BQ #7: Unit V - Derivatives and the Area Problem

In order to find the difference quotient, we must work with secant lines. Now, what is a secant line? A secant line basically intercepts a parabola at TWO different points. In order to find the difference quotient, however, we need a tangent line. We get this tangent line by reducing the space between the two points of the secant line until the practically overlap. In the picture below, we see that "x" is the space between the origin and the first point of intersection between the parabola and the secant line. The space between the first and second points of intersection is called "h". This makes the second point "x+h". 

In Order to find the difference quotient, we use the equation m= (y2 - y1) / (x2-x1). Our values (as shown in the graph are;

y2 = f(x+h)

y1 = f(x)

x2 = x+h

x1 = x

With this in mind, we have a denominator of "x+h-x", which means that we cancel the "x" and are left with "h" in the denominator. We have a denominator of "f(x+h) -f(x)", which leaves us with the difference quotient of "f(x+h) -f(x) / h".

BQ #6 : Unit U

1. What is continuity? What is discontinuity?

-In a nutshell, a continuity is basically a graph which is completely predictable.  These graphs are the types of graphs we have been dealing through almost all of algebra. A continuous graph has no points, holes, jumps, and always ends "landing"in the predicted location. On the other hand, a discontinuity is a graph in which you do NOT always end up in the location that is predicted. This can mean that there is either a point, a hole, or a jump. There are two kinds of discontinuities: removable and non-removable. If you wish to know what this means, then by all means, keep reading.

Removable:
The Point discontinuity shown below is known as a hole, is characterized by the hole that is has in he middle, and is the only Removable discontinuity. This is because the graph is technically continuous. This is because you go from negative infinity to positive infinity.

http://bfreshrize.files.wordpress.com/2012/01/unknown.jpeg?w=500

Non-Removable
The Jump discontinuity shown below is characterized by its similarity to the hole discontinuity. However, upon close inspection, we can see that it different because it resemble a cliff; thus the name "jump discontinuity".

http://upload.wikimedia.org/wikipedia/commons/e/e6/Discontinuity_jump.eps.png

The Oscillating Behavior discontinuity shown below is characterized by its wiggly aspect.

http://webpages.charter.net/mwhitneyshhs/calculus/limits/limit-graph8.jpg

The infinite discontinuity is practically what we know as a vertical asymptote. This discontinuity is known as unbounded behavior and exists when there is a vertical asymptote.

http://web.cs.du.edu/~rjudd/calculus/calc1/notes/dis3.png

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 location that you planned to reach. The limit exists whenever you reach the same location when approaching from both Left and Right. THE LIMIT DOES NOT EXIST!!!!!!!!!1 whenever you do not reach the same location. For example, if Mrs. Kirch and Mrs. Lund intend on meeting in front of the ASB window and they end up in front of the window, then the limit exists. However, if Mrs. Kirch somehow ends up in front of the attendance window while Mrs. Lund waits in front the ASB window, the limit does not exist. This means that in a jump discontinuity, the limit doesn't exist because you reach different locations when going from left and right. In infinite discontinuity, you never reach the intended location because the vertical asymptote is there. In the oscillating behavior, we do not have a limit because you graph never approaches a specific point on the graph. The difference between a limit and a value and the limit is that the limit is the INTENDED location while the value is the ACTUAL location you reach.

3. How do we evaluate limits numerically, graphically, and algebraically?

In order to evaluate the limit numerically, you start out with the intended "height" and fill in the rest of the table. Usually, we have the intended height in the middle, and then three blank boxes that signify where the graph crosses (x axis). In order to solve graphically, we look at the limits we are asked to find. Then, we look at the graph and find the location we are asked to find, then simply use our knowledge of discontinuities to solve. For example, we know that in a jump discontinuity the limit doesn't exist because you approach two heights and in the point discontinuities, the limit would be the hole. In order to solve algebraically, we ALWAYS try directly substituting the given limit. If that doesn't work out, then we factor both the numerator and the denominator (if both are possible) and then just go wild and (in an adrenaline pumping fashion) eliminate those common terms. When we have fully reduced, we plug in the limit to find the answer.  Another way is to us the Rationalizing and Conjugate method. In order to do this, we simply multiply the conjugate of either the top of the bottom (it depends on where the radical is). Then, we multiply the outer and inner parts of the numerator. We NEVER EVER EVER EVER foil out the denominator. This will just give you a problem that is a godzillion times more difficult. Anyways, we simply reduce and then plug in the limit. Keep in mind that we always try to directly substitute.

