Week 11

This week, we continued on in chapter 4, Trigonometric Functions.

On day 1, we looked at Graphs of Sine and Cosine Functions. 

An equation for Sine would look like:

y= a sin b (x-c) + d

a stands for= amplitude: amplitude is the max and min or how high it goes up

b stands for b in period: Period is 2π/b. A period is the length of one cycle. 

c stands for SlOP: Slop is slide opposite and if the number is negative. Move to the right and vice versa.

d stands for vertical shift: which is moving up or down on a graph.

On the second day, we continued on this and looked at how to find all x-intercepts of sine and cosine.

To find the x-intercepts for sine: Let bx+c = nπ
To find the x-intercepts for cosine: Let bx+c= π/2 + n

Below are the graphs of sine and cosine:

The Graphs are very similar but the only difference is where the x-intercepts are and where the graph starts.

On the next day we then looked at Other Trig Functions and we looked at Tangent.

The Tangent Graph is very different from sine and cosine as sine and cosine do not need asymptotes nor does it move upward for infiniti. The Tangent Graph moves upward and only at the origin does it change. Below is what the tangent graph would look like

y=tanθ

The graph makes a sudden twist at the origin and then goes on forever without touching the asymptotes and then more graphs are created to the right and left of each graph. 

the tangent equation is: y = a tan (bx+c) + d

To find the asymptotes of an equation you would use the equation:

π/2 + kπ

or -π/2b and π/2b

to find the period of the equation: π/b

To find the x-intercept you put (bx+c)=nπ

a is the amplitude and the amplitude of the equation is where the turn is created. For the equation above, the amplitude is 1 and so it turns at 1 and -1.

This is what was covered for the tangent graph.

The next trig function is cotangent. 

Cotangent is practically the same thing as tangent but there are a few or multiple differences. One difference is that cotangent goes in the opposite direction. 

For cotangent: the asymptotes of the graph is where the origin of each graph is. 

Below is a picture of cotangent

y= cotθ

What is suggested is that you graph the problem as if it was tangent and then convert it into cotangent

For example: y = 7 cot x

You would graph this as y = 7 tan x and then you convert it to cotangent the Graph is in the switched direction of the tangent graph drawn and the origin or where it meets the x-axis is also the asymptotes for a regular tan graph.

Cotangent is practically the same thing as tangent but switched.

Next is Secant. Secant is the opposite of cosine. 

The equation for secant is y = a sec (bx+c) + d

What makes secant different from all the other graphs is that it does not exactly connect just as tan and cotangent do not.

Below is a secant graph:

y= sec θ

In the blue is the original cosine graph and the walnut color is secant.

Secant practically is just parabolas over the turns of each one.

It is also recommended to sketch the cosine graph prior to actually drawing secant as that is the only way to find it. The asymptotes are where the lines touch the x-axis or the midpoint.

Whenever there is a vertical shift: draw a fake x-axis and draw asymptotes from where cosine touches it. Then you draw the secant graph.

The next trig function is very similar to secant but it is cosecant. It follows the sine graph instead and does exactly the same thing as secant and you would always draw a graph of sine first before putting the cosecant graph up. 

Below is a graph of the cosecant graph:

y = cscθ

It is exactly the same as secant but it is on a sine graph. You would continue to do this until you must stop graphing. This is how you solve a cosecant graph.

Those are the trig functions used and from these many other functions appear. 

This is what we did this week and HAPPY HALLOWEEN

SEE YOU NEXT WEEK!

Week 11: Tangent Graph

This week we covered more Trigonometry and the blog this week is on the Tangent Graph.

The Tangent Graph is very different from sine and cosine as sine and cosine do not need asymptotes nor does it move upward for infiniti.

The Tangent Graph moves upward and only at the origin does it change. Below is what the tangent graph would look like

y=tanθ

The graph makes a sudden twist at the origin and then goes on forever without touching the asymptotes and then more graphs are created to the right and left of each graph. 

the tangent equation is: y = a tan (bx+c) + d

To find the asymptotes of an equation you would use the equation:

π/2 + kπ

or -π/2b and π/2b

to find the period of the equation: π/b

To find the x-intercept you put (bx+c)=nπ

a is the amplitude and the amplitude of the equation is where the turn is created. For the equation above, the amplitude is 1 and so it turns at 1 and -1.

This is one of the trig functions used. All the other trig functions will be mentioned in the next post. 

This is what was covered for the tangent graph.

SEE YOU NEXT POST!

Week 10

This week we started Chapter 4, Trigonometric Functions.

We started on Day on, Angles and their Measurements. This lesson was review from Algebra 2 except for a few elements that was added in.

Trigonometry is practically all about triangesl inside circles and that is what we will be looking at for the next few weeks and months.

