Week 18 – Top 5 things I have learned in Pre-calculus 11

June 18, 2018June 19, 2018 keishan2015

There are a number of new things that I learned this year in Pre-calculus 11 that really stuck with me and made way for those “ah ha” moments. This is a list of the top 5 things that I learned in Pre-calculus 11.

Acronyms: One of the things that i learned that really helped me in Pre-calculus 11 were a number of different acronyms. The acronyms allowed for me to easily remember what steps i needed to take when dealing with specific questions. They made things a lot easier for me. Two of the acronyms that I learned that really helped me are CDPEU, and CAST.

The CAST Rule: Starting in quadrant 4 and moving in a counter clockwise manner, will tell you which trigonometric ratios will be positive in the certain quadrants. This makes it easier for you if you are trying to find if your value will be positive or negative, since it depends on what quadrant it’s in.

Ex. Find what quadrants the terminal arm of the angles could lie.

$tan\Theta= -12$ –> Quadrants: 2 and 4

To find what possible quadrants a terminal arm could lie you use the CAST law. If we know that the tangent angle is negative then we know that it will have to lie in either quadrant 2 or 4. In quadrant one, all angles are positive, and in quadrant three, tangent is positive.

CDPEU (Can Divers Pee Easily Underwater): This short acronym helps you remember the steps that you must follow when dealing with factoring. First you must look for anything that may be Common.

ex.

$2x^2$ +8x [both have a common factor of 2x]
2x(x+4)

Then you check if it is a Difference of Squares. These only occur in binomials, and only work if the terms are separated by a negative.

ex.

$2x^2$ -9 [Because both are perfect squares and have a negative in between, this is a difference of squares]
(x+3)(x-3)

Next you would check if there is a Pattern. The pattern that is being referred to is the pattern of $a^2+bx+c$ , which is very common during factoring. This can tell you if you are dealing with a quadratic or linear function. Which would tell you if you are trying to isolate the variable or if you need to factor in order to find the “zeros”.

ex.

$x^2$ +5x+4 [because this follows the pattern this is quadratic and needs to be factored]
(x+4)(x+1)

The next letter in the acronym stands for Easy. This is when the expression has a factor of one in front of the $x^2$ , this is because we can easily factor the solution without worrying about the value of the $x^2$ .

ex.

$x^2$ -x-20
(x+4)(x-5)

The last letter in the acronym stands for Ugly. Ugly, refers to the expressions where the $x^2$ has a coefficient. This makes the expression harder to factor due to the coefficient since you need to be cautious of it when factoring.

Finding the Discriminant: The discriminant is taken from the quadratic formula and really tells you a lot about an expression. It allows you to find how many “zeros”/root/solutions an expression or equation has as well as if it is extraneous of or not. Finding the discriminant made questions a whole lot easier because I could just find the discriminant as one way of making sure that my answer(s) is correct. If the value of the discriminant is:

Positive: then that means that you will have to solutions/Negative: then that means that there are no solutions/Equal to zero: then that means that there is one solution

ex.

Since the equation is already written in the proper format of ( $ax^{2}+bx+c=0$ ), we need to write down the values for a, b, and c. (a = 2, b = 9, c = -4)
Next, you input the values into the formula.
Use BEDMAS to simplify the equation until you are left with a value.
Since the number you are left with are 113, and it’s positive, then we know that this quadratic equation will have two possible solutions.

Finding the Vertex: Being able to find the vertex of a quadratic function can tell you a lot about it. From the axis of symmetry to even the range, the vertex gives you a lot of information on the function. The vertex can be found by a number of different way. One method that really stuck with me however was completing the square.

Reciprocal Functions: Reciprocal functions are one of the top 5 things that I learned because it was something that was really new to me that I learned how to deal with. At first I found that its concept was really hard for me to grasp, but after a bit of practice they became easier for me to understand. I think that they are something worth talking about and are super interesting.

Ex.