BQ # 4 - Unit T Concept 3

Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill?

Alright, so a big reason for this would be our asymptotes. As we know, the trigonometric functions are tan(x) = sin(x) / cos(x), and cot(x) = cos(x) / sin(x). As we know, tangent has asymptotes whenever cos(x) = 0, and cotangent has asymptotes when sin(x) = 0. To make this simpler, we know that cos(x) = x / r, and sin(x) = y / r. This means that both x and y have to be 0. For tangent, we know the asymptotes are located at 0 pi, 1 pi, and 2 pi, while cotangent has asymptotes at pi / 2 and 3 pi /2. The reason that tangent goes uphill, however, is because in quadrant two, where the graph begins, is because cosine is negative. This places our first 1/2 of the graph in the negative territory. Tangent goes into the positive section in quadrant three, because both the x and y values are negative in the third quadrant, meaning that you divide two negatives, which gives you a positive. The opposite can be said for cotangent. Since our first asymptote for cotangent is on 0 pi, our graph starts in quadrant one. Since both cosine and sine are positive in quadrant one, we start out in the positive section. As we go into the second quadrant, only sine is positive while cosine is negative, which means that cotangent is negative. Then, we run into our asymptote. As we restart our graph in quadrant three, both cosine and sine are negative, giving us a positive graph. In quadrant four, only cosine is positive while sine is negative, making cotangent negative.

BQ #3 Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each of the others?

In the graphs shown below, the highlighted sections are Quadrant I (red), Quadrant II (green), Quadrant III (orange), and Quadrant IV (blue). We know this because in the Unit Circle, Quadrant I is 0 pi to pi/2, Quadrant II is pi/2 to pi, Quadrant III is pi to 3pi/2, and Quadrant IV is 3pi/2 to 2pi. This is because, as we know, the Trigonometric graph is ultimately an outstretched version of the Unit Circle. Sine is red, cosine is green, tangent is orange, and cotangent blue.

DESMOS GRAPH CREATED BY MRS. KIRCH

Tangent:
Tangent is similar to sine and cosine because (first of all) tangent is tan(x)= sin(x)/cos(x). Since both sine and cosine are positive in the first quadrant, tangent is also positive because you end up dividing a positive by a positive. Looking at the second quadrant, we see that sine (the red line) is also positive, while cosine is negative. This means that tangent becomes negative. Looking at quadrant three, we see that both sine and cosine are negative, so tangent is positive. In quadrant 4, sine is negative and cosine is positive, so tangent is negative as well. Also, we know that we have asymptotes when when we have an undefined answer, which is basically when cos=0. Cosine=0 at 0 degrees, 90 degrees, and 270 degrees. In terms or radians, those values are, (respectively), pi/2 and 3pi/2. This means that we have a tangent graph whenever sine touches the x-axis, but not when cosine touches the x-axis.

DESMOS GRAPH CREATED BY MRS. KIRCH

Cotangent:
Being the inverse of tangent, the trigonometric identity of cotangent is cot(x)= cos(x)/sin(x). In quadrant one, both cosine and sine are positive, making the blue cotangent line positive (please ignore the orange cotangent line which I forgot to delete from my graph). In quadrant number 2, since cosine is negative while sine is positive, then cotangent is negative in the second quadrant. Cotangent is positive in quadrant three since both sine and cosine are negative, and cotangent is negative in quadrant four since sine is negative while cosine is positive. As for the asymptotes, we know that we have asymptotes when our answer is undefined. This is only when sin=0, which happens at 0 degrees, 180 degrees, and 360 degrees, which are, in terms of radians, o pi, pi, and 2pi.

DESMOS GRAPH CREATED BY MRS. KIRCH

Secant:
We know that the trig function of secant is sec(x)= 1/cos(x). This means that our graph for secant will from, and be the inverse of, cosine. Knowing this, we can predict that the secant graph will sprout from the cosine graph, and this is verified by the graph above, where both our secant and our cosine graphs touch. Wherever cosine has mountain, a secant valley touches it, and wherever cosine has a valley, a secant hill touches it. Looking at quadrant one, we see that since cosine is positive, secant is positive. In both quadrant two and quadrant three, cosine is negative, which means that secant is also negative. Lastly, when we look at quadrant four, cosine is positive, making secant positive as well. Now, let's talk asymptotes. We know that cosine has to be 0 in order to have the undefined answer that would give us our asymptote. Since we know that cosine is adjacent/hypotenuse, we know that this asymptote happens whenever x is 0, and this happens at pi/2 and 3pi/2. We have asymptotes in these two areas, meaning that the secant graph can't cross those x-values.

