\
p(0)&=-6\
end{align*}
$$

Therefore the graph of $y= p(x)$ intersect $y$-axis at $(0,-6)$.

$b.$ In order to find where does the graph of $y= p(x)$ intersect $x$-axis we will compute,

$$
begin{align*}
y&=p(x)\
0&=p(x) tag{1}\
end{align*}
$$

So, form $(1)$ we get,

$$
begin{align*}
x^{2}+5x-6&=0 \
x^{2}+6x-x-6&=0\
x(x+6)-1(x-6)&=0\
x(x+6)-1(x+6)&=0\
(x-1)(x+6)&=0\
x&=1,-6\
end{align*}
$$

Therefore the graph of $y= p(x)$ intersect $x$-axis at $(-6,0)$ and $(1,0)$.

$c.$ If q$(x) = x^{2}+5x$, then the intercepts of the graph of $y = q(x)$,

$x$-intercept, when $x$ is $0$

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

Thus $x$-intercept of q$(x)$ is $0$.

$y$-intercept, when $y$ is $0$

$$
begin{align*}
q(x)&=y\
q(x)&=0\
end{align*}
$$

So,

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

Thus $y$-intercept of q$(x)$ is $0$ and $-5$.

$d.$ Thus, we sketch the graph,

The graph of q$(x)$ is the graph of p$(x)$ shifted six units up.

$e.$ Now, the value of p$(x)$-q$(x)$,

$$
begin{align*}
p(x)-q(x)&=x^{2}+5x-6-(x^{2}+5x)\
p(x)-q(x)&=x^{2}+5x-6-x^{2}-5x\
p(x)-q(x)&=-6\
end{align*}
$$

Therefore the value of p$(x)$-q$(x)$ is $-6$.

(a) $y=mid xmid$ and (b) $x=mid ymid$ are plotted in the graphs below;

(c) Both are linear and discontinuous and the slopes of all line segments are equal magnitude $pm1$ and lines meet at the origin. Only (a) is a function since (b) does not satisfy the vertical line test. The graph of (a) is restricted to the 1st and 2nd quadrants whilst (b) is restricted to the 1st and 4th.

(d) domain of (a) $xepsilonmathbb{R}$ and range $0leq y<infty$

domain of (b) $0leq x<infty$ and range $yepsilonmathbb{R}$

Given,

$$
begin{align*}
3y-4x&=-1 \
9y+2x&=4\
end{align*}
$$

$a.$ In order to solve this system of equations algebraically we will choose the elimination method,

Now, we will multiply the second given equation by two in order to get opposite coefficients of $x$.

$$
begin{align*}
(9y+2x&=4)times2\
18y+4x&=8tag{1} \
end{align*}
$$

Now we add the first given equation and $(1)$ to get the value of $y$,

$$
begin{align*}
&3y – 4x = -1 \
&18y + 4x = 8 \
&rule{75pt}{0.5pt} \
& 21y = 7\
& y = dfrac{7}{21}\
& y = dfrac{1}{3}\
end{align*}
$$

Put $y=left(dfrac{1}{3}right)$ in $(1)$ , we get

$$
begin{align*}
18left(dfrac{1}{3}right) + 4x &= 8 \\
6+4x &= 8 \\
4x & = 8-6\\
x&= dfrac{2}{4}\\
x&= dfrac{1}{2}\\
end{align*}
$$

$b.$ Thus in order to find where do the graphs intersect, we will sketch the graph first,

Therefore the intersecting points are $left(dfrac{1}{2},dfrac{1}{3}right)$.

$c.$ The relation between the two parts is that the intersecting points of the two given equations are their solutions.

(a) Calculating $Eq1-Eq2rightarrow0=6$ so we have an inconsistent system and therefore the graph will not have any points of intersection.

(b) $2times(Eq1-Eq2)rightarrow0=x^{2}-4x+4=(x-2)^{2}$ so we have the equation of a parabola with two real solutions and therefore the graph will have two points of intersection.

(c) $Eq1-Eq2rightarrow y^{2}-y=2rightarrow y^{2}-y-2=(y-2)(y+1)=0$ so we have the equation of a sideways parabola with two real solutions. Since one of our solutions is $<0$ there are only two points of intersection if we consider $y=pmsqrt{x}$. Otherwise there is only one intersection.

