Wednesday, June 4, 2014

BQ#7 Unit V

Picture Discription of how to get the slope of the secant line
To start of basically you have a secan line drawn, that means it touches two points on the graph. The first distance you have is the (x ) and the one next to it is the (h ). the vertical up to point a is f of x f(x) and then the distance up is f (x+h) because it has to go the disatnce of x and h. The cordinates of a is (x , f(x) ) and of b its ( (x+h) , f(x+h) ).

The math of how to get the difference quotient!
To find how those points and how its revelent, we have to use the slope formula. Since we are finding the slope we can find the slope to help find the difference qoutient. In the end we just cancel out things and we are left with our famous f(x+h)-f(x)/h!!!!!!!

Saturday, May 17, 2014


1. What is continuity? What is discontinuity?

A continuous function means you can draw it without your pencil lifting from the paper. It is predicatble meaning you know whats going to happen. There are no breaks, jumps, or holes. It can be drawn with a single unbroken pencil stroke.  ex: lim as x approaches c of f(x)=f(c)- the limit and value are the same. On the other hand a discontinuous function is UNpredicatble. There are 4 types of discontinuities. A point discontinuity, which is also known as a hole, a jump discontinuity(different left and right ), oscillating behavior (wiggly), and an infinite discontinuity which results in unbounded behavior caused by a vertical asymptote. These continuities from two families, non removable, and removable discontinuities.

2.What is a limit? When does a limit exist? when does a limit not exist? What is the difference between a limit and a value?
A limit is the intended height of the function. A limit exists as long as you reach the same height from the left and the right. A limit does NOT exist if the right and left do not meet.
In the above picture the limit DOES exist because from both the left and right it approaches the same intended height.

In the above picture the limit DOES not exist because of a jump discontinuity and also because the left and right do NOT reach the same intended height.
A limit is the intended height, which means it can or cannot reach it, and the value is the ACTUAL height, or what it reaches.

3.How do we evaluate limits numerically, graphically, and algebraically?
To evaluate limits algebraically you can use three " shortcut" ways. There is the direct substitution way which is basically plugging in the x into the function. When doing this method you can produce 1 of 4 answers. You can get a numerical answer which means you are done. You can get a 0/# answer which means you are done. You can get a #/0(undefined) and you are done. Finally you can get 0/0, or indterminate form which requires more work. If the first method does not work you may use the dividing out/ factoring method. In this case you factor out the numerator and denominator and cancel out common terms to remove zero in the denominator. Then , use direct substitution with the simplified expression. Lastly, there is the rationalizing/conjugate method. If it is a fraction , then multiplying the top and bottom by a conjugate would have helped. The conjugate is where you change the sign in the middle of the term. You use the conjugate of wherever the radical is.
To evaluate a limit graphically we look at the graph and from the left and right with our fingers you trace the limit. This means it has to be the same from the left and the right. Make sure to identify the discontinuities.
To evaluate the limit numerically you use tables. The limit as x will go in the middle box and then from the leftb and right the numbers get closer and closer then on the bottom you graph the function and hit trace on those values.

Saturday, April 19, 2014

BQ #3 Unit T Concepts 1-3

How do graphs of sine and cosine relate to each of the others ? Emphasize asymptotes in your response.


Cosecant is similar to sine.The ratio for cosecant is 1/sin and since sin is y/r cosecant is r/y.In the unit circle wherever sine is positive so is cosecant. So basically, since sin is positive so is cosecant in the first quadrant. Sine does not have any asymptotes because they only occur when there is an undefined answer. However, cosecant does have asymptotes at o and pi.

Secant is related to cosine. Acording to the unit circle cosine is poitive in the first last quadrants and negative in the second and third. Secant follows this as well. The asymptotes or where an undefined answer occurs are at pi over 2  and 3pi over 2. This is where cosine is 0, making the ratio undefined. Because secant is 1/cosine. 

 The identity for cotangent is cosine over sine or x/y. It is very similar to tangent. Cosine and sine are positive in the first quadrant because cosine/sine will be positive In the second quadrant, the sine is positive and cosine is negative, so cotangent is negative because a negative cosine divided by a positive(sign wise) will be negative. For the third quadrant, cosine and sine are negative, making cotangent positive because a negative divided by a neagtive will cancel out the signs and make the answer positive. In the last quadrant cosine isnpositive and sine is negative, making cotangent negative because, again, a neagtive and positive leaves a negative. Since asymptotes occur when 0 is below the divid]sion sine, sine must eqaul zero so that an undefined answer may occur and asymptotes are found. Therefore, we know that sine is equal to 0at 0 and pi. 