DESMOS GRAPH CREATED BY MRS. KIRCH

Cosecant:
As we all know, the trigonometric function of cosecant is csc(x)= 1/sine(x). Just like in the secant graph, our cosecant graph sprouts directly from the sine graph. This means that wherever sine has a mountain, cosecant has a valley. Wherever sine has valley, cosecant has a mountain. In quadrant one and quadrant two, cosecant and sine are positive. In quadrant three and quadrant four, both cosecant and sine are negative. Let's go over those asymptotes. We know that sine has to be equal to 0. This happens at 0 pi, pi, and 2pi. These are the locations which the secant graph will NEVER cross

BQ # 5 - Unit T Concepts 1-3

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

The reason that sine and cosine do not have asymptotes is because their ratios are (y/r) for sine and (x/r) for cosine. As we know from the Unit Circle, the letter "r" is the hypotenuse, which is equal to . This means that we will never divide by zero (or any other number, actually), so we will never have an undefined answer, which, as we know, is the only time we have an asymptote. On the other hand, we have Cosecant, Secant, Tangent, and Cotangent. The ratios for these functions are (respectively): (r/y), (r/x), (y/x), and (x/y) since we can have values of x=0 as well as y=0, we would divide by zero in these four trig functions. This would give un an undefined answer, meaning we have asymptotes.

BQ#2 - Unit T Concept Intro

How do the trig graphs relate to the Unit Circle?

Being near the end of the second semester in this mighty Math Analysis class, we are all familiar with the seemingly endless circle of power that is the Unit Circle. This same Unit Circle can be transformed into the trigonometric graphs we have all come to love greatly by now. We achieve this by simply unwrapping the almighty Unit Circle and placing it in the shape of a simple graph.

-Period? - Why is the period for the sine and cosine 2 pi, whereas the period for tangent and cotangent is pi?
The reason for this is simple. In the sine and cosine graphs, you have your "mountain" that takes up basically quadrants I and II, while the "valley" takes up quadrants III and IV. Since the pattern is technically +|+|-|-|+|+|-|- and repeat for sine, it takes you two negatives to get the back to the double + pattern, thus the 2 pi. In the case of cosine, the pattern is
+|-|-|+|+|-|-|+|. As you can see, it still takes 2 negatives before you can get to 2 positives, thus the 2 pi. In the case of tangent and cotangent, the pattern is +|-|+|-. This means that you only have to get through 1 negative before you get to the positive.

-Amplitude? - How does the fact that sine and cosine have amplitudes of one (and the other trig functions don't have amplitudes) relate to what we know about he Unit Circle? Basically, both Sine and Cosine are (respectively) y/r and x/r, which gives is basically y/1 and x/1, thus setting the the restriction of 1 and -1. On the other hand, Tangent and Cotangent are x/y and y/x. This means that we do not have any restrictions because we divide by numbers other than 1 and -1.

Reflection #1 - Unit Q, Concepts 1 & 5

#1: What does it actually mean to verify a trigonometric function?
-Well before explaining how to solve a trig identity, I'll explain WHAT a trig identity is. A trigonometric identity is a "rule" that is always true no matter what. It's kind of like saying that pi= 3.14 or that the constant of gravity on Earth is 9.80 meters per second squared. So basically, in order to solve the identities, we need to use these rules in order to either simply the problem as much as possible, or just prove that the problem given to you is true (depending on what the problem asks for). That being said, there is always a way to solve the trigonometric functions, and you will not have a no solution. You can have one solution, more than one, but you will always have at least one.
#2: What tips and tricks have you found helpful?
-When solving the functions, I have found that it is always helpful to evaluate the problem before you start on it. Looking at what you have and think about what you can plug into the problem is a good way to start. Once you have a specific identity that you can plug in, you just plug them in and then you start to either combine, cancel, or multiply by the inverse, etc.
#3: Explain your thought process and steps you take in verifying a trig identity.  Do not use a specific example, but speak in general terms of what you would do no matter what they give you?

-As mentioned in question #2, the first thing I do, is evaluate the problem. After I do this, I basically just tend to substitute in as much as possible, and then factor out. By doing this, I make my life easier, because I can just eliminate like terms. Then, you can usually re-substitute some of the identities and eliminate further. After this, you are usually done with the problem.