(d) rearranging system;
$$
begin{bmatrix}2y=4x-10\y=2x-5end{bmatrix}
$$
then noting $Eq1-2Eq2rightarrow0=0$ so we have a linearly dependent system and therefore the graph will have infinite points of intersection.

$$
begin{cases}
x^2+y^2=25\
y=x^2-13
end{cases}
$$

a) When we substitute the expression of $y$ in the first equation we get a quartic equation, therefore we can have a maximum of 4 solutions.

$x_1=-4$

$x_2=-3$

$x_3=3$

$$
x_4=4
$$

$x^2+(x^2-13)^2=25$

$x^2+x^4-26x^2+169-25=0$

$$
x^4-25x^2+144=0
$$

c) We combine the two equations to get an equation in $x$ only:

$y+13+y^2=25$

$y^2+y+13-25=0$

$$
y^2+y-12=0
$$

We combine the two equations to get an equation in $y$ only:

The easiest equation to solve is the one in $y$ because its degree is smaller.

$y^2-3y+4y-12=0$

$y(y-3)+4(y-3)=0$

$(y-3)(y+4)=0$

$y-3=0Rightarrow y_1=3$

$y+4=0Rightarrow y_2=-4$

d) We solve the equation in $y$:

$x^2=13+y$

$y_1=3$

$x^2=13+3=16$

$x=pmsqrt{16}$

$x_1=-4$

$x_2=4$

$y_2=-4$

$x^2=13-4=9$

$x=pmsqrt{9}$

$x_3=-3$

$x_4=3$

We determine $x$:

$$
begin{cases}
x^2+y^2=25\
y=x^2-13
end{cases}
$$

$$
begin{cases}
x^2+y^2=25\
y=x^2+6
end{cases}
$$

$y=x^2+6$

$y>6Rightarrow y^2>36$

$$
x^2+y^2=25Rightarrow y^2leq 25
$$

$$
begin{cases}
x^2+y^2=25\
y=x^2+5
end{cases}
$$

$y-5+y^2=25$

$y^2+y-30=0$

$y^2-5y+6y-30=0$

$y(y-5)+6(y-5)=0$

$(y-5)(y+6)=0$

$y-5=0Rightarrow y_1=5$

$y+6=0Rightarrow y_2=-6$

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

$$
x=0
$$

As $y^2leq 25$ only the solution $y=5$ fits. We determine $x$:

$x=0$

$$
y=5
$$

$$
begin{cases}
x^2+y^2=25\
y=x^2
end{cases}
$$

$y+y^2=25$

$y^2+y-25=0$

$y=dfrac{-1pmsqrt{1^2-4(1)(-25)}}{2(1)}$

$=dfrac{-1pmsqrt{101}}{2}$

$y_1=dfrac{-1-sqrt{101}}{2}approx -5.52$

$y_2=dfrac{-1+sqrt{101}}{2}approx 4.52$

$x^2=4.52$

$x=pmsqrt{4.52}$

$x_1=2.13$

$$
x_2=-2.13
$$

As $y^2leq 25$ only the solution $y=4.52$ fits. We determine $x$:

$x_1=2.13; y=4.52$

$x_2=-2.13; y=4.52$

$$
begin{cases}
x^2+y^2=25\
y=x^2-5
end{cases}
$$

$y+5+y^2=25$

$y^2+y-20=0$

$y^2+5y-4y-20=0$

$y(y+5)-4(y+5)=0$

$(y+5)(y-4)=0$

$y+5=0Rightarrow y_1=-5$

$y-4=0Rightarrow y_2=4$

$y_1=-5$

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

$x_1=0$

$y_2=4$

$x^2=4+5=9$

$x=pm 3$

$x_1=-3$

$$
x_2=3
$$

We determine $x$:

$x_1=0; y_1=-5$

$x_2=-3; y_2=4$

$x_3=3; y_2=4$

$$
begin{cases}
y=3x-5\
y=-2x-15
end{cases}
$$

$3x-5=-2x-15$

$3x-5+2x=-2x-15+2x$

$5x-5=-15$

$5x-5+5=-15+5$

$5x=-10$

$x=dfrac{-10}{5}$

$$
x=-2
$$

$y=3(-2)-5$

$$
y=-11
$$

We determine $y$:

$x=-2$

$$
y=-11
$$

The solution tells us that the two graphs intersect in the point $(-2,-11)$. We check by graphing the equations:

$$
begin{cases}
y-7=-2x\
4x+2y=14
end{cases}
$$

$y=7-2x$

$4x+2(7-2x)=14$

$4x+14-4x=14$

$$
14=14
$$

$(k,7-2k), k$ real

$$
begin{cases}
y=2(x+3)^2-5\
y=14x+17
end{cases}
$$

$2(x+3)^2-5=14x+17$

$2x^2+12x+18-5=14x+17$

$2x^2+12x+13-14x-17=0$

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

$2(x^2-x-2)=0$

$x^2-x-2=0$

$x^2+x-2x-2=0$

$x(x+1)-2(x+1)=0$

$(x+1)(x-2)=0$

$x+1=0Rightarrow x_1=-1$

$x-2=0Rightarrow x_2=2$

$y_1=14(-1)+17$

$y_1=3$

$y_2=14(2)+17$

$y_2=45$

We determine $y$:

$x=-1; y=3$

$$
x=2; y=45
$$

The solution tells us that the two graphs intersect in two points: $(-1,3), (2,45)$. We check by graphing the equations:

$$
begin{cases}
y=3(x-2)^2+3\
y=6x-12
end{cases}
$$

$3(x-2)^2+3=6x-12$

$3(x^2-4x+4)+3=6x-12$

$3(x^2-4x+4+1)=3(2x-4)$

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

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

$x^2-6x+9=0$

$(x-3)^2=0$

$$
x=3
$$

$y=6(3)-12$

$$
y=6
$$

We determine $y$:

$x=3$

$$
y=6
$$

The solution tells us that the two graphs intersect in the point $(3,6)$. We check by graphing the equations:

a) $(-2,-11)$

b) $(k,7-2k), k$ real

c) $(-1,3),(2,45)$

d) $(3,6)$

Question asks us to read off directly from the graph the function values; (a) $f(3)=1$ (b) $f(0)=4$ (c) $f(4)=2$ (d) $f(-1)=5$.

(a) $f(3)=1$ (b) $f(0)=4$ (c) $f(4)=2$ (d) $f(-1)=5$

(a) Observing the graph. the equation $mid x-3mid+1=1$ is satisfied at the coordinates shown in red (3,1);

(b) $mid x-3mid+1leq4$ is satisfied at the shaded locations shown $0<x<6$

(c) $mid x-3mid+1=3$ is satisfied at the coordinates shown in red (1,3) and (5,3)

(d) $mid x-3mid+1>2$ is satisfied at the locations shown; $x4$

Given,

$$
begin{align*}
y&=4left(dfrac{1}{x+5}right)+7\\
end{align*}
$$

In order to graph and completely dec=scribe the given function, we will sketch the graph of the given function first.

To graph the function, we start with the parent function f$_{0}(x)=dfrac{1}{x}$, which we shift five units to the left, then vertically stretch by a factor of four and finally shift seven units up to get $y$.

Thus the graph is mentioned below.

The function’s domain is $(-infty,-5)$ and $(-5,infty)$.

The function’s range is $(-infty,7)$ and $(7,infty)$.

The function has a vertical asymptote which is $x = -5$ and a horizontal asymptote which is $y = 7$.

The function is decreasing on $(-infty,-5)$ and on $(-5,infty)$.

The function is discontinous in $x = -5$.

The $x$-intercept is,

$$
begin{align*}
y&=0\
0&=4(dfrac{1}{x+5})+7\
0&=dfrac{4}{x+5}+7\
-7&=dfrac{4}{x+5}\
-7(x+5)&=4\
-7x-35&=4\
-7x&=39\
x&=-dfrac{39}{7}\
end{align*}
$$

Therefore $x$-intercept is $-dfrac{39}{7}$.

