Week 18 – Precalculus 11

Posted on June 11, 2018June 21, 2018 by shannonf2015

Top 5 things I have learned in Precalc 11

Graphing: Is the number one thing that I learned. This semester, I found graphing straightforward and eyeopening. I learned that once you find the vertex, you just have to follow the 1,3,5.. pattern and multiply the pattern if it has a stretch or compressed value (number in front of the $x^2$ ). I learned that you could find the number of solutions which is also the x-intercepts, as well as the vertex which is a key component to graphing. This made graphing absolute values easy as well as reciprocal graphs. It was really cool to see I could graph all these different types of graphs with just some information given.

2. Substitution: Solving systems with substitution was another thing I learned in precalc instead of drawing the graphs out. I finally knew how to solve systems algebraically which made me feel more knowledgeable about solving systems and it became a useful tool if I didn’t have graph paper to graph out the systems.

3. Completing the squares: I never realized how important completing the squares was. We used it a lot during this course such as changing general form into standard form to get the vertex, to find the axis of symmetry, and to solve an equation. It does take awhile when it comes to solving an equation, however, it still is a useful tool when it’s an equation is unfactorable and I found myself using it a lot when I had to find the vertex.

4. Trigonometry: The most important lessons I got out of this chapter is the Sine Law and Cosine Law. They were very useful when it came to scalene triangles and I found solving sides and angles a lot easier than I thought it would. I am now able to tell when to use Cosine Law and Sine Law (Cosine – when three sides are given or two sides and a contained angle are known; Sine – when two angles and one side are known or two sides and a non-contained angle are known).

5. Square roots : Whenever we were solving an equation and we had to square root both sides to get the variable by itself, there was both a positive and negative sign in front. Before, I always treated a square root as having one answer but now I realized that there are both positive and negative values (only when I add in a square root but if there is already a square root, you take the positive roof of it). Even though it is a simple concept, this was a big eye opener for me.

Week 17- Trigonometry

Posted on June 9, 2018 by shannonf2015

This week we have learned the primary trig ratios (Sine, Cosine and Tangent), how to sketch angles in standard position, as well as sine law and cosine law.

Let’s look at the primary trig ratios:

Before we used SOH CAH TOA to remember the ratios for trig, now we have replaced the opposite angle with y, adjacent with x and the hypotenuse is replaced with r.

Sineθ: $\frac{y}{r}$

Cosθ: $\frac{x}{r}$

Tanθ: $\frac{y}{x}$

We have also learned CAST which helps with determining if the angle is positive or negative, the two triangles used to determine the trig ratios and a diagram that can be used to determine the primary trig ratio when an angle has its terminal point on the x-axis or the y-axis

Example when using the two triangles:

$\tan{150^{\circ}}$

We must first find the reference angle and determine which quadrant the angle is in. It is in quadrant 2 so we must take 180 and subtract 150 degrees which gives us 30 as the reference angle.

If we look back at the two triangles, we can see that the triangle that has 30-60-90 will be the triangle we must use. Looking at the 30 degree point of the triangle, we can start labelling the sides of the triangle to find the ratio.

Remember that tan’s ratio is $\frac{y}{x}$ . y will be 1 and x will be $\sqrt{3}$ . When we look at CAST, we can tell it’s going to be negative because 150 is in the second quadrant (only sin is positive in second quadrant). So the ratio will be $\frac{-1}{\sqrt{3}}$ .

Let’s look at another example using the unit circle for quadrantal angles.

$\cos{130^{\circ}}$

The ratio for cosine is $\frac{x}{r}$ . Look at where angle 180 is which has the point (-1,0), where x=-1 and y=0. r is always going to be positive for quadrantal angles. So the ratio is going to be : $\frac{-1}{1}$ = -1.

We have also learned how to find the reference angle of each quadrant. Let’s look at an example:

If the reference angle is $\cos{43^{\circ}}$

Quadrant 1: It’s going to be the reference angle : $\cos{43^{\circ}}$

Quadrant 2: 180 – 43 = 137

Quadrant 3: 180 + 43 = 223

Quadrant 4: 360 – 43 = 317

To know when to use Sine Law, we have to be given at least one angle and side that is across from each other. For Cosine Law, we need either three sides or at least two sides and an angle.

Sine law: $\frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}$

We have the degree on the top if we are looking for an angle and we have the side on the top if we are looking for a side.

Cosine law: $a^2=b^2+c^2-2bccosA$

We use the cosine law if we cannot use Sine Law but it has to have at least two sides and an angle or three angles.

To use the cosine law and sine law, you plug in the given sides and angles and isolate what you are trying to find.

Week 16 – Numeracy Assessment

Posted on June 3, 2018 by shannonf2015

Last week, we have done some practice with the numeracy test. What I have learned from doing these practice questions was that when we do the written part, we have to be convincing and detailed in our explanation and that it’s not so much about which is the correct answer. It’s more about your reasoning. Another important method I have learned while doing these questions is that you should always think or write down your reasoning behind why you chose your answer even when it’s multiple choice.

