**Captain K**and his crew of math genius' continued their travel through

**in search of the**

*The Land Of Pre-Calculus***When they came across an obstacle in their way of eternal bliss.....**

*Legendary Holy Golden Credit!!***This sea port was the only way they were able to cross the sea without being swept away by its' tides of sinusoidal functions.**

*A SEA PORT!*"What do we do now Captain K?" said

**Craigmyer**.

"Look there's a sign near by, let's see what it says" said

**Melk**

"Trying to get across the sea port? It's not going to be as easy as walking across, first you must solve these problems. Then you shall see what happens next..." read

**the crew**.

**At this sea port, the depth of the water, h meter, at time, t hours, during a certain day is given by this formula:**

h(

a) State the: (i)

b) What is the maximum depth of the water? When does it occur?

c) Determine the depth of the water at 5:00 am and at 12:00 noon.

d) Determine one time when the water is 2.25 meters deep.

h(

*t*) = 1.8sin[2Π (*t*-4.00)/12.4] + 3.1a) State the: (i)

*period*(ii)*amplitude*(iii)*phase shift*b) What is the maximum depth of the water? When does it occur?

c) Determine the depth of the water at 5:00 am and at 12:00 noon.

d) Determine one time when the water is 2.25 meters deep.

"WHAT ARE WE SUPPOSE TO DO!?!?!" yells

**Danny Boy**frantically

"No need to worry Bertman is here! We just have to

*sketch a graph*of this sea and we'll get our answers from there!" says

**Bertman**speaking in a voice strangely close to that of Batman's.

"Easier said then done, but I'll give it a try" says

**Mr. Siwwy**.

"What a minute,

*we don't even need the graph for the first part of the question*!" states

**Kasiaw Cole**(get it, Keyshia Cole, the singer?). She continues, "to find the period don't you have to go

*2Π/B*? But that fraction, that makes it so much harder then usual."

"I got it!" says

**Vinsanity**"you

*just got to multiple it all out first, then this will give you the answer for the period*"

**Vinsanity**shows his work to the rest of the crew:

**2Π X (**

= 2Π/1 X (

= 2Π (

= 2Π/12.4 X (

B = 2Π/period

B = 2Π/12.4

PERIOD = 12.4

*t*-4)/12.4= 2Π/1 X (

*t*-4)/12.4= 2Π (

*t*-4)/12.4= 2Π/12.4 X (

*t*-4)B = 2Π/period

B = 2Π/12.4

PERIOD = 12.4

"Nice! I

*got the amplitude and phase shift*!" says

**Dr. Grey-M**"By

*looking at the formula we get the amplitude to be 1.8 and the phase shift to be 4.00*."

"Awesome, on to the next questions!" says

**Danny Boy**

**NOTE: This is where my story kind of just stops for a bit and I just post the answers to the remaining questions b), c), d). This is taking way longer then I expected and it's already cutting into my sleep time, forget about my studying time =S. Continuing......**

**b) The maximum depth of the water is 4.9 meters. You get this by taking your sinusoidal axis and adding the amplitude to it:**

3.1 + 1.8 = 4.9 meters

To find when it occurs you plug in 4.9 into the formula:

4.9 = 1.8sin[2Π (

1.8 = 1.8sin[2Π (

1 = sin[2Π (

Π/2 = [2Π (

(12.4)(Π/2) = 2Π (

(12.4)(Π/2) / (2Π) =

3.1 + 4.00 =

c) For each time, you just plug them into the formula to find your depth at that time:

h(5 am) = 1.8sin[2Π (5-4)/12.4] + 3.1

h(5 am) = 3.9735 m

h(12 noon) = 1.8sin[2Π (12-4)/12.4] + 3.1

h(12 noon) = 1.6766 m

NOTE: If you were given a time value of 5:00 pm this would be equal to 17 hours, and that would be the value you plug into the formula.

d) To determine one time when the depth of the water is 2.25 meters deep, we just plug 2.25 meters into the formula:

2.25 = 1.8sin[2Π (

Let Θ = [2Π (

2.25 = 1.8sinΘ + 3.1

-0.85 = 1.8sinΘ

-0.4722 = sinΘ

-0.4918 = Θ

-0.4918 = [2Π (

(12.4)(-0.4918) = 2Π (

(12.4)(-0.4918) / 2Π =

-0.9706 + 4 =

3.0294 =

3:02 am

3.1 + 1.8 = 4.9 meters

To find when it occurs you plug in 4.9 into the formula:

4.9 = 1.8sin[2Π (

*t*-4.00)/12.4] + 3.11.8 = 1.8sin[2Π (

*t*-4.00)/12.4]1 = sin[2Π (

*t*-4.00)/12.4]Π/2 = [2Π (

*t*-4.00)/12.4](12.4)(Π/2) = 2Π (

*t*-4.00)(12.4)(Π/2) / (2Π) =

*t*-4.003.1 + 4.00 =

*t*

7.1 =*t*

7:06 am =*t*c) For each time, you just plug them into the formula to find your depth at that time:

h(5 am) = 1.8sin[2Π (5-4)/12.4] + 3.1

h(5 am) = 3.9735 m

h(12 noon) = 1.8sin[2Π (12-4)/12.4] + 3.1

h(12 noon) = 1.6766 m

NOTE: If you were given a time value of 5:00 pm this would be equal to 17 hours, and that would be the value you plug into the formula.

d) To determine one time when the depth of the water is 2.25 meters deep, we just plug 2.25 meters into the formula:

2.25 = 1.8sin[2Π (

*t*-4)/12.4] + 3.1Let Θ = [2Π (

*t*-4)/12.4]2.25 = 1.8sinΘ + 3.1

-0.85 = 1.8sinΘ

-0.4722 = sinΘ

-0.4918 = Θ

-0.4918 = [2Π (

*t*-4)/12.4](12.4)(-0.4918) = 2Π (

*t*-4)(12.4)(-0.4918) / 2Π =

*t*-4-0.9706 + 4 =

*t*3.0294 =

*t*3:02 am

**BACK TO THE STORY:**

"Wow, that took a long time but we finally figured it out" says

**Richard S.**

Just then they saw a magical bridge appear out of no where allowing them to cross the sea safely and on towards their next challenge while in search of....the

*Legendary Holy Golden Credit!!*"Hey how did you get here before we did Samus?" says

**Dino**

**"Didn't you guys see that perfectly safe bridge right down over there?"**

**Samus**points towards the bridge.

"WHY DIDN'T YOU TELL US!!??!!??"

**the crew**screams at Samus.

"You never asked? DUH!" says

**Samus**

**PART TWO OF THE STORY:**

**Transformator And His Evil Pre-Test..... NOTE: I'm going to use past colours again, there's too many people in our class and not enough colours =S**

So after getting past the sea port and it's puzzling questions, our adventurers ventured onward along the

**not knowing that they were about to encounter the**

*"yellow brick road"*

*EVIL TRANSFORMATOR DUN DUN DUN!*"Why are we walking on this yellow road anyways?" says

**KaDeeM AbDul Ali JaBBar**confused.

"Captain K told us to, he said this road would lead us to......the Legendary Holy Golden Credit!!" says

**Tim_MATH_y**

**"But if it lead us to....the Legendary Holy Golden Credit!! I really wish that'd stop happening everytime we mention....the Legendar Holy Golden Credit!! Sigh, anyways, if it lead us to that, then wouldn't the road be gold instead of yellow? Guys? Guys?" says**

**KaDeeM AbDul Ali JaBBar**

He turns around only to see the whole crew staring at the biggest transformation ever seen on the face of

**,**

*The Land Of Pre-Calculus*

*TRANSFORMATOR!!!**"MUAHAHAHAHA!!!! I'm Transformator and I'm here to phase and vertically shift your functions to doom! Then when I'm done with that, I'll stretch and squash your amplitude until you scream for your MAMA!! MUAHAHAHA!!!" says the hideous Transformator*

**"Ewww, this thing is so ugly it makes me sick. What are we going to do? Stupid Transformator we just want to get the Golden Credit. Oh my gosh leave us alone." says**

**Jeng-Lo**

*"Never! You're trespassing on my territory there's nothing you can do now to escape from my parameters A, B, C, and D!" says*

"Alright then, I purpose a challenge to you Transformator. I bet our whole crew together can solve any transformation problem you throw at us. If we win then you let us go." says

**Transformator****Sandy**softly but bravely.

*"And if you lose?" says*

**Transformator**

"Then we have to stay here forever allowing you to do whatever you want to us" says

**Sandy**

*"Alright then, you have a deal! Here's a Pre-test chalk full of challenging problems, lets see if you can get out of this one Captain K! MUAHAHAHA!" says*

The crew recieves the test and starts to begin solving the questions:

**Transformator**in delight.**1. f(x) = 2x² - 3, where x is greater than or equal to 0, then a function g that will have domain and range that are both different from those of function f is:**

a) g(x) = f(-x)

b) g(x) = -f(x)

c) g(x) = f-¹(x)

d) g(x) = kf(x), k greater than 0

a) g(x) = f(-x)

b) g(x) = -f(x)

c) g(x) = f-¹(x)

d) g(x) = kf(x), k greater than 0

"Oh! The answer is

**C)!**" says

**Aichelle the Incredible**

"The answer wouldn't be

**A)**because

**f(-x)**will only result in a

**different domain**, and it wouldn't be

**B)**because

**-f(x)**only results in a

**different range**. For

**C),**

**f inverse of (x) means your x coordinates would be your y coordinates, and your y coordinates would be your x coordinates**therefore giving you a

**different domain and range**!" continues

**Aichelle the Incredible**

*"Argh, that's correct, next question" says*

**Transformator**angered**2. The graph of a function f is a parabola opening upward, with its vertex on the x-axis. The graph of a new function g, where g(x) = 2f(x), will have:**

**a) the same domain and the same range as f**

b) the same domain but a different range than f

c) a different domain but the same range as f

d) a different domain and a different range than f

b) the same domain but a different range than f

c) a different domain but the same range as f

d) a different domain and a different range than f

"Jojo Rocks!" yells

**Jojo**

"The answer has to be

**A)!**Since it's

**a parabola, going 2f(x) wouldnt affect the range because the parabola would be going up both sides to infinity**so you could

**trash B) and D)**already. It

**can't be C)**either because the

**vertex is on the x-axis, therefore the domain will be left unchanged**so the answer has to be

**A)!"**continues

**Jojo**

*"ARGH! Right again! Next one, you won't get this!" grumbles*

**Transformator**

**3. Given the graph of f(x) below, sketch 1/f(x) in the space provided:**

**Johnny Johnson**

"First things first

**find your invariant points**, on

**this graph there's two. One at (-1,-1) and the other at (1,1).**Then

**find your asymtotes**, it just so happens that

**the asymtotes are the x and y-axis**. After that just

**remember Dr.Suesus' version of math when drawing the actually reciprocal graph in (smallering and biggering).**Also

**remember to have arrows where neccesary and dots that shows an end on the graph**where necessary. In this case

**the graph ends at (-5,-1)**but

**continues on going in the other direction**." states

**Johnny Johnson**in what seems like an eternity to explain.

*"NO!! RIGHT AGAIN!! THERE'S NO WAY YOU'LL GET THESE LAST TWO!" screams*

**Transformator****4. Given f(x) = cube√3x² -4, write the equation for its inverse f-¹(x).**

"They just keep getting easier and easier! I'll show you the work, no sweat." says

**Bond, Robert Bond**

*"THIS CAN'T BE HAPPENING!! THIS LAST QUESTION IS IMPOSSIBLE, NOT EVEN I WAS ABLE TO SOLVE IT!!!" yells*

**Transformator**, now shaking the ground with his mighty power.**And this is where I'm going to stop for tonight. Sorry guys I put out more then I could chew, but this final question will be up by tomorrow! Good luck to all on the test in the AFTERNOON. Scribe as you all know already is Mark. =D**

## 8 comments:

Haha, that was a funny scribe post. I like the way you implemented as a story. You kinda left me hanging there, I hope you'll continue the scribe post for the pre-test tomorrow. Good job!

GOOD JOB! It's so funny because it is a journey for all of us!

haha indeed lol it's quite entertaining ! good job !

i like it, its funny :D

thanks guys! sorry i'm taking so long to complete EVERYTHING making a story out of our two classes was harder and longer then I thought. I'm trying to include you all into the story, I hope no one minds if I do that sorry never really asked for permission =S And if no one minds I hope nobody minds if they don't happen to be in the story, I can only fit so many people without dragging the story and making it boring. Thanks again guys!

Great scribepost. I like the way you used humor to make the post fun to read. Great math review aswell.

Thanks for scribing.

Mr. Harbeck

Sargent Park School

Hey Danny! This is a great scribe post, I love how your post isn't actually just a scribe, but is a story with all of us too. And though humourous, you still integrated what we learned in class into the story. Well done!

that is awesome oh my gosh Aichelle the Incredible! haha woo!

you added more things! =)good job Danny

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