SP#7: Unit Q Concept 2: Finding All Trig Functions When Given One Trig Function and Quadrant

Please see my SP 7, made in collaboration with Brian Chuong, by visiting their blog here. Also, be sure to check out the other awesome posts on their blog.

I/D #3: Unit Q - Pythagorean Identities.

Inquiry Summary Activity:

1) sin²x + cos²x = 1 is derived from the Pythagorean Theorem, which is, x² + y² = r².


When we divide r² from both sides, we get "1" on the right, and (x² / r²) + (y² / r²) = 1. We can make this into (x / r)² + (y / r)² = 1. We know that x/ r is sine, and y/r is cosine, so therefore we have sin²x + cos²x = 1.

2)

In the first picture, you divide cosine from each side, and end up with cos²x / cos²x + sin²x / cos²x = 1/ cos²x. this simplifies into 1 + tan²x = sec²x, and the second one is divided by sine, and you end up with 1 + cot²x = csc²x

Inquiry Activity Reflection

1) I connections that I see between Units N, O, P, and Q so far are that they all relate to one another. We start off by using the unit circle, then we basically derive the identities using the proofs in concept Q.

2) If I had to describe trigonometry in THREE words, they would be: Just Keep Practicing.

WPP #13/14: Unit P Concepts 6-7 - The Laws of Sine and Cosine

Please see my WPP 13-14, made in collaboration with Brian Chuong, by visiting their blog here. Also, be sure to check out the other awesome posts on their blog.

BQ#1 Unit P Concept 2 and 4

Law of Sines:

-Why do we need the need the Law of Sines?

We need the Law of Sines because it lets us solve non-right triangles.

-How is it derived from what we know?

When we are given something other than a non-triangle, we are faced with the challenge of being able to solve it. Because of that, we have to use the Law of Sines

The first step towards deriving the Law of Sines is to draw a line perpendicular to side b. Since we don't have a right triangle, we have to make one if we want to derive the law.

Now that we have two right triangles, we need a trigonometric function that involves the opposite of either A or C, and the hypotenuse. This being the Law of SINES, the function just so happens to be sine.

Since Sine is Opposite/Hypotenuse, we have sin C = h/a and sin A = h/a

In order to find h, we need to multiply c and a to both sides (of their respective equations). When we do this, we have a*sin*C = h and c*sin*A = h. Since both of the equations are equal to side "h", we can set them equal to each other.

When we set them equal to each other, we can divide both sides "c" and "a" in order to cancel them out and get the Law.

After you divide "c" and "a", you are left with sinC/c = sinA/a.

Area of an Oblique Triangle

In this triangle, you draw a line perpendicular from angle B to line b. In order to get your right triangle in order to find the area of the oblique triangle.

Like in the Law of Sines,  you take the Sine of C, in order to derive the are of the oblique triangle.

Once you take the Sine, then you multiply a to both sides, and get h = a*sin*C. Since you have an answer to what "h" is, then you plug h into the original area equation. The original equation is A = 1/2*b*h, then you have A = 1/2 b*(a*sin*C).

How does it Relate to the Original Equation:

This equation relates to the original equation, because the only thing that changes, is that you find the value of one of the variables, and then plug it into the original. 

WPP #12: Unit O Concept 10 - Solving Angle of Elevation and Depression Word Problem

Mark is spending a glorious weekend away from school, and is indulging in music. At a local venue, Mark's band was invited to play with multiple other local bands. When they arrive at the venue, Mark can't help but realize that the building is quite tall. 

A) If Mark is standing 10 feet away from the building, how tall is the building itself if the angle of elevation is 50 degrees?

B) Mark is now standing on top of the building. He looks across the street, and sees that the restaurant across the street has tables outside. In the starry night, his gaze is captured by a rather extremely attractive girl who appears to be his age and is sitting the equivalent of 30 feet away from him if he were to zip line over to her. Being the geek he is, the first thing that comes to Mark's mind is "How far away might this lady be?" (angle of depression is 40 degrees)

The Solution

I/D2: Unit O - How can we derive the patterns for our special right triangles?

Inquiry Activity Summary

1) 30-60-90 Triangle

In order to derive the 30-60-90 from an equilateral triangle whose sides all equal 1 (in this case), we start out by cutting it straight down the middle (vertically) in a way that the triangle becomes two identical triangles. Original, the triangle is made up of all 60 degree angles. By cutting it in half, the top angle become 30 degrees, and either the right or left angle (depending on which one you use) is the 60 degree angle, thus giving you the 30-60-90 Triangle.

