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.

 –> 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.

• +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.

• -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 , 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.

• +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 , this is because we can easily factor the solution without worrying about the value of the .

ex.

• -x-20
• (x+4)(x-5)

The last letter in the acronym stands for Ugly. Ugly, refers to the expressions where the 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.

1. Since the equation is already written in the proper format of (), we need to write down the values for a, b, and c. (a = 2, b = 9, c = -4)
2. Next, you input the values into the formula.
3. Use BEDMAS to simplify the equation until you are left with a value.
4. 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.

 

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*

New and improved Pythagoras:

Sine Law: Works with any triangle

 (for calculating side length) OR (calculating angles)

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

 (for calculating side length) OR  (calculating angles)

Example 1:

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

 –> 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:  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: , 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.

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

*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.

• –> 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
• –>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:

• 
• Total = sugar + water
• Part = sugar
• 20% sugar –> –>
• x = amount of sugar needed

Equation based on information:

• =
• –> 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.

 

 

 

This week in Pre-Calculus 11 we continued to learn more about rational expressions. One new thing that I learned this week was how to add and subtract rational expressions.

Recall:

• Lowest Common Multiple (LCM):  the smallest number that two or more numbers divide evenly into. ex. The LCM of 2 and 3 is 6.
• What you do to the numerator you need to do to the denominator (what you do to the top you need to do to the bottom).
• In order to add and subtract fractions they need to have a common denominator.

Adding and Subtracting rational expressions:

• When adding and subtracting rational expressions it is important that they have a common denominator. This can be achieved by finding the LCM between the denominators of the fractions.

Example 1:

• First, you need to see if you can factor anything in the expression. In this case you would leave it as is since it can not be factored any further.
• Next, you look at the denominator and check if they are the same, if not you need to make it so that they have a common denominator. In this case since one denominator is 3x and the other is 5x, we know that the x’s are common but not the coefficient in front, we can fix this by multiplying each expression by the opposite coefficient. You would multiply the expression with 3x as the denominator by 5 and the one with 5x in the denominator by 3.
• After multiplying across the numerator and denominator, both expressions will have a common denominator. Due to this, you can write you expression as one big fraction since the denominators are now the same. You do this by combining the numerator of each expression (writing it as one continuous expression).
• Next you simplify the numerator if possible. In this case you simplify it by adding the two numbers in the numerator together.
• After you have completely simplified the expression, you need to make sure that you find the non-permissible values since the denominators contained variables. To do this you take a look at the factored form of the expression. In the factored form, the denominators were 3x and 5x, since this is the case we know that x≠0, otherwise the expression would be undefined since the denominator can never be equal to zero.

Example 2:

• First, you need to factor the expression if possible. If not then you would leave the expression as is.
• Next, you need to find the common denominator between the two expressions. In this case one expressions has a denominator of (x+2)(x-2) and the other has a denominator of (x+2), since both have x+2 in common, we know that the expression with the denominator of only (x+2) will need to be multiplied by (x-2) to make the denominators of the two expressions the same.
• Next, since we have multiplied to make the denominators the same, we can re-write the expression as one big fraction, by combining the numerators. Note, since this involves subtraction, the numerator of the expression that is being subtracted needs to have brackets around it. This is because of the negative in front that needs to be distributed in.
• After distributing the negative, simplify the expression if possible.
• After completely simplifying the expression, you need to find the non-permissible values, which you can find by taking a look at the factored form of the expression and the denominators of the factored expressions. In this case x≠ -2,2.

This week in Pre-calculus 11 we started a new unit that is all about rational expressions. At first we learned about equivalent rational expressions and then we learned how to multiply and divide them. What stuck with me was how to multiply and divide the rational expressions.

What to know:

• Rational Expression: The quotient of two polynomials ex.
• Non-Permissible Values: Similar to restrictions, Permissible values tell you what numbers x can not be equal to. These are only written if there are variables in the denominator, this is because we need to make sure that the denominator does not equal to zero since zero can’t be the denominator of a fraction. The non-permissible values are found using the zero product law. ex. x+8 = 0, x=-8 (when x = -8, the equation is equal to zero).

