BQ#7- Unit V

Where does the formula for the difference quotient comes from?

The difference quotient formula comes from the secant line that touches the graph twice meaning there's two points. The first point is ( x , f(x)) and the second point is (x+h , f(x+h)), from there we find the slope of the secant line which is y2-y1/x2-x1. We substitute the y and x with the given points from the secant line and will turn out to be f(x+h) - f(x)/x + h -x. Then we see that there are variables that cancel each other -x and +x from the denominator then just leaving "h". That's how you get the difference quotient formula f(x+h) - f(x)/h. 

BQ#6- Unit U Concept 1-8

1) What is continuity?
Continuity is predictable and it has no breaks, no holes and no jumps. A continuity function can be drawn with a single, unbroken pencil stroke for example the limit as x approaches a number of f(x) is equal to a number, the intended height. 

What is discontinuity?

Discontinuity is the opposite of continuity meaning it's interrupted and the limit does not exist. There are two families of discontinuity, removable and non removable. Removable has point discontinuity meaning there's a hole in the graph. Non removable has jump-it jumps from one graph the other-,oscillating-the graph is wiggly meaning it never reaches a point- and infinite-when it has vertical asymptotes due to unbounded behavior. 

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 a function when a value is the actual height of a function. The limit exist when the left and right behavior are the same, when there's no unbounded behavior and when there's no oscillating behavior. A limit doesn't when the left and right bahavior are different, there's an oscillating behavior and there's unbounded behavior. 

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

We evaluate limits numerically, verbally and algebraclly. Numerically means its set on a table to find the intended the height of a function. Verbally is stateting the limit statement as the limit as x approaches a number of f(x) is equal to a number. Algrebraclly has three methods substitution, dividing & factoring out and rationaling & conjugate. 

Direct substitution is when it's substituted by approaching number and has three answers numerical-2-, zero-0/2-, and undefined-1/0. 

When its inderteinate meaning its 0/0 then the dividing & factoring method is use. In this case we factor out both the numerator and denominator to cancel out common terms. Then substitute and get either numerical-2-, zero-0/2-, or undefined-1/0. 

In rationaling & conjugate is when we conjugate the numerator or denominator depending where the square root is at. After common terms are cancel then we substitute to get the answer. 

Work cited 

SSS packet

BQ#4 - Unit T Concept 3

Tangent goes uphill because its positive in quadrants one and three and negative in quadrants two and four. Its positive in zero uphill and pi uphill etc. The asymptotes are in zero, pi/2, pi, 3pi/2 and 2pi because x is zero. The tangent graph starts at negative and goes uphill to be positive and to go to the next quadrant.
Cotangent goes downhill because its postive in quadrants one and three and negative in quadrants two and four, just like tangent. Its negative downhill pi/2 and 3pi/2 etc. The asymptotes are in zero, pi, and 2pi because y is zero. The cotangent graph starts positive and goes downhill to be negative its like the reciprocal of tangent.

BQ#3 - Unit T Concepts 1-3

First of all sine and cosine dont have asymptotes when the other graphs do have asymptotes. Tangent related to cosine when cosine is zero because it becomes undefined and the asymptote are in pi/2 and 3pi/2. For cotangent related to sine when sine is zero because it becomes undefined and the asymptotes are in 0 and pi. For secant when cosine is zero then secant is undefined so there's going to be asymptotes at pi/2 and repeat every pi unit. Finally for cosecant when sine is zero then cosecant is undefined like secant, so there's asymptotes at pi and continue every pi unit.

BQ#2 - Unit T Intro

The quadrants from the unit circle, when placed horizontally in numerical order, create the basis for the trigonometric graphs. Since sine is positive in quadrants A and S and negative in quadrants T and C, when unwrapping it starts positive and positive then negative and negative and start again in another revolution.  For cosecant its positive in  quadrants A and C and negative in quadrants S and T, when unwrapping it starts at positive then negative then negative and end positive. That's how trig functions relate to the unit circle. 

The reason sine and cosine is 2pi because it needs all four quadrants to repeat unlike tangent and cotangent needs only half that why it's just pi. 

Sine and cosine has an amplitude of 1 because it can only be between -1 to 1 unlike the others there's no limit so their amplitude can greater than 1. 

Unit P concept 3-5

3. Law of Cosines
Why do we need it? How is it derived from what we already know?

We need law of cosines to find the third side when knowing two sides and an angle (SSA) or knowing all three sides (SSS) to find the angles.

We got a triangle label ABC that's not a right angle. Angle A's point is (0,0), angle B's point is (ccosA,csinA), ccosA is the x value because cosA equals d/c and csinA is the y value because sinA equals h/c. Angle C's point is (b,0), b is the distance across from angle B between angle A and C. Draw a line down angle B to make h, h is the drawn height of the triangle. With that it makes a 90 degrees angle. Then label side c and d also ccosA. Side c is the distance across from angle C between angle A and B. Side d is the distance across from angle B in just the right triangle. Side a is the distance between angle B and C (see picture 1). Use the Pythagorean Theorem in triangle CBD but substitute h to csinA and r to ccosA. Foil to get b^2-2bccosA+c^2cos^2A (see picture 2). Put cos^2A + sin^2A in parenthesis, since since cos^2A + sin^2A equals one. Finally get the formula (see picture 3), it can also be used to produce other letters statements (see picture 4). That's how we derive law of cosines.

Picture 1

Picture 2

Picture 3

Picture 4

Refrence:
http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

5. Area formulas
 Draw out a right triangle with angle A equal to 35 and angle B equal to 65. Side b equal to 4. Use the law of sines to get side b and c. Use the area of an oblique triangle, and Heron’s area formula to get both areas (see picture 5).

Picture 5

 Work cited

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm