Unit 7. Radical & Rational Functions

Hey guys, it's Manpreet.

I'll be blogging about what we did in your class today. First we finished the Radical functions using transformation part of the unit, where we learned how to determine a radical function from a graph.

With only a graph of a trasformated radical fuction, we can determine the traformation by first drawing the original function (y=√x), then using one of the two methods, comparing vertical distance (vd), or horizontal distance (hs). Vd is used to determine the value of (a) in the function y=a√(bx-h)+k, and hd is used to determine the value of (b).

Then we went to the next part of the unit, which is The Square Root of a Function. This is when you convert a any function to a radical function. 

There are some rules that are very importent when converting and graphing square root function:

          1) f(x)<0  √f(x) does not exist

          2) f(x)=0 

          3) 0<=f(x)<=1  √f(x) will be above f(x)

          4) f(x)>1  √f(x) will be below f(x)

This one is fairly straight forword, there are a couple of rules the you should follow. You should, when mapping the number for the original function, try to find the value of x where y will equal 1 and/or 0 so it will make it a lot easier in find the invariant points.

This one deals with quadratic functions, which as first might seem difficult to understand but practicing this will help in using certain methods to complete it. The most importent thing to do first is to find the x and y intercepts. This will help you in graphing the square root function.

There is one more thing, Mr. P told us to find the reason why we can't use f(x)>0 to get the restrictions, and I couldn't find anything on it but I tried to solve it anyways. So, I found a way to get the answer but in a really weird way, atleast to me.

   

First I took the equation and arranged it in the proper order, then I factored it out. I wrote the equation twice, one to solve (-1/2x+1)>=0 and the other to solve (x+2)>=0. The reason that you can solve for either one, is because if 0 is divide by any value you get 0, so then you can solve for either. For (-1/2x+1)>=0 I got x<=2 and for (x+2)>=0 I got x>=-2. Then putting them together I got -2<=x<=2. 

*I don't know if this works on any other equation.. So far worked on this. If you are having trouble graphing the function try it.

shameless plug: https://www.youtube.com/user/themanpreet23/videos

The End.

HEY, ITS JASDEEP

In the morning class we started a new unit on Radical & Rational Functions. Its two units put together that are pretty similar to each other but are different from the rest of the units.

In the first lesson we learned basic radical functions by getting to know how to graph a basic function

Ex:

                     

We did a question in graphing radicals by using a transformation

Ex:

In the afternoon class we had a class assignment on exponential and logarithmic functions 

WE HAVE A TEST TOMORROW :( BOOO!! 

HI,

This is Jasdeep.

Logarithmic Functions

Hi there ! This is Jaspreet (:

Don’t worry this blog is going to be really short and to the point . I don’t know too much about the logarithmic functions but whatever I found out , I thought it would be useful to share it with you guys (:

Mr.P asked us 3 questions:

1) Why x>0 ?!?!?!?

2) Why c>0 ??!?!?!

3) Lastly why c≠1!?!?!?

This is a logarithmic function that we were looking at in class today,

                                                   f(x) = log_c^x

1) Logs are undefined for negative x or for x = 0. Therefore x have to be greater than 0.

2 &3) “C” is any value greater than 0, except 1 because when c=1, the graph is undefined.

 I'm sorry I cant explain everything completely but that’s all I got to know for now, so it’s all on Mr.P to add on and explain it better (:

 

Unit 5- proving identities

Hi everyone. This is manveer Today I will be blogging about the things that we went over in class but very briefly. I don't like to talk to much so I will go straight into it.

Proving Identities :

Know these strategies :  
1) Use known identities i.e. ( sinx/cosx = tanx)
2) Use the Pythagorean Identity or one of its alternate forms
3) Rewrite the equation/expression using sin and cos
4)Multiply the numerator and denominator by the conjugate
5) FACTOR

So let's begin with the first question .

To determine the non-permissible values we need to know when sinx ≠ 0 .

x ≠ 0 ,180 , 360

The general solution for this equation would be x≠180n , where nEI

To verify the equation we need to substitute our x values. In this case x=30

1-sin²x  = sinxcosxcotx

L.H.S    = 1-sin²(30)

             = 1-(1/2)²

             = 1- 1/4

             = 3/4

R.H.S   = sin(30)cos(30)cot(30) 

            = (1/2) ( √3/2) (√3)

            = 3/4

So now according to the results we can tell that L.H.S = R.H.S

Lastly , to prove that the identity is true for all the permissible values of x we need to simply the equation.
1-sin²x  = sinxcosxcotx
cos²x = sinx * cosx * cosx/sinx
cos²x = cosx*cosx
cos²x = cos²x

That's it !

Since this was a tricky question and it was a TRAP for all of us, I thought of reminding you all again.

tan(π/2 - θ ) = cotθ

All we need to do is REPLACE θ with π/2 - θ

tan(π/2 - θ) = sin(π/2 - θ) / cos (π/2 - θ)

                   = cosθ / sinθ

                   = cotθ

There are a lot of steps for this question but I have 6 easy steps that could help you get the answer.

1) Break sin3x into sin2x + sinx

2) α=2x & β=x

3) Replace sin2x => 2sinxcosx

4) Replace cos2x => 1-2sin²x

5) Multiply inside the brackets

6) Combine like terms

MAKE SURE YOU ADD THE BRAKCETS !



here are the double angle identities don't forget them :) 

ok bye

Unit 5 - Trigonometric Identites

Hey everyone ! This is Jaspreet (:

First of all , 43 days left till the provincial exams ! I know Mr.P reminds us this everyday so I thought of reminding you again :)

Today this blog is going to be about Trigonometric Identities. So far everything is going alright but I know its going to get harder . Before I begin lemme tell you all that I am not a very good person when it comes to explaining things like this, unless I actually do it but I'll try my best! (:

Let's start off with something cute :)

Here is a brief review of the things that we did yesterday ! (Its a lot of formulas!)