The $y$-intercept is,

$$
begin{align*}
x&=0\
y&=4(dfrac{1}{x+5})+7\
y&=dfrac{4}{0+5}+7\
y&=dfrac{4}{5}+7\
y&=dfrac{4+35}{5} \
y&=dfrac{39}{5} \
end{align*}
$$

Therefore $y$-intercept is $dfrac{39}{5}$.

The general form of the absolute value function is:          $f(x)=a|x-h|+k$

From the given figure, $(h, k)$ is the point $(-3, 4)$ and one of the lines passes the point $(-1, 0)$

Substituting in the general form equation,

$0=a|-1+3|+4$

$2a=-4$

$a=-2$

The equation is:

$$
f(x)=-2|x+3|+4
$$

$$
f(x)=-2|x+3|+4
$$

Given,

$$
begin{align*}
a. y&=sqrt[3]{x}\
b. y&=9x^{5}-x-9 \
c. y&=4x^{3}+8x^{7}\
end{align*}
$$

$a.$ Now,

$$
begin{align*}
y&=sqrt[3]{x}\
y&=x^{tfrac{1}{3}}\
f(x)&=x^{tfrac{1}{3}}\
f(-x)&=-x^{tfrac{1}{3}}ne f(x) \
-f(x)&=-x^{tfrac{1}{3}}ne f(-x) \
end{align*}
$$

Thus the given function is neither odd nor even.

$b.$ Now,

$$
begin{align*}
y&=9x^{5}-x-9\
f(x)&=9x^{5}-x-9\
f(-x)&=9(-x)^{5}-(-x)-9\
f(-x)&=-9x^{5}+x-9\
f(-x)&=-(9x^{5}-x+9)ne f(x)\
-f(x)&=9(-x)^{5}+x-9ne f(-x) \
end{align*}
$$

Thus the given function is neither odd nor even.

$c.$ Now,

$$
begin{align*}
y&=4x^{3}+8x^{7}\
f(x)&=4x^{3}+8x^{7}\
f(-x)&=4(-x)^{3}+8(-x)^{7}\
f(-x)&=-4x^{5}-8x^{7}\
f(-x)&=-(4x^{5}+8x^{7})ne -f(x) \
end{align*}
$$

Thus the given function is odd .

$a.$ Nor odd neither even

$b.$ Nor odd neither even

$c.$ odd

Given,

$$
begin{align*}
&a. y=sqrt[9]{m} \\
&b. dfrac{2}{g^{7}}\\
&c. dfrac{-3}{b^{-5}}\\
end{align*}
$$

In order to rewrite each expression in an equivalent form first, we will simplify them.

$a.$ Now,

$$
begin{align*}
y&=sqrt[9]{m}\
y&=m^{tfrac{1}{9}} \
end{align*}
$$

Thus the equivalent form is $y = m^{tfrac{1}{9}}$.

$b.$ Now,

$$
begin{align*}
&=dfrac{2}{g^{7}}\\
&=2g^{-7}\
end{align*}
$$

Thus the equivalent form is $2g^{-7}$.

$c.$ Now,

$$
begin{align*}
&=dfrac{-3}{b^{-5}}\
&=-3b^{5}\
end{align*}
$$

Thus the equivalent form is $-3b^{5}$.

Assuming Jamal’s cupcake and Dinah’s sandwich flew from flew from the point $(0, 0)$.

The path of Jamal’s cupcake can be represented with a quadratic equation, where the vertex of the parabola is the point $(10,9)$

$j(x)=-a(x-10)^2+9$

Substituting for the point $(0,0)$

$0=-a(0-10)^2+9$

$0=-100a+9$

$a=dfrac {9}{100}$

$$
j(x)=-dfrac {9}{100}(x-10)^2+9
$$

The path of Jamal’s cupcake can be represented with a quadratic equation, where the vertex of the parabola is the point $(12,6)$

$d(x)=-a(x-12)^2+6$

Substituting for the point $(0,0)$

$0=-a(0-12)^2+6$

$0=-144a+6$

$a=dfrac {1}{24}$

$$
d(x)=-dfrac {1}{24}(x-12)^2+6
$$

To determine how tall is Harold, we solve both equations.

