Category: Math 11

Pre calc 11 – Top 5 things I Learned This Semester

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This semester in math we learned many new important things that were fundemental for my learning. We added on lots from last year, as well as learned some completely new concepts.

  1. Factoring – we learned factoring last year, however it was this year that it really solidified with me. We used factoring in almost every unit we did, and added on to it from last year. Firstly, I learned the box method and the cross method. Personally, I prefer the cross method as it works best for me and I find it to make more sense visually. We used to solve for X in the quadratics unit, and used it in rational expressions. We used it to simplify expressions when evaluating or solving, and when finding the discrimiants. Without a solid understanding of factoring, completeing the course would of been torture and nearly impossible.
  2. Completeing the square – With factoring, we learned another concept called ‘completeing the square.’ This is an alternate way of solving a quadratic equation that does not factor. After learning this concept, it continued to be used in other units. We used completing the square when turning an expression in general form, to vertex form, and had the option between factoring, the quadratic formula, and completing the square on tests. This understanding of completeing the square was crucial to my learning, even though at first I found it tricky. Now, I love it and will continue to use it this year.
  3. Quadratic formula – Out of all the methods to solve/evaluate a quadratic equation/expression, completeing the square was my favourite. I caught onto it right away, and unsuprisingly, we continued to use it throughout the course. I found it to crucial for my learning, as we used it for finding the discriminant and any other quadratic that cannot be factored.
  4. Rationalizing denominators – With the radicals unit this year, we learned how to rationalize a radical expression. We multiplied by the conjugate and treated the denominator as a complete the square question. This understanding is super important as you cannot leave a radical in the denominator. We rationalized denominators for multiple units, including quadratics that had radicals in their denominator. At first the concept confused me, but now I find it straight forward, which is super important as it is a necessary understanding for pre calc 11.
  5. Sine Law – This year during the trig unit we learned the sine law and cosine law when solving non-right angle triangles. This understanding is super important because most triangles that you deal with are not perfect right angle triangles. Now that I can comfortably use the sine law, I feel so much more confident about trig than I did last year, and prefer non-right angle triangles to regular right angle ones that we did last year.

 

Core competencies – pre calc 11 – Aidan S

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For my reflection I chose critical thinking cc. Since the start of math this semster, I have bettered by my critical thinking by analyzing and interpreting the material that I was taught in pre calculus 11. I had be a detective in some cases to understand how certain formulas work, or why a graph looks the way it does etc. I also had to think critically about how I was going to prepare myself for exams/evaluations so that I could get the best possible grade.

Here is an example of some visual learning that was done on the whiteboards.

Week 17 – Pre Calc 11 – Sine law

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This week in math we learned how to use and apply the sine law. This understanding is important because in combonation with the cosine law, you can tackle any triangle no matter what type. This understanding is also apart of the foundation of our math skills for the future, and is extremely useful. I chose this topic because I find that it is a simple way to deal with complicated looking triangles.

Lets take a look at some examples.

Example 1.

Step 1. Refering to the sine law which has been written at the top of the paper, appy the appropirate sine law to use in order to determine the length of b. Once you have determined which sine law you are to use, plug in your values to the formula.

Step 2. After plugging in your values, we can cross multiply to help solve this equation and get rid of the fractions.

Step 3. Since we are looking for side b, we want to isolate it. I divide both sides of the equation by sine 80 degrees to isolate b.

Step 4. Finally we can plug this all into our calculator to get our value for b. Remeber, round to the nearest tenth of a cm.

Example 2.

(In this example the shape/sketch of the triangle is given to us already)

Step 1. Apply the appropriate sine law: We are looking for angle J and we have all the info for sine H so we use these two. After determining your sine law, plug in your values to the formula.

Step 2. Cross multiply to get rid of fractions.

Step 3. Since we are looking for angle J, we want to isolate. I divide both sides of the equation by 7 to do so.

Step 4. Calculate : my answer is 80 degrees, however, since the diagram was given to use it shows that angle J is much wider than 80 degrees, so we can reject this solution. Being a detective, we know that questions with sine law can have two answers. In quadrant one and in quadrant two.

Step 5. Lets use our understandings to calculate angle J in quadrant two. 180-80 gives us 100 degrees.

Glossary :

Sine Law : the ratio of side length to the sine of the opposite angle

Week 16 – Pre Calc 11 – Special Triangles

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This week in math we learned about so called “special triangles” which are triangles that have reference angles of 60, 30, and 45. This understanding is important to triganomatry because it sets the foundation for trig in the future, and encorporates other things we have learned in math in the past. I chose “special triangles” for this blog post because I find that once you have identfied the triangles as “special”, the math is quite straight forward and manageable.

Lets look at some examples…

Step 1. The first step is to get a visual of your angle by drawing it on your x, y axis.

Step 2. Now we want to identify our principle angle, to do so, I add 360 (full rotation) as many times as needed until I reach a positive value, which happens to be 120.

Step 3. Now that we have identfied our principle angle, lets identify our reference angle. I subtract 120 from 180 to get 60. Since 60 degrees is one of our special angles, we do not need a calculator for this problem.

Step 4. Draw the triangle and label its sides.

Step 5. Determine the exact values of 3 primary trig ratios. Recall the formulas for sin, tan, and cos. Since the instructions call for exact, leave the answer as a fraction.

Example 2.

Step 1. Draw angle on standard x, y axis.

Step 2. Find principle angle – I subtract 360 from 1305 until I get my principle angle of 225 degrees.

Step 3. Find reference angle. I subtract 180 from 225 and get my reference angle of 45 degrees.

Step 4. Since 45 degrees is a special triangle, draw out and label all sides of the special triangle for 45 degrees.

Step 5. Determine the exact values of 3 primary trig ratios. Recall the formulas for sin, tan, and cos. Since the instructions call for exact, leave the answer as a fraction.

Glossary :

Reference angle : the positive angle between the terminal side of θ and the x-axis.

Principle angle – the counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position.

 

 

Week 15 – Pre calc 11 – finding missing side, angles, and rotation of a triangle

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This week in math we reviewed our trig understandings of last year, and added some more information that as a collective we can use to solve more advanced questions. This understanding of trig is important because it is fundemental for our future in math and quite often, trig is an understanding that can be very helpful in day to day life. I chose this topic because I am a very visual learner, and being able to sketch and lable my own triangles is very satisfying and important to my general understanding.

Lets take a look at some examples…

Example 1.

Step 1. I am given some information regarding a right angle triangle, but I am not so sure what all of it means so lets take a look. We are told that point (5,8) is on the terminal arm if an angle in standard position.

Point (5,8) means that this point is 5 units right from the vertex (0,0) on the x axis, and 8 units up the y axis.

The terminal arm refers to the arm of an angle in standard position that meets the initial arm at the origin to form an angle.

Finally, standard position means that an angle’s vertex is located at the origin and one arm is on the positive x-axis.

Lets sketch this triangle… we know that the base of the triangle is 5 units long, and it stands 8 units tall.

Step 2. Determine distance from origin (0,0) to P – in other words find the missing side which is the hypotoneuse. To do this we need to use pythagorean theorem. As seen in the image, this means that our hyptoneuse is \sqrt{89} units.

Step 3. Find the angle – it is important to note that finding the hypotoneuse was not necessary in finding our angle, but since this post is to explain each step of completeing a right angle triangle, it is still important.

I can use either TAN, COS, OR SIN to find my angle. As seen in the image, I used Tan to get an angle of 58 degrees.

Example 2.

Point (-7,5) is on the terminal arm of an angle in Quad 2.

Step 1. Draw the triangle – you may notice that this triangle is negative, as it is 7 units to the left from the vertex and 5 units up the y axis.

Step 2. Lable sides – I know the adjacent and the opposite but I am unsure of the hypotenuse. The hypotoneuse is the longest side of a right triangle.

Step 3. For the sake of this example, we are going to find the hypotoneuse using pythagoreum theorum. This gives me my hypotoneuse of \sqrt{74} units.

Step 4. Now we find our angle – In my example I used TAN again, but we can use TAN, COS or SIN. This gives me my angle of 36 degrees.

Step 5. Since my triangle is negative, we want to determine the rotation. The Rotation is measured from the initial side to the terminal side of the angle. Since we roared 180 degrees, we take 180 and subtract our angle of 36 to get our rotation of 144 degrees.

Week 14 – Pre calc 11 – solving linear rational expressions

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This week in math we learned how to solve linear and quadratic rational equations, testing out different methods. For my blog post, I chose solving linear rational expressions because I find it satisfying to isolate x at the end. I also like that I can make a connection with other subjects I have learned in math this semester, and in years before. This undersyanding is important because it is a subject that will continue to advance, and having the foundational understandings of how to solve a linear rational expression is necessary in the success of your future math self.

Lets look at some examples…

Example a.

Step 1. My first step is to identify an LCM. Since one term has a denominator of x, and another of 9, the LCM which must include a 9 and an x can only be 9x.

Step 2. Identify non-pernissible values. Since our LCM is 9x, we need to identify a number that would make 9x = 0. Our non permissible value = 0; 9(0) = 0.

Step 3. Make the denominator common – while we know our LCM, we need to multiply each denominator by whatever term sets it to 9x and do the same to the numerator. In our first term, 9 is the denominator, so we multiply both the denominator and numerator by x. In our second term, we multiply the top and bottom by 9. Since everything is over the same denominator, there is no need to include the denominator in your work.

Step 4. Reorder the terms – now that we have distributed and gotten our new values for the numerator, we need to re order the terms so that the variable is on one side, and the constant on the other. I want to keep x positive, so I move x to the left with 4x making 3x.

Step 5. Now we can isolate for x – I divide everything by 3 to get rid of the coefficent infront of x, leaving me with x = 3. It is not our non permissible value so there is no need to reject the equation.

Example b.

Step 1. As always, we must determine our LCM. Looking at this equation I see there is a 4a, a 12, and 3a in our denominator’s place. I know that 12 includes 4 and 3, and since there is an a, this means our LCM is 12a.

Step 2. Now we must identify our non-permissible values, so we dont end up with a solution that is extraneous. a = 0, because if you multiplied each denomintaor by 0, you will end up with zero.

Step 3. We must now multiply each denominator by a value that will let it equal 12a. We must multiply the numerator by the same values as the denominator. I distribute the 3 into (a + 5), the a to the 11, and the 4 to the 2. Since each denominator is the same, we do not need to include it in our work.

Step 4. Reorder terms; Since both a values are on the left handside, i can keep them there and add them together. I move the 15 to the right hand side becoming 15 + 8 which equals -7.

Step 5. Isolate for X; I divide both sides by 14 to isolate x, leaving me with a negative fraction as my value for a. a = \frac {-1}{2}

Example C.

Step 1. I notice right away that one of my denominators contains a x^2 so I will try to factor it. Luckily, it factors leaving us with (x + 4)(x + 3 ). The other terms are already in their simplest forms.

Step 2. Now I can find my non permissible values, look at the value being added to x and determine what would make that value equal to zero. Since I have a 4 and a 3, the non permissible values are -4 and -3.

Step 3. Now we identify an LCM; looking at this I know that our LCM is (x + 3)( x + 4)

Step 4. Make denominator equal to (x + 3)(x + 4) . I multiply my first term top and bottom by (x +4), my second term by (x+3) and my third term already contains both parts of the denominator. Now I can distribute and write out my values.

Step 5. Now I can add my like terms; 2x and 3x = 5x, and 8+9 = 17.

Step 6. Rearrange, I want to keep my x value positive, so I shift the constant to the right.

Step 7. All thats left to do is isolate; I divide everything by 5 to isolate x, giving me the answer x = -2.

 

 

Week 13 – Pre Calc 11 – Adding Rational Expressions

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This week in math we learned how to add rational expressions with monomial denominators. I chose this topic because I like how it encorporates everything I have learned about fractions so far in math, and combines it with my understanding of factoring. This skill is important because as questions become harder, having the foundations of factoring, and evaluating fractions is extremely important and useful.

Lets look at some examples…

Example a.

Step 1. Are the denominators the same? Yes! we are lucky because there is no need to find a LCM.

Step 2. Is our denominator factorable? Yes! We always want to see if we can factor our denominator when adding a rational expression.

Step 3. Add 3h-8h above our factored denominator.

*Side note* Use the factored denominator to find our non permissable values; h cannot be 1 and h cannot be -4.

Example b.

Step 1. Are the denominators the same? No! This means you need to find the LCM.

Step 2. Find LCM; our LCM needs to include all parts of both of our denominators.

Step 3. Remeber… what you do on the bottom you must do to the top.

Step 4. You can go ahead and add together your numerators and put them over your LCM.

Glossary :

LCM: The least common multiple of two numbers is the smallest number that is a multiple of both of them.

 

Week 12 – Pre Calc 11 – Identifying Non-Permissible Values of Rational Expressions

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This week in math we learned how to identify the non-permissible values of a rational expression. This understanding is important because it builds off of the foundation we have developed from past units of math, such as factoring and quadratics, and will continue to be seen as our math levels advance. I chose this subject because I liked the connection that it has to factoring, as that was one of my favourite units we have done to date.

Lets check out an example…

Example a.)

In this example, our denominator is already in factored form, which makes determining our non-permissible values much easier.

Our first and only step is to find the values that make our denominator = to zero. To do this, plug in values for x1 and x2 that cancel out and make zero.

Since we have (x +4) and (x+6), we would plug in a -4 into our first set brackets and -6 into our second set of brackets; making two zero pairs.

Example b.

In this example we want to find a value that we can plug into x that will cancel out -24.

To do this, we use our math skills to come up with two numbers that when squared, can multiply with 6 to equal 24.

The first number I chose was 2; because 2^2 = 4; I know that when multiplied by 6 the product will be 24.

24-24 = 0 which means that our first non permissable value is 2.

What about the second value?

Since we are plugging in a value for x, and that number is in brackets, we can assume that -2 would be our second non-permissible value.

Lets prove this… (-2)^2 = 4.

4×6 = 24 which means that -2 works as our second permissible value.

Example C.