We began at standard position, which is the initial side on the positive x-axis.

We then looked at coterminal angles, which are angles that have the same terminal line.

ex: 960°

this could be 240° or even -120° 

Then Minutes and Seconds 

1°= 60' (minutes)

1°=3600" (seconds)

We then looked at Radian Measures.

θ= S/r

S= Sector 

θ=angle

r=radius

To convert degrees to radians:

You times the degree by π/180

radians to degrees: 

180/π

We then looked at complementary and supplementary

For Complementary:

you add 90°

or add π/2

ex:  θ= 74.23°

74.23 + x = 90

You then solve for x and you find the complementary angle.

For Supplementary:

you add 180°

or add π

ex: π/3 radians

π/3 + x = π

You then solve for x 

Lastly, we looked at arc lengths: Arc lengths.

Degrees has to converted to radians beforehand to be solved

The equation is: S=rθ

Almost all of this is review from Algebra 2 except for a few, but they are mostly self-explanatory.

On day 2, we looked at Sine and Cosine Functions. 

We had a look again at the identities of the Sine and Cosine Functions from last year. 

Just one thing before we start: COS= x    Sin= y.

Below is a graph of what of a triangle and how sine and cosine works.

On the side are a few of the Identities:

sin θ = opp/ hyp         csc θ = hyp/ opp

cos θ = adj/ hyp         sec θ = hyp/ adj

tan θ = opp/ adj         cot θ = adj/ opp

other identites: 

tan θ= sin θ / cos θ

cot θ= cos θ / sin θ

csc θ= 1 / sin θ

sec θ= 1 / cos θ

An important identity is the pythagorean thereom: a^2+b^2= c^2

It practically is 

x^2+y^2=r^2

x^2+y^2=1

this can translate to 

cos^2 θ + sin^2 θ = 1

This is called the Pythagorean Identity.

After this we looked at the acrynom: All Students Take Calc.

From this acrynom, it states where the functions are postive. in the A box, all the functions are positive. In S, Sine and cosecant are positive. In T, tangent and cotangent are positive. In C, Cosine and Secant are positive. 

Next we looked at reference angles: To find the reference angle, you just drop a line from the angle and you find the measurement from the angle. 

Some More Identities:

cos(-t) = cos t

sin(-t) = -sin t

cos(π/2 -t)=sin t           sin(π/2 -t)= cos t

cos(t + π) = -cos t        sin(t+π) = -sin t

cos(π - t) = -cos t        sin(π - t) = sin t

Now to solve problems: 

 ex 1: 

If you were looking for sin θ: cos θ= -2/5

Point, P in II Quadrant

1. you first sketch it on a graph

2. sinθ =O/H

a^2+ (-2)^2 = 5^2

a^2 + 4 = 25

a^2 =21

a^2 = √21

sinθ = √21 / 5

ex 2: Angle: -7π/4   Find the Reference Angle

1. Drop a line to the x-axis from the angle

It would then be 9π/20

This is what we looked at this week. We reviewed after this and then took a quiz on the last day.

SEE YOU NEXT WEEK!

Week 10: Sine and Cosine Functions

This week, we looked at Sine and Cosine Functions. 

We had a look again at the identities of the Sine and Cosine Functions from last year. 

Just one thing before we start: COS= x    Sin= y.

Below is a graph of what of a triangle and how sine and cosine works.

On the side are a few of the Identities:

sin θ = opp/ hyp         csc θ = hyp/ opp

cos θ = adj/ hyp         sec θ = hyp/ adj

tan θ = opp/ adj         cot θ = adj/ opp

other identites: 

tan θ= sin θ / cos θ

cot θ= cos θ / sin θ

csc θ= 1 / sin θ

sec θ= 1 / cos θ

An important identity is the pythagorean thereom: a^2+b^2= c^2

It practically is 

x^2+y^2=r^2

x^2+y^2=1

this can translate to 

cos^2 θ + sin^2 θ = 1

This is called the Pythagorean Identity.

After this we looked at the acrynom: All Students Take Calc.

From this acrynom, it states where the functions are postive. in the A box, all the functions are positive. In S, Sine and cosecant are positive. In T, tangent and cotangent are positive. In C, Cosine and Secant are positive. 

Next we looked at reference angles: To find the reference angle, you just drop a line from the angle and you find the measurement from the angle. 