The identity for tangent is sine/cosine or y/x.  If sine or cosine are negative, then tangent will be negative because a negative and positive sign make a negative answer. If sine or cosine are positive or  both negative, tangent will be positive because a negative and negative sign will be positive, as they cancel out. In the first quadrant, sine and cosine are positive, so tangent is positive. In the second quadrant, cosine is negative, but sine is positive so it is negative.In the third quadrant, both sine and cosine are negative, making tangent positive as they both cancel out each others signs. The fourth quadrant is negative because cosine is positive, but sine is negative. Tangent has asymptotes when cosine is equal to 0 because cosine in the denominator. They are at pi/2 and 3pi/2

BQ # 5 Unit T Concepts 1-3

BQ #5

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.

Asymptotes only occur when it is undefined. What that means is the answer from a trig ration product has to be undefined. Zero must be in the denominator. The trig ratio for sine is y/r , cosine is x/r , cosecant is r/y , secant is r/x , tangent is y/x , cotangent is x/y. ZERO HAS TO BE IN THE DENOMINATOR FOR THE ANSWER TO BE UNDEFINED. the only trig functions that produce a undefined answer are all but sine and cosine. Sine is y/r and r=1 therfore there cannot be an undefined answer. Cosine is x/r and r=1, again it cannot be undefined. Cosecant is r/y and y can be 0 on a certian place on the unit circle so it can be defined. The same for secant with x in the bottom. Tangent with x in the bottom and cotangent with y in the bottom.

Thursday, April 17, 2014

BQ #4 Unit T: Concept 3

Why is a “normal” tangent graph uphill, but a “normal” tangent graph downhill? Use unit circle ratios to explain.

To start off, tangent and cotangent graphs have asymptotes. The graph cannot touch the asymptotes. To have asymptotes sin has to be 0 so that it is undefined and cosine has to eqaul 0 so that tangent is undefined. In the examples shown below the period must occur between then asymptotes and CAN'T TOUCH IT. Beacuse the asymptotes are in different marks, tangent being at 90 and 270 and cotangent being at 180 and 360, in order for the graph to not touch the asymptote, tangent must go up and cotangent must go down.

Wednesday, April 16, 2014

BQ #2 Concept:intro

BQ #2 !!!!!!!!!!!!!!!

First question

First off, a period is how long it takes for a pattern to occur fully in. For sine quadrant 1 is positive, quadrant 2 is positive, qaudrant 3 is negative, and quadrant 4 is negative. Therefore, the pattern is + - + - . Since it takes the whole unit circle to complete this pattern it takes 2pie. For cosine qaudrant 1 is positive , 2 is negative, 3 is negative, and 4 is positive. So the pattern that occurs is + - - + . Again for the pattern to repeat , since there is no repeating unit like sine, it is 2pie. The first qaudrant of tangent is positive the second negative the third positive and the 4th negative. The pattern in + - + - . They pair + - repeats twice so the period is pie because it takes pie units to complete the cycle.

Question #2 
Amplitudes are half the distance between the highest and lowest points on the graph. Sine and Cosine graphs have these. Amplitudes are like restrictions. When taking the sine or cosine inverse , the number for these two trig identies are between -1 and 1 because they take the x/r and y/r. The other trig identies dont have restrictions at all and thats why they dont have amplitudes.

Thursday, April 3, 2014

Reflection#1: Unit Q Verifying Trig Identies

        1. It just meas to break it apart. Essentially that is what you are doing. They both mean the same thing in the end. There are different ways of doing it. Verifying is not done by some specific way. There are so many different thought processes to do it. Some take a lot of time while othere take minimal time. Pay careful attenton to your work. Memorize the identities.
        2. Memorize!!!!!!!! Once you memorize the trig functions, everything else comes together. When your stuck just convery everything to sin and cos that helps. Take everything apart step by step. Another trick that is helpful is to multiply by the conjugate when its a fraction and you don"t know what to do. NEVER divide by an identiy. Most likley when you have the urge to do that just multiply by the conjugate. Verifying is the easiest because you already know the answer, all you need to do is find a way there. Try different identities. Break it apart.
        3. When i first look at a problem the first thing I think of is sin and cos. They are easiest to deal with when they are in sin and cos. HOWEVER, always check if theres an identity there. For example any of the pythagoreom theroems can be hidden etc. Next i hope that things start to cancel or if there is a fraction multiply by conjugate. Most likley if things get messier you are not doing it right. I feel like most of it is trial and error. You plug and chug and eventually you get yur answer. Patience is key