Week 17 in PreCalc 11 – Sine Law

This past week in PreCalc 11, we learned more Trigonometry concepts, and unlike the previous week, we began the second part of Trig which does not involve graphs as much. 

Additionally, this week in PreCalc 11, we expanded our learning as we began to work with new topics which were not discussed in other years. In other words, this means that we learned about something known as the Sine Law. 

Before learning about the Sine Law, Ms. Burton explained to us when to use Sine Law. Sine Law is to only be used in non-right triangles (triangles with no angle of 90 degrees), and can only be used when we are given a certain number of sides and angles. Also, Sine Law has its own formula, which we can see that, with this image.

 

With the formula of the Sine Law, the little “a”, “b”, and “c” can be described as the matching side, and has another name of side AB (little c).  If we had a question like this triangle, where we had to find side little “b”, we would easily be able to do that using the sine law. First we would go \frac {b} {sin 48} = \frac {10.8} {Sin 81}. Then we can isolate “b”, by moving the sin 48 to the other side, where it would multiply. b = \frac {sin 48 * 10.8} {Sin 81}. By using a Scientific Calculator, we can plug these values in, and easily get an answer of 8.13 cm.

As we used a Scientific Calculator, and were able to recognize and memorize the Sine Law formula, we were able to take this non-right triangle, and quickly figure out the missing side. The Grade 10 version of me would have been so lost as I did not know the Sine Law! Thankfully, I know it now!

Week 16 in PreCalc 11 – CAST Rule

“Hey class, can you guys believe we only have 2 more weeks left of Pre-Calc 11? For some of you this may even be your last Math class at Riverside”, says Ms. Burton. “And as Ms. Burton says this, the only thing I can think of is my Grade 12 year, where I will have both Pre-Calc 12, and Calculus 12”. Regardless of this, I must reflect on what I learned this week, as there will soon be a test on Trigonometry. 

Furthermore, this past week I was taught about different concepts which are slightly new to me, and I will need to know the meaning of, for the next test. In addition, this week I learned about the CAST rule

As I started to learn about the “Grade 11 version” of trigonometry, we were first shown some elements which are often used with graphs, and are helpful to visualize different things. Firstly, Ms. Burton taught us to use something which is called the “CAST” rule. This stands for “All Students Take Calculus”, despite it starting with a “C”, and there is an explanation for this. When we look at a graph, we read the Quadrants of the graph starting with the “Top Right”. This Quadrant is known as Quadrant 1. From there, we go to the “Top left” which is Quadrant 2, then “Bottom Left” (Q3), and lastly Quadrant 4, in the bottom right. With this, we can now see why the “CAST rule” is “All Students Take Calculus”.

Now, with this knowledge, we are able to apply it to what the rule means, and be able to easily visualize which letters, go with which quadrants. Additionally, the CAST rule is very helpful, as it shows us where each “Trig” ratio is positive. This means the “A”, means all trig ratios are positive in that quadrant. It would also mean where the “S” is, only Sine is positive, and so on. This rule has been very helpful for me lately, as it helps me visualize and predict if a rotational angle will have a positive or negative Cosine ratio, for example. 

In conclusion, this week in Pre-Calc 11 I have learned about a very helpful rule and tool for me to “make life easier” during practice questions and tests, and I can say right now I will definitely be using the CAST rule often. 

Week 18 – My TOP 5 things from PC 11

“YO I cannot believe that our school year is already going to be over after this Friday! This year has been a blast, and gone by so fast, but I cannot wait for the summer!” , says Gui to his friends. Pre-Calc 11 has been a good experience, and I am ready to reflect on the top 5 things I have learned this semester. 