In some practice questions, I have learned how to determine which graph represents the situation the best, as well as how to read a graph. When we did some questions that involved making our own plan, I learned it’s helpful to write down ideas in a chart to lay out my ideas clearly then explain each one in detail and why I chose that specific amount and choice. I also learned that keeping units is extremely important when doing these types of questions. For example, we did one question where we had to plan our water use to a certain amount of litres per week. It was helpful to draw out a chart so that I could put in my plan for the high-efficiency appliances and fixtures and do some adjustments on my plan if needed before started to do my explanation. I also remembered to keep the units in when I was writing my explanation when I kept referring to the chart.

Lastly, I have figured out that there can be more than one answer and that we should show all our work when doing calculations as it’s not as simple as typing in a bunch of numbers in the calculator. These questions makes you think more and you’ll have to analyse the information and images carefully. When I got stuck with a question that included calculations, I drew an image which helped me see the situation more clearly. We also had to do some other questions where they would give a list of different strategies a family decides to do in order to use less water and you would have to check off the ones that are unreasonable. Analysing each choice carefully and the chart given helped me eliminate choices and I was able to come to a conclusion on which one(s) were the least reasonable. I also tried to think if what they planned to do was even possible which also helped me.

Week 15 – Solving Rational Equations

Posted on May 26, 2018May 26, 2018 by shannonf2015

This week I have learned how to solve when there is two fractions by cross multiplying and how to solve by finding the common denominator by factoring.

Let’s start with an example using cross multiplying:

First step is to write down the restrictions. The restriction is 2.

Next is to draw the butterfly, which is basically circle the numerator of one fraction with the denominator of the other fraction. Once you got those circled, those values circled together indicate that it will be multiplied together. Write them down making sure to keep the equal sign as you are solving. Now foil in the 3 in the brackets and isolate s. You get 2. Since we wrote down that the restriction of s cannot be 2, this is extraneous.

Let’s look at another example where we have to find the common denominator instead:

In this example, it looks too complicated to do cross multiplying. The first thing we should do is write down the restrictions which is x cannot be 0 or -1. Next, we can see in the first fraction, the denominator is factorable. We can take out a 2x and get 2x(x+1). By looking at both denominators of the fractions, the common denominator would be 2x(x+1) which allows us to cancel out the denominator of the first fraction, as well as the second fraction by multiplying both fractions with the common denominator. You will then have 2x(x-2) as the nominator for the second fraction which you will distribute the 2x in the bracket. Next, move the 6 over so you can factor. Start by finding out what is common which is 2 then start factoring. The answer is -1 or 3 but since we said the restriction is 0 and -1, -1 is not a solution; therefore, the answer is 3.

Week 14 – Rational Expressions

Posted on May 21, 2018 by shannonf2015

This week, I have learned what are non-permissible values and how to simplify rational expressions.

Non-permissible values are the values that x cannot be. The values are determined by the denominator; it’s the values that results in 0 as the denominator.

Let’s look at an example:

First step is to factor.

Once you factor everything out, list out the non-permissible values. After, you can simplify by crossing out what is common on the nominator and denominator which basically equals to one.

Let’s look at another example of multiplying and dividing rational expressions.

When it comes to dividing, we start by listing the given non-permissible values. The first one that we can see is 4.

Since it’s division, we must flip the second fraction (reciprocal) and then we can list the other non-permissible value which is 5. Next is to factor then list out any other given non-permissible values.

The other non-permissible value is -5. Now we can cancel out values diagonally which will give us the simplified expression.

Week 13 – Graphing Reciprocals

Posted on May 12, 2018 by shannonf2015

This week we have learned how to graph reciprocals for both linear functions and quadratic functions.

For graphing reciprocal functions, we have to determine the asymptotes.

The x-axis is a horizontal asymptote and a vertical asymptote is when x has a certain value, a vertical line that the graph approaches but never reaches.

For all reciprocal functions, y cannot be 0 since $\frac{1}{f(x)}$ can never be 0.

When we take the reciprocal of a number, the only numbers that stay the same are -1 and 1. Therefore, when we graph, we find the points that line up to where y=1 and y=-1. These are the invariant points.

Let’s look at an example of a linear function.

$\frac{1}{-3x+9}$

First step is to graph the line.

Next is to locate the invariant points.

Next is to find the asymptotes. Horizontal asymptote is always y=0.

Once you have the line graphed, now you must draw a curved line that goes only through the variant points while approaching the asymptotes but not actually touching it.

let’s look at another example but with quadratic functions.

Example:

y = $2x^2-4x-6$ and y = $\frac{1}{2x^2-4x-6}$

First is to graph the parabola. Start by factoring to find the x-intercepts.

Now that you have the x-intercepts, you can add them together and divide it by two to find the axis of symmetry and plug that and a point in standard form to find q for the vertex.

When you draw out the parabola, circle the invariant points and draw in the asymptotes.

Draw a curved line through the invariant points.

If the parabola has one x-intercept, then it will be divided into 4 zones and will have two invariant points:

If the parabola has no x-intercepts, then it will divided in two zones and it won’t have any invariant points or a vertical asymptote:

Week 12 – Absolute Value Functions

Posted on May 5, 2018 by shannonf2015

This week, I have learned how to graph the absolute value of a linear function, the absolute value of a quadratic function and how to write it in piecewise notation.