First, we graph the linear function as we normally would if it wasn’t reciprocated. To do this we would write down everything we know about the original function and plot it on the graph. Since the original function is 2x+6, we know that the function will be positive and have a slope of 2, we also know that the y-intercept will be 6.
Next, you circle your invariant points which will be when y is at both 1 (_,1) and at -1 (_,-1). To do this go to where 1 is on the y-axis and then slide your finger vertically along the graph until you touch the linear function. You would do the same thing for -1.
Next you draw in the asymptotes. Since the horizontal asymptote will always be y=0, we draw a horizontal broken line along that part of the graph. Also, since the vertical line is half way in between the two invariant points, we know it will be when x= -3. Another way to find what the vertical asymptote is without looking at a graph is by making the original linear function equal to zero and solving for x.
After drawing in the asymptotes, you can now draw the hyperbolas. To do this you would take of the points along the linear function, reciprocate it and plot its reciprocal. You would then draw the hyperbola through the invariant and the new reciprocated points.

Note: when writing the domain or range x and y can be any real number except they can’t be equal to the corresponding asymptote.

Special Triangles: I think that the special triangle were very helpful. There are two special triangles , the 30°-60°-90° and the isosceles right triangle (45°-45°-90°), and since we know the sides of these triangles, we know their trigonometric ratios. If you come across any of these angles you will be able to refer to these two special triangles for your ratios instead of trying to calculate. Note: If one of the sides was multiplied by 2, you would need to do the same to the rest of them.

Week 17 – Trigonometry

June 10, 2018June 11, 2018 keishan2015

This week in pre-Calculus we learned all about Trigonometry and built upon what we had learned in math 10. I feel as though I really learned a lot in this unit from rotation angles to how to use the sine and cosine laws.

Terms to know:

Rotation angle: formed between the initial arm and terminal arm. A counter clockwise rotation will result in a positive (+) angle whereas a clockwise rotation will result in a negative (-) angle.
Initial Arm: The initial point in where you started measuring from. Your starting point of the arm.
Terminal Arm: the area where the measurement of the angle stops. Your ending position of the arm.
Reference Angle: Formed between the x-axis and the terminal arm.
Co-terminal: Angles with the same terminal arm
The CAST Rule: Starting in quadrant 4 and moving in a counter clockwise manner, will tell you which trigonometric ratios will be positive in the certain quadrants.

Special Triangles: There are two special triangles , the 30°-60°-90° and the isosceles right triangle (45°-45°-90°), Since we know the sides of these triangles, we know their trigonometric ratios. If you come across any of these angles you will be able to refer to these two special triangles for your ratios instead of trying to calculate. Note: If one of the sides was multiplied by 2, you would need to do the same to the rest of them.

Quadrantal angles: These are the angles where the terminal arm is located on the x or y-axis, ex. 0°, 90°, 180°, etc. r = 1, always.

Formulas To Know:

The new and improved SOH CAH TOA: *r = radius*

$sin\Theta=\frac{o}{h}\rightarrow\frac{y}{r}, cos\Theta=\frac{a}{h}\rightarrow\frac{x}{r}, tan\Theta=\frac{o}{a}\rightarrow\frac{y}{x}$

New and improved Pythagoras:

$x^{2}+y^{2}=r^{2}$

Sine Law: Works with any triangle

$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$ (for calculating side length) OR $\frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}$ (calculating angles)

Cosine Law: Works for all triangles, can re arrange to fit which parts of the triangle your are looking for

$a^{2}=b^{2}+c^{2}-2bcCosA$ (for calculating side length) OR $CosA=\frac{b^{2}+c^{2}-a^{2}}{2bc}$ (calculating angles)

Example 1:

Q: Find what quadrants the terminal arm of the angles could lie.

$tan\Theta= -12$ –> Quadrants: 2 and 4

To find what possible quadrants a terminal arm could lie you use the CAST law
If we know that the tangent angle is negative then we know that it will have to lie in either quadrant 2 or 4. In quadrant one, all angles are positive, and in quadrant three, tangent is positive.