5. Firstly, I want to reflect on Solving Quadratic Functions using factoring. This topic makes it into my top 5, as factoring has been something which we have used throughout over 60% of this course, and is a way we have to used to solve and simplify many different equations and expressions throughout Pre-Calc 11. Although solving quadratic equations is different from factoring, the way of approaching equations and the techniques used are the same. If you had the equation x ^ {2} + 15x + 56 = 0, you could factor it into \left ( x \times 8 \right) \left ( x \times 7 \right) = 0. We know that it factors into this because the product of 56, is equal to 15. This means that 7 * 8 = 56, and 7 + 8 = 15. This tells us that our factoring has been done correctly, and we are able to move onto the next step of finding the two solutions. Since we are left with \left ( x \times 8 \right) \left ( x \times 7 \right) = 0, we can now use the method of inspection, and solve for the two solutions. The method of inspection teaches us to use mental math to solve the equation from here, and tells us the two solutions are x = -7, and x = -8.

4. Next, I believe it is crucial to reflect on predicting solutions, when using the Quadratic Formula. For me this is something very significant which we have learned for a few reasons. For starters, we have used the quadratic formula quite a bit during the second half of this course, and I found it to be a key learning. Also, by predicting solutions, I have been sometimes able to only do half of the question, with this piece of learning. In addition, if we had : x^{2} -4x + 4 = 0. By plugging it into the quadratic formula, we get : x = (4) \pm \frac {\sqrt {(-4)^{2} -4(1)(4)}} {2(1)}. Then, we simplify it and get a result of x = 4 \pm \frac {\sqrt {16-16}} {2}. As we can see, it would result in a discriminant of 0, meaning we would only have 1 solution rather than two solutions. 

3. Furthermore, I found Perfect Square Trinomials to be in the Top 5 things that I have learned this year. Perfect Square Trinomials was a cool concept to learn, as it gave us yet another way to solve unfactorable questions. This method was also used in Graphing Quadratic equations, which to me makes it important, as we used it in various units. In the “Completing the square” method, it requires you to divide the second term (8), by two (8/2 = 4), then square it, 4^ {2} = 16, leaving you with x^ {2} + 8x + 16 = 11, since you need to add 16 to both sides of the equation. From here, you can factor the trinomial x^ {2} + 8x + 16, which would be \left ( x + 4\right) ^{2} = 11. Then, you can square root both sides, \sqrt {\left (x + 4\right) }^ {2} = \sqrt {11}. This would then give you the final answer of x = -4\pm \sqrt {11}

2. Moreover, Inequalities and Systems was yet another topic which stood out to me. This concept stood out as it expanded on my learning from previous years, involved methods to solve them from other units/connected with other units (Factoring), and as I felt it was one of the more challenging units this year. As we take a look at the graph of x < 2, we can recognize how it has a dashed line going through x = 2, rather than the usual solid line. This happens as they are not the same symbol (<, and \leq), and is used to show a difference in the two inequalities, and that there cannot be a solution which involves “2”, as the “x” value. Something which I also picked up from this unit, was the ability to see where all the possible solutions appear, by viewing the graph. 

1. Lastly, two variable inequalities also makes it onto my Top 5 list, and is at the number one spot. My reasoning for this is because it again expanded from my learning of inequalities previously, was more challenging for me, and because it also connected with our quadratic functions unit. In addition, it was also quite different from what I had seen before, as this time with inequalities, I could see that the solutions can be to different sides of the equation, unlike I had seen before. This was something that caught me by surprise! Two variable inequalities can include more than one side of the inequality shaded, unlike one variable inequalities.

Another difference, is that before when x < 2, the point just had to satisfy the “X” value, but now if it is y < x^ {2} + 1, the coordinate has to satisfy both the “X”, and “Y” value.

 

To summarize, Pre-Calc 11 has been pretty challenging, yet fun, and I can definitely say I benefited significantly from having taken the course, as I learned a substantial amount of new concepts!

Week 8 in Pre-Calc 11 – Predicting Solutions

This week in Pre-Calc 11 we expanded our knowledge of Quadratic Equations, and using the Quadratic formula to find the solutions of different equations. 