$j(x)=d(x)$

$-dfrac {9}{100}(x-10)^2+9=-dfrac {1}{24}(x-12)^2+6$

$-dfrac {9}{100}(x-10)^2+9=-dfrac {1}{24}(x-12)^2+6$

$-dfrac {9}{100}(x-10)^2+3-dfrac {1}{24}(x-12)^2$          (Subtracting 6 from each side)

$-9 cdot 6(x-10)^2+3 cdot 600=-25(x-12)^2$          (Multiply each side by 600)

$-9 cdot 6(x-10)^2+3 cdot 600=-25(x-12)^2$          (Multiply each side by 600)

$-54(x^2-20x+100)+1800=-25(x^2-24x+144)$

$-54x^2+1080x-5400+1800=-25x^2+600x-3600$

$29x^2-480x=0$

$x=0$ or $x=dfrac {480}{29} approx 16.55$

$j(16.55)=-dfrac {9}{100}(16.55-10)^2+9$

$j(16.55)=5.14$

Harold is 5.14 feet.

The below is the graph of both functions $j(x)$ and $d(x)$. The intersection point $(16.55, 5.14)$ determines that the tall of Harold 5.14 feet.

$a_1=52,000$

$d=3,000$

$b_1=36,000$

$$
q=1+0.11
$$

$a_n=a_1+(n-1)d$

$a_n=52,000+(n-1)(3,000)$

$a_n=52,000+3,000n-3,000$

$$
a_n=3,000n+49,000
$$

a) We determine the general term of the arithmetic sequence $a_n$:

$b_n=b_1q^{n-1}$

$b_n=36,000(1+0.11)^{n-1}$

$$
b_n=36,000(1.11)^{n-1}
$$

We determine the general term of the geometric sequence $b_n$:

Letting x be the cost of the chocolate and y the cost of the caramel we set up the system
$$
begin{bmatrix}5x+2y=4.25\2x+8y=3.5end{bmatrix}
$$
then solving by elimination and substitution; $4times Eq1-Eq2rightarrow(20-2)x+(8-8)y=17-3.5$. Then solving for x; $18x=13.5$ so $x=$0.75$ and $5(0.75)+2y=4.25$ so $y=$0.25$

Solving graphically by plotting $y=dfrac{1}{2}(4.25-5x)$ and $y=dfrac{1}{8}(3.5-2x)$ and reading off the intersection coordinates as $(x,y)=(0.75,0.25)$;

$$
(x,y)=($0.75,$0.25)
$$

Let, weight of a cylinder be $x$ ounces and weight of a prism be $y$ ounces then below system of equations is obtained:
begin{align*}
4x+5y&=32 & textrm{Equation (1)}\
x+8y&=35 & textrm{Equation (2)}\\
intertext{Subtracting $4 times$ Equation (2) from Equation (1): }
4x+5y-4(x+8y)&=32-4(35)\
4x+5y-4x-32y&=32-140\
-27y&=-108\
y&=4
intertext{Putting $y=4$ in Equation (1):}
4x+5(4)&=32\
4x+20&=32\
4x&=12\
x&=3
end{align*}
Hence, weight of a cylinder is 3 ounces and weight of a prism be $4$ ounces

Weight of a cylinder is 3 ounces and weight of a prism be $4$ ounces

$$
(a)
$$

$$
5-3(dfrac{1}{2}x + 2)=-7
$$

$$
12=3(dfrac{1}{2}x + 2)
$$

$$
4=dfrac{1}{2}x + 2
$$

$$
x=4
$$

$$
(b)
$$

$$
5(sqrt{x-2} + 1)=15
$$

$$
sqrt{x-2} + 1=3
$$

$$
sqrt{x-2}=3-1
$$

$sqrt{x-2}=2$ Let’s raise both sides on 2nd power:

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

$$
x-2=4
$$

$$
x=6
$$

$$
(c)
$$

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

$dfrac{2x}{3} + x=10$ Now let’s multiply with 3

$$
2x + 3x=30
$$

$5x=30$, so $x=dfrac{30}{5}$

Finally: $x=6$

$$
(d)
$$

$$
-3(2x+1)^3=-192
$$

$$
(2x+1)^3=64
$$

$$
(2x+1)^3=4^3
$$

$$
2x+1=4
$$

$$
2x=3
$$

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

(a) $x=4$, (b) $x=6$, (c) $x=6$, (d) $x=3/2$

$$
2x^2 + 5x -3leq x^2 + 4x + 3
$$

Let’s check if it works for $x=-1$ and $x=5$

$$
2(-1)^2 + 5(-1) -3 leq (-1)^2 + 4(-1) + 3
$$

$$
2 -5 – 3 leq 1 -4 + 3
$$

$-6leq 0$ – CORRECT

$$
2(5)^2 + 5cdot5 -3 leq 5^2 + 4cdot5 + 3
$$

$$
50 + 25 -3 leq 25 + 20 + 3
$$

$72 leq 48$ NOT CORRECT

$x=-1$ is a solution of given equation but $x=5$ is not

(a) Plotting $y=mid xmid$ then shading the region where the inequality $yleqmid xmid$ is satisfied;

(b) Plotting $x=mid ymid$ then shading the region where the inequality $mid ymidgeq x$ is satisfied;

$$
f(-x)=3(-x)^{3}+2=-f(x)+4
$$

(a) is neither even nor odd since $f(-x)ne f(x),-f(x)$

$$
f(x)=f(-x)
$$

(b) is an even function since it satisfies the relation $;$

It is given that ,
$$begin{aligned}
text{Water required by the family}&= 1000 text{cubic feet}\
text{ Monthly service fee}&= $ 12.70 \
end{aligned}$$

|Water used$(ft^3)$ |Charge per 100 cubic feet |
|–|–|
|300 |$ 3.90|
|300 |$ 5.90 |

Writing a function using the table:
$$begin{aligned}
f(x)&= 12 + 3.90 cdot dfrac{x}{100} , 0 leq x leq 300rightarrow(1)\
f(x)&= 12 + 3.90 cdot dfrac{x}{100} , 0 leq x leq 300 + 5.20 cdot dfrac{x – 300}{100}, x > 300rightarrow(2)\
end{aligned}$$

$b$.
The above – sketched graph is a function because every value of $x$ there is a value of $y$.

$c$.
Domain $rightarrow [0, infty)$.
Range $rightarrow [12.70, infty).$

$d$.
If the monthly service fee is decreased to 10.20, the graph would be shifted down by $12.70 – 10.20 = 2.50$ units.

For x-intercept $y=0$

$0=x^2+2x-80$

$(x+10)(x-8)=0$

$x+10=0$ or $x-8=0$

$x=-10$ or $x=8$

x-intercepts are :          $(-10, 0)$ and $(8, 0)$

For y-intercept $x=0$

$y=(0)^2+2(0)-80=-80$

y-intercept is :          $(0, -80)$

x-coordinate for the vertex $=dfrac {-b}{2a}=dfrac {-2}{2}=-1$

y-coordinate for the vertex $=1-2-80=-81$

The vertex is:          $(-1, -81)$

The graphing form of the equation is:

$$
y=(x+1)^2-81
$$

x-intercepts are :          $(-10, 0)$ and $(8, 0)$

y-intercept is :          $(0, -80)$

The graphing form of the equation is:          $y=(x+1)^2-81$

$$
2x^2+5x-3=x^2+4x+3
$$

$x_1=-3$

$$
x_2=2
$$

b) We label the graph by checking one point on each of them: for example the $y$-intercepts (the function $y=2x^2+5x-3$ has the $y$-intercept $y=-3$, so it is the green one):

$$
xin[-3,2]
$$

c) We determine the values of $x$ for which $2x^2+5x-3leq x^2+4x+3$ by finding those $x$ for which the green graph is below the red graph:

$$
xin(-infty,-3)cup(2,infty)
$$

e) We determine the values of $x$ for which $2x^2+5x-3>x^2+4x+3$ by finding those $x$ for which the green graph is above the red graph:

(a) For the inequality to be true; $4mid x+1mid-2>6$ so there must be two boundary points $x>1$ and $x<-3$. Since x=1 and x=-3 are the smallest values that make the inequality false they are therefore the boundary points and are denoted by unfilled circles