Example 2: Sine Law

Q: $\angle C$ to the nearest degree

First, since we are trying to find the angle, we use the sine law formula that has the angles on the top and the side on the bottom of the fractions
Next, take all the information you know and put it into the formula. Note: the lower case letters represent the sides, the sides are named after the angle that is opposite of it.
If you take a look, you can see that one of the fractions has no information known about it, due to this we can take it out of the equation and use the other two fractions to help us find angle E. Now you rearrange the equation to isolate angle E. To do this you would multiply both sides by the denominator of the fraction that contains SinE; you would then need to get rid of the sin attached to the E, to do this you multiply the other side by the inverted sin.
Next, solve and round your answer to the nearest degree.

Example 3: Cosine Law

Q: To the nearest tenth of a cm, find side a

First, since we are trying to find one of the side lengths we would use the following formula: $a^{2}=b^{2}+c^{2}-2bcCosA$ , since this is written in the form that we need it in, we would leave it the same.
Next input all information you know about the triangle into the formula and the solve.
Take that answer that you get in your calculator and square root it to find what the value of a is.

Week 16 – Application of Rational Expressions

June 3, 2018June 4, 2018 keishan2015

This week in pre-calculus 11, our rational expressions unit came to an end. One thing I learned from that unit was the application of rational expressions in real life situations.

Things to know:

Word problems are filled with a lot of valuable information that is easy to look past one you read it onece.t is very important to read a word problem more than once, making sure to read slowly and carefully. It helps when you highlight or underline the important information in the question. Be sure to take note of key words such as less than, greater than, equal to, the difference of, the sum of, the product of, etc.
Due to the fact that it is a word problem, make sure that you write your final answer in a sentence always making sure to include the right units if necessary.
You should always check to make sure that your solution ‘makes sense’.

Example 1:

Q: Jake rows a boat 6km upstream in the same time that it takes the boat to travel 12km downstream. The average speed of the current is 5km/hr. What is the average speed of the boat in still water?

Based on the word problem we know the following information:

*Let x = the speed of the boat in still water

	Downstream	Upstream
Distance	12 km	6 km
Speed	X + 5	X – 5
Time	$\frac{12}{x+5}$	$\frac{6}{x-5}$

*we know that because the boat is moving up stream the speed of the boat will be a bit slow considering it is working against the current, hence the -5. We know that the speed of the boat going down stream will be a bit faster considering that it is moving with the current, hence the +5. We also know that time is equal to when distance is divided by speed.

After reading the word problem over again, we know that the two times are equal. Due to this information we can now write an equation.

$\frac{12}{x+5}=\frac{6}{x-5}$ –> cross multiply (fraction = fraction)
12(x-5) = 6(x+5) –> distribute the coefficient in front of the brackets
12x-60 = 6x + 30 –> Re-arrange so that the variable is on one side
6x = 90 –> isolate x by dividing both sides by 6
$\frac{6x}{6}=\frac{90}{6}$ –>Divide
x = 15 –>Rewrite as a sentence
The average speed of the boat in still water is 15km/hr.

Example 2:

Q: How much sugar should be added to 6L of water to make a solution that is 20% sugar?

We know:

$\frac{part}{total}$
Total = sugar + water
Part = sugar
20% sugar –> $\frac{20}{100}$ –> $\frac{1}{5}$
x = amount of sugar needed

Equation based on information:

$\frac{part}{total}$ = $\frac{1}{5}$
$\frac{1}{5}=\frac{x}{6+x}$ –> cross multiply
6+x = 5x –> Move variable to one side
6 = 4x –> Divide each side by 4 to isolate the variable
x = 1.5 –> Write answer in a sentence
1.5L of sugar is needed to make the solution 20% sugar.

Keisha's Blog

Month: June 2018

Week 18 – Top 5 things I have learned in Pre-calculus 11

Week 17 – Trigonometry

Week 16 – Application of Rational Expressions