Example 1:

• First what you need to do is factor each fraction so that it is easier for you to simplify the expression.
• Next, simplify the expression. First what you want to do is need to check to see if there is anything in the numerator and denominator that matches (any pairs?), if so then you would cross them off since they cancel each other out, then once you have found all of the pairs you would multiply across both the numerator and denominator.
• After finding all the pairs in the question, you would re-write the fraction and see if their is anything that can be further simplified. Note: ex. expressions similar to this wouldn’t be simplified any further; this is because though it may be tempting to get rid of both 3’s, the x in the denominator is connected to the 3 by the +, this means that you would need to have something in common with both the x and the 3 to simplify it further.
• Next, since there are variables on the bottoms of the fractions, you find the non-permissible values. To do this we take a look at the original question (or its factored form) and use the product law and make the parts that involve variables equal to zero so that we can isolate and find x, these values will tell you what x can not be equal to.

Example 2:

 

• First, since this is a fraction being divided by a fraction we need to make it easier for us to solve. To do this you need to turn it into a multiplication problem, which is done by reciprocating the second fraction in the question.
• After reciprocating the second fraction, we can treat this like a multiplication question; so therefore you would factor the question, look to see if the numerator and denominator have anything in common (pairs) which you would then cross out since they cancel each other out, and then multiply across both the numerator and denominator.
• Next you would re-write the simplified expression and see if it can be further simplified.
• Now, if there are variables that are located in the denominator you need to find the permissible values. One way that dividing rational expressions differs from multiplying them is the non-permissible values. This is because when writing the non-permissible you need to take into account that both halves of the fraction that was reciprocated were the denominator at one point, therefore, you need to find the non-permissible values of both the numerator and denominator of the fraction that you reciprocated as well as the denominator of the fraction that you didn’t reciprocate.

This week in Pre-calculus 11 we learned more about absolute values and reciprocal functions. One thing that I learned that really stuck with me was when we learned about reciprocal functions and how to graph them.

Things to know:

• A positive reciprocates to a positive (+ –> +)
• A negative reciprocates to a negative (- –> -)
• When a big number is reciprocated it turns into a small number and vise versa.
• When both 1 and -1 are reciprocated they stay the same (called invariant points)
• When 0 is reciprocated it becomes undefined (you can’t have a 0 on the bottom of a fraction)
• Asymptote – horizontal/vertical line that separates the graph. The graph will approach both lines but never touch them. (the vertical asymptote is located halfway in-between the two invariant points/Horizontal will always be y=0).
• Hyperbola – two curves similar to each other, indicated by the similar points. Opposite of each other. Curved lines on the graph.
• The two hyperbolas will never touch due to the vertical asymptote

Example:

   

   

• 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.

This week in Pre-calculus 11 we had our chapter 5 unit test and started unit 8. Something that I learned this week was how to graph the absolute value of both linear and quadratic functions, as well as how to write them in piecewise notation.

Recall:

• When finding the absolute value of something it is asking how far away a number is from zero. The absolute value will always be a positive number.

Need to know:

• The critical point (also called the point of inflection) is the point along the x-axis where the graph of the function changes direction.
• Piecewise notation, the first part of piecewise notation is used to state which parts of the original line are positive and didn’t need to be reflected, uses signs (≤ or ≥). The second part is used to describe the part of the function that had to be reflected upwards, uses signs (< or >).

Example 1:

• First, you graph the function as if there were no absolute value symbols around it. When graphing it normally we know that the y-intercept is 4 and that the linear function has a slope of 2.
• Next we graph the absolute value of the function. Since we already know that absolute values will always be positive we take the part of the function that runs past the x-axis and reflect it upwards. Reflecting it upwards will put it above the x-axis, making it positive (it’s the same as multiplying the negative points by -1). The part of the line that is already in the positive stays the same.
• After graphing the absolute value of the function you need to write the piecewise notation. first we write a notation for the portion of the function that stays the same. Since the right side of the graph doesn’t change we would write the notation as, 2x+4, x ≥ -2 (-2 is the critical point and doesn’t change). For the second part we write about the part of the graph that was reflected, when reflecting the values, it is the same as multiplying by -1. we would write the notation as -(2x+4), x < -2 (we only use less than because the critical point never changed).

Example 2:

•  Graph the function as if there were no absolute value symbols around it. When graphing it normally we know that the vertex is (3,-2) and that the quadratic function has a scale of 2-6-10.
• Next we graph the absolute value of the function. To do this we take the part of the parabola that is in the negative and reflect it upwards so that it is positive. To do this all you need to do is change the sign of the y in the vertex to a positive –> (3,2). The part of the function that is in the positive stays that same.
• After graphing the absolute value of the function you need to write the piecewise notation. First we write the notation about the part of the function that remained the same, we know that , x ≤ 2, x ≥ 4 (we write about the sides of the function since they didn’t change). We then write about the part of the function that was reflected upwards (multiplied by -1). We know that , 2 < x < 4 (in between those two points is where the function changes).

 

 

This week in Pre-calculus 11 we learned all about inequalities. Something that I learned that really stuck with was how to solve linear inequalities and graph them.

Signs and symbols to know:

• > greater than / < less than
• ≥ greater than or equal to / ≤ less than or equal to
• An open dot is used to indicate that the boundary point isn’t apart of the solution / A closed dot is used to indicate that the boundary point is a part of the solution
• A dotted line is used to is used when the values along the line aren’t included in the solution [> or <] / A solid line is used when the values along the line are included in the solution [≥ or ≤]

Example 1:

• First, you need to make it so that one side is equal to zero.
• Next, you factor the inequality and find its zeros. You do this using the zero product law. In this case the zeros would be: -5 and 1.
• Next you would write these on a number line and pick three test numbers. You pick one number from each of the three sections of the number line. (In this case you would pick a number less than -5, one in between -5 and 1, and one greater than 1).
• after doing that you test the points by putting them into the inequality. This will help you write your solution. Out of the three test points there will be some that prove to be true and others that will not. You take the one(s) that are true and use them to write your solution. (In this case our test point in between -5 and 1 proved to be true).
• Write your solution. Since we know that the value of x is in between -5 and 1 we would write it as {-5<x<1}.

Example 2:

*Note: another method used to solve is a sign chart*

• First you need to factor the expression. In order to do this you need to make sure that one side is equal to zero. To do this you would just rearrange the inequality and move things from one side to the other.
• After factoring you find the zeros of the inequality. (in this case they would be -2 and 6).
• Next you would create a sign chart. To do this you create two number lines with the zeros written on them. You would then write one of the binomials beside one of the number lines and the one beside the other and circle the zero that corresponds with that binomial. For the first number line we would write in each section of the line if the solution would be + or -. If we know that putting -2 into the binomial (x+2) = 0, then we know that anything less than it would result in a negative and anything greater than it would result in a positive. For the second line we do the same thing. If we know that when we put 6 into the binomial (x-6) = 0, then we also know that any value less than 6 would result in a negative and anything greater than 6 would result in a positive.
• Next you multiply the signs that are across from each other. We know: [(-)(-)= +/ (+)(-)= -/ (+)(+)= +]. You then compare your inequality to the new values and use the sections that prove the inequality to be true to write your solution. (in this case you would compare the inequality to + – +)
• Then you write the solution. We know that the solution is -2≤x≤6 because the only time that zero is greater than or equal to x is when x is in the middle section (in between -2 and 6).

 

This week in Pre-Calculus 11 i learned a number of different things about quadratic formulas and graphing. One thing that really stuck with me was how to convert from the general form to standard form and so on. Knowing how to convert between the two formulas can help you find more information about a graph.

Example:

• Note: When going from the general form to the standard form, you need to use the difference of squares method [creates a set of zero pairs])
•  First what you would do is check and see if there is a coefficient in front of the , if there is a coefficient in front then you would need to get rid of it to make the question easier. To get rid of the coefficient all you would need to do is divide the first TWO terms by the coefficient. *Do not divide the third term by the coefficient*
• You would then put brackets around the first two terms and bring the coefficient of the first term in front.
• Next, you create the zero pairs and add them inside of the brackets. To do this you take the term in the second part of the bracket (bx, ie. 4x), divide it by two and then square it. You then write it inside the brackets using +/- to make it a zero pair. *The first three terms inside the brackets should  create a perfect square trinomial*
• Then next step is to get the 4th term in the brackets outside of the brackets and simplify the perfect square trinomial [write in format of: ]. All you need to do to get the 4th term out of the brackets is to multiply it by the coefficient that is in front of the brackets and add it to the number at the end of the equation.
• To change the format of the perfect square trinomial you take the third term and divide it by two, you then write it into the formula where the variable, p is. *make sure that the sign in between the x and p matches the one in the middle term of the perfect square trinomial*

Example:

• In order for you to change the standard form to the general form all you need to do is expand the equation.
• First, to make things easier, you would rewrite the equation so that the squared binomial is side by side and not written in a condensed form. After rewriting it out you multiply the brackets together.
• Next you add up all of the like terms, and your equation should be written in the general form.