More specifically, we learned to predict how many solutions would result in a equation, by looking at the discriminant of a quadratic formula. As we have learned, each equation in Math has 1 solution, but Quadratic equations can have two different solutions, or none at all, which we will explore how it happens. 

For starters, let’s be reminded of the quadratic formula which we use: x = -b \pm \frac {\sqrt {b^{2} -4ac}} {2a}. As we can now see, this is the quadratic formula which we can use to solve non-factorable quadratic functions, and has one piece of it which is the most significant. This piece is the discriminant of the formula, which is the same as b^{2} -4ac, and can help us determine how many solutions will result from this function/equation.

Now, let’s see how this works, and why the discriminant can determine the number of solutions: x^{2} +3x -2 = 0. With this unfactorable equation, we can plug it into the quadratic formula: x = -(3) \pm \frac {\sqrt {(3)^{2} -4(1)(-2)}} {2(1)}. Just by getting the product of the discriminant we can see it would result in x = -3 \pm \frac {\sqrt {17}} {2}. With the discriminant resulting in a positive number, we can tell that this equation will give us two different solutions, indicated with the \pm

In addition, if we had : x^{2} -4x + 4 = 0. By plugging it into the quadratic formula, we get : x = (4) \pm \frac {\sqrt {(-4)^{2} -4(1)(4)}} {2(1)}. Then, we simplify it and get a result of x = 4 \pm \frac {\sqrt {16-16}} {2}. As we can see, it would result in a discriminant of 0, meaning we would only have 1 solution rather than two solutions. 

Lastly, if we had : x^{2} -2x + 6 = 0, and plugged it into the quadratic formula. x = (2) \pm \frac {\sqrt {(-2)^{2} -4(1)(6)}} {2(1)}. Then further simplify it: x = 2 \pm \frac {\sqrt {4-24}} {2}. After solving the discriminant, we can see it would result in a negative number. Since we are not able to solve the square root of a negative number in Math, it would result in a undefined answer, and our equation would have 0 solutions. 

To conclude, I will be reviewing and practicing what we learned in week 8 of Pre-Calc 11, as well as everything else learned in the previous weeks to ensure I am prepared for the Midterm in the week to come. 

Week 15 in Pre-Calc 11 – Reviewing Rational Expressions

This week in Pre-Calc 11 we reviewed all the different concepts learned in the Rational Expressions and Equations unit, in preparation for the Unit test, which was this past Thursday.

In addition, we practiced our skills with this unit by working on many different Quizizz’s with different partners. The concepts which we reviewed with the Quizizz included: Dividing and Multiplying Rational Expressions, how to Simplifying Rational Expressions, and has us practicing stating the non-permissible values. 

Firstly, we reviewed Simplifying Rational Expressions and stating non-permissible values, as a warm-up, and to remind us of basic rules when dealing with rational expressions. This meant that we were working on questions such as \frac {x ^ {2} -16} {x ^ {2} + 6x +8}. For us to simplify this expression, we would start out by factoring it: \frac {\left ( x-4 \right) \left (x+4 \right) } {\left (x+4 \right) \left (x+2 \right)}. From this step, we are able to state which values x is not allowed to be: x \neq -4, -2. This is the case as the two values would result in one of the brackets equaling 0, and a final denominator of 0. The next step from here would be do simplify the common denominators and get the final answer: \frac {\left (x-4 \right)} {\left (x+2 \right)}

Then, we began to work on dividing and multiplying rational expressions. With the basic rules of dividing and multiplying fractions in mind, I am now able to easily tackle questions which involve dividing rational expressions, as there is only one small difference which we have to keep in mind. Take an expression such as \frac {2x ^ {2} -x -1} {x ^ {2} +2x -3} \div \frac {2x ^ {2} -13x -7} {4x ^ {2} +28x +48}. First, we change this question from a dividing question, to a multiplying question by flipping the fraction after the divide sign: \frac {2x ^ {2} -x -1} {x ^ {2} +2x -3} \times \frac {4x ^ {2} +28x +48} {2x ^ {2} -13x -7}. Then, we can factor all the different trinomials as they are factorable, and will help us get to make the question look more simple:

We can then simplify it further and divide out factors which are common, and get us our final answer:

 

 

 

In conclusion, this week in Pre-Calc 11 we reviewed already learned concepts in preparation for the test, and I felt that this was what we spent the most time on, and was more significant, which is why I felt it was important to do a recap on it.

Week 14 in Pre-Calc 11 – Rational Expressions Part 2

This week in Pre-Calc 11 we have learned many new and different concepts, which I will be constantly studying for next week’s test. To prepare for next week’s test, I will review and write about what I have learned, so I have one more resource to study off of.

Furthermore, this past I week I learned various different concepts in Pre-Calc 11, all of which I will be reviewing starting this weekend, so make sure I do not study on the last night before the test. Also, I learned how to multiply rational expressions, and divide rational expressions, which is an addition from last week’s lessons.

As we began to learn about dividing radical expressions, we were first reminded of a simple rule when dividing fractions: When dividing to fractions, flip the second fraction of the expression, and make it into its reciprocal. \frac {3} {4} \div \frac {5} {8}. As its reciprocal, it then becomes: \frac {3} {4} \times \frac {8} {5}. By flipping the reciprocal, we are allowing for the question to change into a multiplying question, and makes it easier to work with. We can now cross divide, and reduce the 4, and 8, as they are both divisible by 4. \frac {3} {1} \times \frac {2} {5}. Now it becomes easier to multiply the fractions, as we have a smaller numerator and denominator to multiply across. By multiplying across as we would with any other fraction, we get the final answer of \frac {6} {5}.

With these basic rules in mind, I am now able to easily tackle questions which involve dividing rational expressions, as there is only one small difference which we have to keep in mind. Take an expression such as \frac {2x ^ {2} -x -1} {x ^ {2} +2x -3} \div \frac {2x ^ {2} -13x -7} {4x ^ {2} +28x +48}. First, we change this question from a dividing question, to a multiplying question by flipping the fraction after the divide sign: \frac {2x ^ {2} -x -1} {x ^ {2} +2x -3} \times \frac {4x ^ {2} +28x +48} {2x ^ {2} -13x -7}. Then, we can factor all the different trinomials as they are factorable, and will help us get to make the question look more simple:

We can then simplify it further and divide out factors which are common, and get us our final answer:

 

 

 

 

To conclude, this week has been a slightly more challenging, but also slightly more interesting week in Pre-Calc 11, as we learned about new concepts, rather than reviewing previous learned concepts. 

 

Week 13 in Pre-Calc 11 – Rational Expressions Part 1

“Hey class! Can you guys believe we only have six more weeks of class left before the summer break?”, as Ms. Burton says to her class in a excited tone. And while Ms. Burton says that, all that is in the back of my mind is the traveling I’ll be doing, the amount of times I’ll be able to appreciate the great outdoors and go outside, and the good memories which will be made. But enough of that, as now I have to persevere during the last few weeks, and study well to do well in my final exams. 

In addition, this past week I learned loads of new concepts in Pre-Calc, all of which I will have to know well for, as it will be on the final exam. Also, this past week I learned to simplify rational expressions, factor rational expressions, and learned about what the non-permissible values of rational expressions are. 

By learning about Rational Expressions, we were first reminded of a simple rule which states that 0, cannot be in the denominator of a fraction. This is the case, as it would result in a undefined answer, meaning it is not allowed. It also meant that we would be having to state the non-permissible values of the variable which is in the denominator, and would tell us which values would result in the denominator being equal to 0. \frac {5} {x + 5}. x \neq -5.  