In the last three pictures, you can see the way that the 30-60-90 triangle is formed. We know that each leg should be equal to 1, but we have one side that has to values of 1/2 when it is split. This is simply to show that when we split the triangle down the middle, you end up with 2 separate 30-60-90 triangles with legs being equal to 1, 1/2, and the b leg is equal to radical 3/2. The way you find the value of leg b is shown in the next picture.

Here, we have the Pythagorean Theorem. As you can see, we have 1/2 to the second power. This gives us 1/4, which we subtract from 1 on the other side. we are left with radical 3/4 which then simplifies to radical 3 divided by 2. 

Here, we have the 30-60-90 triangle, but we have placed "n" in each value to show that there is a proportion in the triangle. The values of each leg form a ratio, which remains the same regardless of what value you use for n. Since you multiply fairly, you multiply the same amount for "n", thus ending up with the same ratio you started with.

2) 45-45-90 triangle

In order to obtain this triangle from a square, we have to cut the square diagonally. Since we start out with a square, all sides are equal to 1 (the given value), and we simply need to find the value of the hypotenuse.

In these two last pictures, we see the transformation from the square to the triangle. Using what we know of the 45-45-90 triangle, we know that by cutting the square diagonally, your 90 degree angles in two of the corners will split in half to give you the 45 degree angles that we need. Then, we know that the legs directly opposite to the 45 degree angles are "n", which means that n=1 (since legs=1). We know that the leg opposite to the 90 degree angle is n radical 2, which means that the hypotenuse is radical 2.

This picture goes over the way to get the hypotenuse using Pythagorean Theorem. You get the same answer, but have a couple extra steps.

This picture shows the original square, with substituted into the values. Like stated in the 30-60-90, we use the letter "n" to show the ratio between all the legs, and just like in the 30-60-90 triangle, this ratio is the same regardless of what value you substitute for "n".

Inquiry Activity Reflection

Something I never noticed before about special right triangles is that you can derive them from either a square or a triangle of same values.

Being able to derive these patterns myself aids in my learning because I can quickly derive the Special Right Triangles during a test if I forget all else, and possibly pass the test.

I/D1: Unit N - How do SRT and UC relate?

Inquiry Activity Summary:

The activity we did in class basically used our knowledge of geometry to derive the Unit circle from Special Right Triangles of 30, 45, and 60 degrees. We basically had to set our hypotenuse equal to zero. Then, we had to do the same to the other two legs (simplifying the triangle legs fairly). Then, we labeled the triangle with the hypotenuse being r, the vertical leg being y, and the horizontal leg being x, and then placed the triangles in a coordinate plane with the the origin located at the angle measured ( to get the triangle in Quadrant 1. Finally, we labeled the vertices and found the ordered pairs.

1)

http://upload.wikimedia.org/wikipedia/commons/1/15/Triangle_30-60-90_rotated.png

This triangle above is the 30 degree special right triangle. The leg opposite to 30° is the y value (vertical), the leg opposite to 60° is the horizontal value (horizontal), and the leg opposite to the 90° angle is the r value (hypotenuse). In order to find our ordered pairs, we had to set the hypotenuse equal to 1. In other words, we had to divide by 2x. If you divide 2x from the hypotenuse, then you need to divide all of them by 2x. When you do this you are left with x=√3∕2, y=1∕2, and r=1. The ordered pair is (√3∕2 , 1∕2)

2)

http://mathcountsnotes.blogspot.com/2012/05/special-right-triangles-30-60-90-and-45.html

The triangle above is the 45° special right triangle. The leg opposite to the lower 45° angle is the y value (vertically), the leg opposite to the higher 45° angle is the x value (horizontal), and the leg opposite to the 90° angle is the r value (hypotenuse). Once again, we need the hypotenuse to be equal to 1, so we need to divide by √2x. When we do this all legs, the all the Xs cancel out, but you are left with radicals on the bottom so you need to rationalize. In order to do this, you multiply √2 to both the top and the bottom. When you multiply the bottom, the radicals cancel out, and you are left with √2∕2. This means that the values end up being x=√2∕2, y=√2∕2, and r=1. The ordered pair is (√2∕2 , √2∕2)

3)

http://mathcountsnotes.blogspot.com/2012/05/special-right-triangles-30-60-90-and-45.html

This triangle above is the 60 degree special right triangle. The leg opposite to 60° is the y value (vertical), the leg opposite to 30° is the horizontal value (horizontal), and the leg opposite to the 90° angle is the r value (hypotenuse). In order to find our ordered pairs, we had to set the hypotenuse equal to 1. In other words, we had to divide by 2x. If you divide 2x from the hypotenuse, then you need to divide all of them by 2x. When you do this you are left with x=1∕2, y=√3∕2, and r=1. The ordered pair is (1∕2, √3∕2)

4) This activity helps us derive the unit circle because these are basically the three triangles found in the unit circle. All that you need to do is move the triangles around into the other quadrants in order to find the rest of the unit circle.

5) All the triangles in this activity lie in Quadrant I. If we were to transfer the triangles into the other Quadrants, only the signs of either the X or Y values would change. In other words, instead of having a positive X, you would have a negative X, and so on.

http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_45.gif

In this triangle we see that the ordered pair has remained almost the same for the 45° angle. The only thing that has changed is that the ordered pair is no longer (x , y) but rather (-x , y) due to being located in Quadrant II.

http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_34.gif

In this triangle we see that the ordered pair has once again changed slightly from original form for the 30° angle. Instead of having an ordered pair whose signs are (x , y), we have and ordered pair whose signs are now (-x , -y) due to being in located in Quadrant III.

http://02.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_21.gif

In this triangle we see that the ordered pair has (like in the last two) changed slightly for the 60° angle. instead of having the ordered pair of (x , y), we have the ordered pair being (x, -y) due to now being located in Quadrant IV.


Inquiry Activity Reflection:
1) The coolest thing I learned from this activity was the Special Right Triangles. I personally do not remember learning this or the Unit Circle in my past two years of taking math.
2) This activity will help me in this unit because I can draw out the triangles in Quadrant I, label the ordered pairs, and write out the "ALL STUDENTS TAKE CALCULUS" and I'll have the whole Unit circle.
3) Something I never realized before about Special Right Triangles and the Unit Circle was that they existed.

WORK CITED:

http://upload.wikimedia.org/wikipedia/commons/1/15/Triangle_30-60-90_rotated.png

http://mathcountsnotes.blogspot.com/2012/05/special-right-triangles-30-60-90-and-45.html

http://mathcountsnotes.blogspot.com/2012/05/special-right-triangles-30-60-90-and-45.html

http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_45.gif

http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_34.gif

http://02.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_21.gif

Real Wold Application - Ellipses

Ellipse

1)Definition: an ellipse is the set of all points such that the sum of the distances to two fixed points (known as the foci) is constant.

2)Algebraically, an ellipse's equation looks like (x-h/a)^2 + (y-k/b)^2 = 1 or (y-k/b)^2 + (x-h/a)^2 = 1. You have to keep in mind that both terms are squared, and x always goes over a and goes with h, while y goes over b and goes with k.
Here you can see the ellipse being drawn out

http://upload.wikimedia.org/wikipedia/commons/6/6a/Ellipse_construction_-_parallelogram_method.gif

3)Graphically: the ellipse is similar to an oval. It is egg shaped, and consists of 2 vertices, 2 co-vertices, 2 foci, 1 major axis, and 1 minor axis.Here, you can see these aspects

http://www.mathsisfun.com/geometry/ellipse.html

Real World Application: Ellipses can be found everywhere we look. From a glass of water to the way that planet and other things orbit in space, you can find an ellipse.
http://britton.disted.camosun.bc.ca/jbconics.htm

If you wish to learn more about ellipses and what make them up, you can click the link below
https://www.khanacademy.org/math/algebra/conic-sections/conic_ellipses/v/conic-sections--intro-to-ellipses

Work cited:
http://upload.wikimedia.org/wikipedia/commons/6/6a/Ellipse_construction_-_parallelogram_method.gif

http://www.mathsisfun.com/geometry/ellipse.html

http://britton.disted.camosun.bc.ca/jbconics.htm
https://www.khanacademy.org/math/algebra/conic-sections/conic_ellipses/v/conic-sections--intro-to-ellipses

WPP#10 Unit L Concept 9-14- Basic Probability, Events With and Without Replacement, Mutually Exclusive and Non-Mutually Exclusive, Combinations, and Probability

Create your own Playlist on LessonPaths!

WPP#9 Concepts 4-9- Probability

Create your own Playlist on MentorMob!