Some More Identities:

cos(-t) = cos t

sin(-t) = -sin t

cos(π/2 -t)=sin t           sin(π/2 -t)= cos t

cos(t + π) = -cos t        sin(t+π) = -sin t

cos(π - t) = -cos t        sin(π - t) = sin t

Now to solve problems: 

 ex 1: 

If you were looking for sin θ: cos θ= -2/5

Point, P in II Quadrant

1. you first sketch it on a graph

2. sinθ =O/H

a^2+ (-2)^2 = 5^2

a^2 + 4 = 25

a^2 =21

a^2 = √21

sinθ = √21 / 5

ex 2: Angle: -7π/4   Find the Reference Angle

1. Drop a line to the x-axis from the angle

It would then be 9π/20.

This is what is Sine and Cosine Functions are like.

SEE YOU NEXT WEEK!

Week 9

This week, we decided to review the material from last week, for the class did not do the greatest on the test and Miss V wanted to view the mistakes made. We also took the PSAT this week rendering us to only have the class three times a week only. We took a quest: a bit harder than a quiz and not as hard as a test, to recuperate and to raise our scores from the test taken last week.

This is what we did during week 9

See You Next Week!!!

Week 8: Summary of Chapter 3

This week, we had one more lesson to look at till the end of chapter 3, Polynomial and Rational Functions, however we decided to not look at it until a later time due to it being connected to Calculus BC. Since we had a three-day weekend just recently, we only had three days of mathland. We reviewed this week mainly as we had a test on Friday on chapter 3.

To summarize Chapter 3, We began with Polynomial Functions. 

This lesson was simply review of Algebra 2 of End Behavior and Multiplicity. We also looked over even, odd and neither fuctions. 

The most important was multiplicity as it shows how many zeros are there of that one number. Usually this occurs if there is a quadratic function which creates the zero. 

We then looked at Division of Polynomial Functions. 

In this lesson, it is quite obvious we looked at how to divide polynomial functions. To divide you would have the the f(x) and divide it by d(x). To divide it is exactly just as you would divide regular numbers in long division but using x’s and exponents. Below is an example:

Just as shown above the equation’s last exponent on the right side ix x^3. To solve an equation, you would need to put 0’s to fill in the left over pieces. What is needed is shown in red above. You would divide with the number of exponents and number needed to divide. You times the number and put it under the exponent number and then make the numbers negative and you continue till you cannot divide anymore. Another type of division is Synthetic, which is shown below.

You can only divide if the polynomial x has one exponent. ex: x+4

You then take the numbers and plug in any 0’s needed and you divide down and then times the number at the bottom and put that number in the next column and repeat the same process to the end. If the last number is 0 then it is a factor and it has no remainder.

Sometimes you are looking for a remainder and you put the number from the divisor and plug it in for zero. That is the remainder thereom to find the remainder. 

We then looked at Zeros and Factors of Polynomial Functions.

In this one you are looking for the factors and the zeros of polynomial functions. You would divide to find all the zeros. The first polynomial’s exponent tells how many zeros there are. You then find all zeros and tell if they have multiplicity and also the factors. Sometimes the zeros would be imaginary and you would use the quadratic formula to find them. 

After this we looked at Real Zeros of Polynomial Functions.

What would happen if there are no zeros provided to find the rest of the zeros? You use this method which is P/S or Factors of Constant over factors of leading coefficient.

1. You find all the possible zeros from the P/S

2. Test the zeros using remainder thereom

3. Hint- not for test- use graph to see the possible zeros

We then looked at Approximating Zeros.

1. Divide the interval [a,b] in half by finding its midpoint, 

m=a+b/2

2. compute f(m)

3. if f(a) and f(m) have opposite signs, then f has a zero in teh interval [a,m]

if f(m) and f(b) have opposite signs then f has a zero in the interval [m,b] 

if f(m)= o then m is a zero of f. 

You also use the error equation to find out if you are close enough to a zero

error 1/2 (b-a)

the error would need to be  .0005 to be accurate enough to be a zero. 

After this lesson we looked at Rational Functions. 

This lesson was also review from Algebra 2 which consists of vertical asymptotes, horizontal asymptotes and holes. The new thing looked at is slant asymtotes.

To find a vertical asymptote:

set the denominator =0 and find x

To find the horizontal asymptote: 

fx)=anx^n/bmx^m

n=m   y=an/bm

n<m   y=0

n>m   DNE

n= degree of numerator

m= degree of denomator

an= LC of numerator

bm= lc of denominator

also if the HA does not exist (DNE) then there is a slant asymptote. 

To find Slant Asymptote:

Divide the denominator by the numerator and ignore the remainder.

y= answer

Holes:

 Factor the equation and cancel anything possible. What cancels out is a hole, to find the y-int of the hole you plug back in the x to find (x, y)

You would also need to find x-intercepts and y-intercepts:

 Just put 0 for the opposite number to find the answer.

To look more in-depth look at the past posts to understand it better. That is a summary of chapter 3.

See You Next Week!