As we are now aware of this, we can see it in action as we simplify rational expressions. Take an example such as \frac {x ^ {2} -16} {x ^ {2} + 6x +8}. For us to simplify this expression, we would start out by factoring it: \frac {\left ( x-4 \right) \left (x+4 \right) } {\left (x+4 \right) \left (x+2 \right)}. From this step, we are able to state which values x is not allowed to be: x \neq -4, -2. This is the case as the two values would result in one of the brackets equaling 0, and a final denominator of 0. The next step from here would be do simplify the common denominators and get the final answer: \frac {\left (x-4 \right)} {\left (x+2 \right)}

In conclusion, this week has been a very interesting and jam-packed week for learning new concepts in Pre-Calc 11, and I am ready for a week where we will expand on the concepts mentioned in this blog post and learn more.

Week 12 in Pre-Calc 11 – Two Variable Inequalities

“No way! We just started learning about inequalities this past week, and we are already done with it!”. This was me, shocked as I did not believe this would be such a short unit. Nevertheless, I learned more concepts over this past week, which I found super cool! Also, I find I could benefit by reviewing these concepts, to further prove my understanding and ensure I am very familiar with this before moving onto the next unit. 

By building onto what I had learned and reviewed last week, this week I was able to expand my learning, and learn to graph inequalities with 2 variables

As we did a recap on what was learned the previous week, and began to learn the next components, I could recognize that there were some similarities between the one variable inequalities, and the two variable inequalities. By recognizing the similarities, I was able to tell that the two variable inequalities still had a dashed line when it included a “>” or “<“, and that there would still be a section shaded to show the possible solutions.  In contrast, two variable inequalities can include more than one side of the inequality shaded, unlike one variable inequalities. Another difference, is that before when x < 2, the point just had to satisfy the “X” value, but now if it is y < x^ {2} + 1, the coordinate has to satisfy both the “X”, and “Y” value. We can see these similarities and differences with the graphs shown. 

To conclude, it has been enjoying to revisit the concept of solving and graphing inequalities in Pre-Calc 11, and I am excited for this week to learn about the new unit!

Week 9 in Pre-Calc 11 – Midterm Review

“Oh no! The midterm is scheduled for next week, and I am panicking! I’m going to need to study a lot for this so I am confident when I write it, so I need to start studying right away!”. Currently I have not started studying for the midterm, but to start studying for it, I am going to write my blog post to review everything I have learned so far. 

The midterm is going to include elements from Unit 1, Unit 2, and Unit 3, so I am going to reflect on the most important components I learned from each unit. These components will include: fractional exponents, multiplying radicals, and perfect square trinomials

The semester began with learning about Fractional exponents in our Unit 1, and in my opinion it was the most important concept I took away from that unit. Before this Unit, when I saw Fractional exponents I would skip over the question, or be frustrated that I did not how to approach the question. Now, I know how to deal with fractional exponents like 64 \tfrac {2} {3}. With a fractional exponent like this one, the denominator defines the root in which the base will be under, and the numerator defines the power the base will be raised to. This tells us 64 \tfrac {2} {3}, is really just \sqrt [3] {64} ^ {2}. Now that we know this, we can simplify it to 4 ^ {2}, since 4 * 4 * 4 is 64. From here, we know 4 * 4 = 16, meaning the final answer is just 16. By knowing this, it made it much easier for me to approach fractional exponents, and made them easier than I thought they were two years ago.

In the second unit, we learned about radical operations, which would include dividing, multiplying, adding, and subtracting radicals. Right now we will focus on multiplying radicals, as I feel it was the operation which we used the most during this unit. As we learned about the multiplication of radicals, there were some previously learned rules I could apply to the multiplication of radicals, and would benefit me by easily being able to understand this new component. An example would be if you were trying to determine the answer of a binomial like \sqrt {2} + \sqrt {3}. With this example, you would first set up the question as \left ( \sqrt{2} + \sqrt{3} \right ) \left ( \sqrt{2} + \sqrt{3} \right ), then apply the FOIL (First Outer Inner Last), technique to distribute and multiply into the brackets. After proceeding with FOIL, you would be left with \sqrt {4} + \sqrt {6} + \sqrt {6} + \sqrt {9}. Once simplified, you would get your final answer of 5 + 2\sqrt {6}, since \sqrt {4} is equal to two, and \sqrt {9} is equal to 3, allowing you to combine like terms and get 5 + 2\sqrt {6} as the final answer. By applying this previously used technique of FOIL, I was able to get through this unit in a breeze, and understand at a high-level.