(b) The solutions are contained in the open intervals $I_{1}=left( -infty,-3right)$ and $I_{2}=left( 1,inftyright)$. We need to check $x0$.

$$
2x^2+5x-3<x^2+4x+3
$$

$f(x)=2x^2+5x-3-x^2-4x-3$

$$
f(x)=x^2+x-6
$$

a) Instead of solving the above inequality, we can move all terms on one side and solve an inequality of the form:

$f(x)<0$.

$$
xin(-3,2)
$$

To solve $f(x)<0$ we find all $x$ for which the graph is under the $x$-axis:

$x^2+x-6=0$

$x^2+3x-2x-6=0$

$x(x+3)-2(x+3)=0$

$(x+3)(x-2)=0$

$x+3=0Rightarrow x_1=-3$

$x-2=0Rightarrow x_2=2$

$$
(-infty,-3),(-3,2),(2,infty)
$$

$x=-4Rightarrow f(-4)=(-4+3)(-4-2)=6>0$

$x=0Rightarrow f(0)=(0+3)(0-2)=-60$

We take one value in each interval and check if $f(x)$ is positive or negative:

$$
(-3,2)
$$

Since we are interested in $f(x)<0$, the interval that fits is (the same as the one got with another method):

$$
x^2-3x-10geq 0
$$

$$
xin(-infty,-2]cup[5,infty)
$$

We find all $x$ for which the graph is above the $x$-axis:

(a) If the system
$$
begin{bmatrix}y=4mid x+1mid-2\y=6end{bmatrix}
$$
is plotted we can obtain the solutions by looking at the region where $Eq1>Eq2$

(b) Observing the plot we note that the regions to the left of the boundary point $x=-3$ satisfy the inequality and so to do those to the right of the boundary point $x=1$

$$
begin{cases}
ygeq 2x^2+5x-3\
y<x^2+4x+3
end{cases}
$$

$2(-3)^2+5(-3)-3=0leq 0checkmark$

$(-3)^2+4(-3)+3=0not>0$

$2(-1)^2+5(-1)-3=-6leq 1checkmark$

$(-1)^2+4(-1)+3=0not>1$

$2(1)^2+5(1)-3=4leq 5checkmark$

$$
(1)^2+4(1)+3=8>5checkmark
$$

We check if the points$(-3,0),(-1,1),(1,5)$ make both inequalities true:

So the point $(1,5)$ belongs to the solution.

$$
{(x,y)|xin [-3,2],2x^2+5x-3leq y<x^2+4x+3}
$$

b) The solution of the equation is an interval for $x$, while the solution for the system is a set of pairs $(x,y)$.

c) The solution of the system is represented by all pairs $(x,y)$ for which the point $(x,y)$ is above the graph of $2x^2+5x-3$ and below the graph of $x^2+4x+3$:

$$
{(x,y)|xin [-3,2],2x^2+5x-3leq y<x^2+4x+3}
$$

$$
dfrac{1}{3}(x+3)(x-4)<dfrac{4-x}{2}
$$

$$
begin{cases}
yleq dfrac{1}{3}(x+3)(x-4)\
y>dfrac{4-x}{2}
end{cases}
$$

$$
begin{cases}
y>dfrac{1}{3}(x+3)(x-4)\
y<dfrac{4-x}{2}
end{cases}
$$

$$
begin{cases}
yleqdfrac{1}{3}(x+3)(x-4)\
y>dfrac{4-x}{2}\
xgeq 0
end{cases}
$$

$$
3x+2geq x-6
$$

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

$$
2x+8geq 0
$$

$2x+8=0$

$2x=-8$

$$
x=-4
$$

$$
[-4,infty)
$$

$$
2x^2-5x<12
$$

$$
2x^2-5x-12<0
$$

$2x^2-5x-12=0$

$2x^2-8x+3x-12=0$

$2x(x-4)+3(x-4)=0$

$(x-4)(2x+3)=0$

$x-4=0Rightarrow x_1=4$

$2x+3=0Rightarrow x_2=-dfrac{3}{2}=-1.5$

$$
(-1.5,4)
$$

a) $[-4,infty)$

b) $(-1.5,4)$

$$
|2x+3|<5
$$

$$
-5<2x+3<5
$$