Week 7

This week we looked again in Chapter 3, Polynomial and Rational Functions.

On Day 1, instead of starting another lesson, we looked again on what we looked at last week which included Polynomial Functions, Division of Polynomial Functions and Synthetic Division, How to Find Zeros of Polynomial Functions and Finding the Real Zeros of Polynomial Functions.

We reviewed on problems to see how they work and if we are doing the problems correctly.

On Day 2, We looked at Approximating Zeros. Some Zeros of Graphs are not Solid Numbers like 1, 2 or 3 but are instead decimals or fractions like -0.481923 or 3/47. In this lesson, we were trying to figure out how to find the zeros to the most approximate number as possible.

The steps to find them is:

1) Divide the interval [a, b] in half by finding its midpoint, 

m=a+b

     2

2) Compute f(m)

3) If f(a) & f(m) have opposite signs, then f has a zero in the interval [a, m]

If f(m) and f(b) have opposite signs then f has a zero in the interval [m,b] 

If f(m)=0 then m is a zero of f and the problem is done.

When solving the problem at the end, you would want the 

error< 0.0005

You find if you are done and find the error with the equation 

1/2 (b-a)

Below is a picture of how you would solve it. 

The problems may take a long time and you would do it until you find a number with 0.0005.

Sometimes you would find it with 1 dec which is < 0.05

2 dec < 0.005

3 dec < 0.0005

4 dec < 0.00005

However the most common one is 3 dec.

That is what we did on day 2.

On day 3, we looked at 3.6 which is Rational Functions. We looked at this in Algebra 2 and it is when you use Vertical and Horizontal Asymptotes. 

Usually the equations would look like 

f(x)=P(x)/Q(x)

To find the Vertical Asymptotes of a problem you would set the denominator=0 and the answer would be x= the answer.

To find Horizontal Asymptotes of a problem, you look at the degree of the first equation:

It would look like

f(x)=anX^n/bmX^m.

n= degree of numerator

m= degree of denominator

an= L.C. of numerator

bm= L.C. of denominator

It would be like n/m

To find the horizontal asymptote though, there are three possibilites

When the degrees are both equal. n=m you would use the coefficient of both answers as the answer. If its 2/2 then it would be y=1

Then n<m the answer would be y=0

n>m then the answer DNE.

However, something we have not seen before is Slant Asymptotes which is only possible if the horizontal asymptote is n>m.

f(x)=m(x)/d(x)

The answer from the vertical asymptote is used to divide numerator of the equation. For example x=2 is the VA. You would divide it by its numerator, for example x^2+4. y= answer.

For all equations, there will be x-intercepts and y-intercepts. You just plug in zero for the opposite to find it. If looking for the x-intercept, make y=0 and solve and vice versa.

To find the holes in a problem. You factor the equation and see if anything can be canceled out. Whatever cancels out is the answer and it would usually look like (x,y). If nothing cancels out, there is no hole. 

This is how Rational Functions work.

On day 4, we took a quiz on 3.4-3.6 to test if we knew the information. We then reviewed quickly on some of the material covered last week and the week before. 

That is what we did on Day Four. Next week we will have a test on Friday and will cover one more lesson before that. 

See You Next Week!

Week 7: Rational Functions

The blog topic for this week is Rational Functions. We looked at this in Algebra 2 and it is when you use Vertical and Horizontal Asymptotes. 

Usually the equations would look like 

f(x)=P(x)/Q(x)

To find the Vertical Asymptotes of a problem you would set the denominator=0 and the answer would be x= the answer.

To find Horizontal Asymptotes of a problem, you look at the degree of the first equation:

It would look like

f(x)=anX^n/bmX^m.

n= degree of numerator

m= degree of denominator

an= L.C. of numerator

bm= L.C. of denominator

It would be like n/m

To find the horizontal asymptote though, there are three possibilites

When the degrees are both equal. n=m you would use the coefficient of both answers as the answer. If its 2/2 then it would be y=1

Then n<m the answer would be y=0

n>m then the answer DNE.

However, something we have not seen before is Slant Asymptotes which is only possible if the horizontal asymptote is n>m.

f(x)=m(x)/d(x)

The answer from the vertical asymptote is used to divide numerator of the equation. For example x=2 is the VA. You would divide it by its numerator, for example x^2+4. y= answer.

For all equations, there will be x-intercepts and y-intercepts. You just plug in zero for the opposite to find it. If looking for the x-intercept, make y=0 and solve and vice versa.

To find the holes in a problem. You factor the equation and see if anything can be canceled out. Whatever cancels out is the answer and it would usually look like (x,y). If nothing cancels out, there is no hole. 

Below is one example of how to solve a rational function.

This is how Rational Functions work.

See You Next Week!