Moreover, the third unit was a bit more challenging, but was the Unit I felt I learned the most new concepts in, and knew I was going to be using a lot in the future. In this unit, the concept I found the most helpful and interesting, was Perfect Square Trinomials. I found it very unique as it allowed me to factor trinomials which might not have been factorable by taking the product of the “C” value in a trinomial, x^ {2} + bx + c. Now let’s see this in action with a trinomial such as x^ {2} + 8x + 5 = 0. So, you first move the third term in the trinomial over to the right side of the equation, which be defined as “c”. This would bring you to x^ {2} + 8x + 0 = -5. Then, the second step requires you to divide the second term (8), by two (8/2 = 4), then square it, 4^ {2} = 16, leaving you with x^ {2} + 8x + 16 = 11, since you need to add 16 to both sides of the equation. From here, you can factor the trinomial x^ {2} + 8x + 16, which would be \left ( x + 4\right) ^{2} = 11. Then, you can square root both sides, \sqrt {\left (x + 4\right) }^ {2} = \sqrt {11}. This would then give you the final answer of x = -4\pm \sqrt {11}. I found this a super interesting concept, and as the name says, I found it “perfect”!

In conclusion, my blog post will help me a lot to review these concepts, and the focus from this point on is to study, study, study for the midterm to guarantee myself a good grade!

Week 10 in Pre-Calc 11 – Quadratic Function and equations forms

“Week 10 in Pre-Calc 11? Already!?”. This was my reaction as I came back to school, the following day after having arrived from my trip. It shocked me how it felt so long since I was in a Pre-Calc 11 class, and I was prepared to adapt myself from spending time with my grandmother and speaking Portuguese, to thinking about Quadratic Functions, and thinking about the language of Math. 

Furthermore, we had only started learning about Quadratic Functions when I left for my trip, and this meant there was much more to learn from Unit 4. This week taught us about the different forms that Quadratic Functions can be written in, and what the domain and range for each function is.

Although we had learned about domain and range in previous years, we had only seen it with linear equations, and this time we would be taking a look at how they are different when it comes to Quadratic Functions.By analyzing a graph such as the one shown, one is able to identify how x \in R, meaning “x belongs to real numbers”. This happens because with an equation like y = x^ {2} + 3, you could plug in any value for “x”, and get a valid output. Another explanation for this can be explained by observing the following graph. As one can see, the parabola opens up to the left and the right, making an infinite curve shape which would allow “x” to be any value on the line. In contrast, it is evident that it is different with the “y” value, where there is a limit to what “y” can be, as the maximum/minimum value of the function (vertex), sets that and is different with every function. In this case, we can see that “y” would be y \geq 3.

In addition, we saw that there are three different main forms that a Quadratic function can be written in, being the Vertex/standard form, the general form, and the factored form. The most commonly used one that we learned about is the vertex form, but it was still very important that we learned the other two forms too. The vertex form can be seen as y = a(x-p) ^{2} +q, and is the most helpful form in terms of collecting data from. The general form is written as y = x^{2} + ax +x, and can easily be identified as it is the same formula as the trinomials from previous chapters. Lastly, there is the factored form which was also used previously, and looks like this: y= a \left (x - x_ {1} \right ) \left (x - x_ {2} \right). Once again, the vertex is form is most useful form, as you are given the stretch value, vertex, and the line of symmetry, where as the other two forms only give you the stretch value and the y-intercept (General form), or the roots of the equation (factored form).

To conclude, I have learned such a significant amount of new concepts this week, and will be studying them a lot to guarantee a good mark on next week’s test.