In this example, the denominator looks a bit different… it is not in factored form.

This means that we must factor our denominator so it results in two sets of brackets.

Assuming we already know how to factor, we see which values we need to add to make zero pairs.

For our x1 value, we plug in 3; 3-3 = 0

For our x2 value we plug in 2; 2-2 =0

this means we have found our two non-permissable values.

Glossary:

Non permissible value: values of the variable(s) that make the denominator of a rational expression equal to zero.

Rational expression – just an advanced fraction

 

Week 11 – Pre Calc 11 – Solving one variable inequalities

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This week in math we covered solving one variable inequalities. This understanding is important because it ties in the aspects we learned earlier in the chapter and will continue to be put to use. I chose this topic because I understood how to solve this type of equation right away, as I was able to use the skills I have previously learned in this chapters.

Lets take a look at some examples…

Example a.

Step 1. My first step is to change the inequality into an equation with an ‘=’ sign so I can solve.

Step 2. I must now distribute the 2 to the terms inside the bracket.

Step 3. Now we rearrange so that the x values are to the left, and the constants to the left.

Step 4. Since I have a leading coefficent of 2 infront of my x value, I need to isolate the x by dividing both sides of the equation by 2. Finally it is important to substitute your inequality sign back in to get your final answer.

Example b.

Step 1. Our first step is to change the inequality to an equation.

Step 1. Since this example is written as a fraction, we want to eliminate the fraction by multiplying both sides of the equation by the denominator of 7. On the lefthand side, both 7s cancel out and on the right -3 is multilpied by 7 to get -21.

Step 3. Since the coefficent of x is one, we do not need to do any isolating; simply reearange the terms. X values to the left and constants to the right. Finally remember to change your ‘=’ sign back to an inequality.

Example C.

Step 1. As always, start by converting the inequality to an equation.

Step 2. Since in this example, both sides of the equation are written as fractions with different denominators we must multiply both sides of the equation by 21.

Step 3. Now that we have eliminated our fractions, we need to distribute the values outside our brackets to the inside terms.

Step 4. Finally we can rearrange. The x values to the left and constants to the right. Since the coefficent of x is 1, there is no need to isolate. Lastly, convert the equation back to an inequality.

Vocab :

Equation – contains an = sign.

Inequality – Compares two values, showing if one is less than, greater than, or simply not equal to another value.

Week 10 – Going from general to standard form – Pre Calc 11

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This week in math we learned how to convert equations in general form, where you can identify the y intercept, into standard form (or vertex) form. This is an important skill because in order to be able to graph a quadratic function you need the vertex, whether there is a stretch, and if the parabolla is facing upwards or down. I chose this topic because I like how similar it is to our last unit where we learned to complete the square.

Lets look at some examples…

Example A.

Step 1. In my first step I divide my B term AKA the middle term by 2 and square the product of that division. After dividing the term and squaring I am left with 9. To keep the equation in balance I add positive 9 and negative 9.

Step 2. My next step, the negative 9 can leave the brackets and there is no stretch that it needs to be multiplied with so its value remainss the same. Now we can add our two constants of negative 9 and one together leaving us with our terms in the brackets and negative 8 outside the brackets.

Step 3. My last step is to combine the inside terms. To do so; put x and b/2 into brackets and square it, with our constant of -8 remaining on the outside. To find the vertex simply take the value inside of the brackets and change its value which would give us -3 as our x value of the vertex. The Y value is our constant on the outside, -8. It remains negative.

Lets take a look at an uglier example…

Example B.

Step 1. My first step is to get rid of the leading coefficent infront of my A value. To do this out the A and B values in brackets and divide them by 2.

Step 2. My next step is to divide the B term by 2. Since my B term is a fraction over 2, multiply the denominator by 2 giving the new product after being divided. Square this term and add it into the brackets, as well as its negative value to balance out the equation. To get rid of the extra negative value we added, multiply it by the coefficent 2 we factored out so it can leave the brackets.

Step 3. Now we may simplify. Combine X plus the divided b term in brackets and square it. For our constants on the outside of the brackets, we can first divide -50/16 by 2 giving -25/8. Because we are combining -3 with -25/8; the denominators need to be the same so we must multiply -3/1 by 8. This leaves us with our final inside terms and our constant of -49/8. To find the vertex, take the value inside our bracket and change its sign to negative giving us our x value. This means there is a horizontal shift to the left. Our why value is the constant outside the brackets. This means there is a vertical shift downwards. The 2 outside the brackets tells us there is a stretch of 2.

Glossary :

Horizontal shift : inside changes that affect the input (x−) axis values and shift the function left or right.

Vertical shift : The outside changes that affect the output (y−) axis values and shift the function up or down.

Vertex/standard form : y = a ( x − h ) 2 + k

Stretch : (narrowing, widening) of the parabolla