Sum and Difference Identities

Don't worry the formulas on top are on your formula sheet ! (:

 

 

 

To write an expression as a single trigonometric function, you need to know "α" & "β" and the correct formula.

 

In this example ;

 α = 45°

 β = 17°

Therefore, sin48°cos17° - cos48°sin17°

                 = sin(α - β)

                 = sin (48°-17°)

                 = sin31°           

 

                           

Moving on to today's lesson. Today we talked about Double Angle Trigonometric Identities.

 

Double Angle Trigonometric Identities

 

Double angle identities:



 

 

      

 

Sine & Cosine of a Double Angle

 

       Since,  sin2α = sin (α + α)

                   sin2α = sin α cos α + cos α sin α
Therefore,  sin2α = 2sin α cos α

The cosine formula is just as easy:

       Since,  cos2α = cos (α + α)
                   cos2α = cos α cos β - sin α sin β
Therefore,  cos2α = cos²α -sin²α



There are two other formula's for cos2α

 

1.) cos2α = cos²α − sin²α

We can replace "cos²α" with " 1-sin²α" because according to the Pythagorean identity , cos²θ + sin²θ = 1.

 

Therefore ,  sin²θ = 1 - cos²θ

                               &

                   cos²θ = 1 - sin²θ

 

Moving back to cos2α = cos²α − sin²α

                          cos2α = 1 - sin²α - sin²α

                          cos2α = 1- 2sin²α

 

2.) cos2α = cos²α − sin²α

Instead of replacing "cos²α" , we are going to replace "sin²α".

            [ Remember to add the brackets!!!! ]

 

                  cos2α = cos²α − sin²α

                  cos2α = cos²α − (1 − cos²α)

                  cos2α = cos²α − 1 + cos²α

                  cos2α = 2cos²α − 1

 

 

 

To write an expression as a single trigonometric function using double-angle identities you need know your " α "

 

For this example ;

 

  2tan76°                         α = 76°

 1-tan²76°

 

  2tanα°             

 1-tan²α°   = tan2α

                 = tan 2 * 76°

                 = tan154°

 

 

This was probably the hardest part of this unit so far , which confused most of us at first but eventually we got it ! {THANKS TO MR.P } (: Although I'm not a 100% confident about this part but I will try to explain the best I can .

 

For this example ;

Part A ) We need to find the permissible values for the expression

Part B ) We need to simplify the expression to one of the 3 primary trigonometric functions { sinx , cosx , tanx }

 

I have clearly shown all the steps in the picture but I will give a brief explanation of what needs to be done.

 

So for the first part, we need to find all the values where sinx=0 

 sinx = 0 , π , 2π , .......

 x= πn , where n∈I

 

Then we need to find all the values where cosx=0

cosx = π/2 , 3π/2 , ........

x=π/2 + πn , where n∈I

 

To get general solution for this equation we need to combine the two solutions together.

 General solution : x= π/2 n , where n∈I

 

∴ All the permissible vales for the expression 1-cos2x  are the real numbers except

                                                                            sin2x

    when x= π/2 n , where n∈I

 

 

 

Last but not the least , determining exact trigonometric value for angles .

Note : When finding exact vales KEEP YOUR CALCULATORS AWAY !

 

 

 

Finding the exact value is probably the easiest thing so far.

 

To find the exact value for : 2sinπ/12 cosπ/12                    [ α=π/12 ]

= 2sinαcosα

= sin2α

= sin2 * π/12

= sin π/6

=1/2

 

Well that's all for today ! We have learnt a lot so far and there's more to go >.< (45-20 = 25 more pages to go ! ) I hope this blog helped everyone who is away and also those people who missed today's class .

Good luck and goodnight <3

 



Hi everyone ! It's Ella.

I'll be summarizing what we have learnt today. The topic was translation of sine and cosine functions. To start it off, let me recall about what Mr. Piatek discussed last week.

Transformations of some and cosine functions
The formulas are:
y = asinbx and y = acosbx

In vertical stretches :
• amplitude changes from the basic of 1 to |a|.

Amplitude is equal to maximum - minimum divided by 2.

•if a < 0, the function is reflected through the horizontal middle axis of function.

In horizontal starches:

• period changes from the basic of  2p to 

•if b < equal to 0, the function is reflected in the y - axis.

Now going to translation of sine and cosine functions.

y = asinb(x-c)+d and y = acosb(x-c)+d

Horizontal translations

•if it's negative, the function shifts c units to the right. 

•if it's positive, the function shifts c units to the left.

•it's called the phase shift.

vertical translations

• called vertical displacement. It is the result of change in the middle axis.

•if it's positive, the function shifts d units up.

•if it's negative, the function shifts d units down.

Order for applying transformations:

•perform all horizontal stretches and reflections.

•perform all vertical stretches and reflections.

•perform all translations.

Here's an example:

Now going to graphing the tangent functions

The graph of the tangent function, y = tanx, is periodic, but it is not sinusoidal. Periodic, meaning that it will repeat itself over regular intervals (cycles) of it's domain. Not sinusoidal, meaning that it does not fluctuate back and forth like a sine or cosine graph. 

Here's an example of tangent graph :