Example of a linear function:

y = I3x-5I

Let’s start by graphing this.

We know the y-intercept is -5, and the slope is 3. Start by plotting the given information on the graph for y = 3x-5

Now, to be able to place the absolute value, we must remember that absolute values cannot be negative. So the line that goes past the x-axis will instead change to positive values which will make the line bounce upwards after touching the x-axis.

To write this in piecewise notation, we can’t really tell what the x-axis is here, so we can use the equation to determine the x-intercept.

Let’s look at an example of a quadratic function:

Let’s start by graphing this. We know the vertex is (-1, 1) and it’s congruent to $y=x^2$ which follows the pattern 1,3,5,7,9…

Once you have it graphed without the absolute value symbol, you now can start graphing the equation when it’s in the absolute symbol. To do this, all the negative values must switch to positive values. Therefore, the outsides of the parabola will flip up.

Now that you have this, we can use this graph to determine the intercepts and domain and range. The x-intercept is -2 and 0 and the y-intercept is 0. The domain is always the element of all real numbers (xER) and the range is y ≥ 0 as the values of y must be +.

To write this in piecewise notation, we must first look at the x-intercepts. We can see that the outer part changes but the inner part does not whether it’s in the absolute value symbol or not. So y = $-(x+1)^2+1$ is if -2 ≥ x ≥ 0 (The inner part of the graph; we are including the + values and 0).

For $y=-(-(x+1)^2+1)$ if x < -2 and x > 0 (The outer parts of the parabola; all the negative values).

So the piecewise notation is:

Week 11 – Graphing Inequalities and Systems of Equations

Posted on April 28, 2018 by shannonf2015

For the past week, I have learned how to write solutions for quadratic inequalities and also how to graph them.

Let’s start with an example:

Example 1: Given an equation

$2x^2-8x-10<0$

First step is take out what is common then factor to be able to find the x-intercepts.

Common: 2

2( $x^2-4x-5$ )<0

2. Factor

2(x+1)(x-5) < 0

x-intercepts: -1 and 5

Now that you have the x-intercepts, we can draw these on a number line.

By looking at the equation, we can tell the parabola will be opening up as the coefficient next to $x^2$ is positive. So there is a +, then a – , then a + as the parabola will dip past the x-axis since it has two x-intercepts.

Looking at the inequality symbol, it’s using <. This means they are asking for the numbers that are less than 0, so negative numbers.

The negative numbers are in the middle so the solution would look like this:

-1 < x < 5

The circles are not coloured because it is not including “or equal to”. This equation is just using less than.

Now let’s look at another example where we are given a graph and we must write an inequality.

Let’s go over some important things:

≤ and ≥ : has a solid line

< and >: has a dash line

Example:

First step is to find the y intercept, which is -1. Next is to find the slope. We can tell by the points that it’s going up by 4 and moves to the right once. So the slope is 4. This linear graph is tilted up so it’s positive.

The equation so far looks like this:

y ☐ $4x^2-1$

To find the inequality symbol, we must test a number from the shaded section that is not a point on the line and determine which symbol to choose which will result in a correct answer.

Let’s use (5,0)

Plug in 5 in x and 0 in y:

0 ☐ $4(5)^2-1$

0 ☐ 4(25)-1

0 ☐ 100-1

0 ☐ 99

To make this true, we must use ≤ because 99 is greater than 0 and the line is solid.

So the equation will be:

y ≤ $4x^2-1$

Week 10 – Radicals, Arithmetic Series and Infinite Geometric Series

Posted on April 21, 2018 by shannonf2015

During review, I have recalled some concepts I have forgotten.

One of them is when a value or variable has an exponent that is a fraction. When it’s a fraction, the denominator represents the root. We can remember this as Flower Power.

Example:

$64^\frac{1}{3}$ —> $\sqrt[3]{64}$ —> 4

Another simple method I recall is adding an arithmetic series that is finite. The first step is taking the first term and the last term in the series and adding them together. Next, count how many terms there are in total and divide the amount by two. Multiply the terms divided by two with the sum of both first and last terms.

Example:

10 + 5 + 0 -5 -10 -15

10 + (-15) = -5
Number of terms in total: 6
$\frac{6}{2}$ = 3
(-5)(3)
Sum = -15

Another concept I did not talk about last time was infinite geometric series. We can tell if they converge or diverge by looking at the value of the common ratio.

Diverges: r > 1; no sum

Converges: -1 < r < 1

Formula:

Image result for infinite geometric series formula

Example:

2 + 3 + 4.5 + 6.75 +…

r = $\frac{3}{2}$

The common ratio is greater than 1, so it is infinite (diverges); no sum.

-0.5 – 0.05 – 0.005 – 0.0005 – …

r = $\frac{-0.05}{-0.5}$ = 0.1

The common ratio is greater than -1 and smaller than 1 so it is finite (converges); has a sum.

Let’s look at an example of how to use the formula with a geometric series.

8 + 2 + 0.5 + 0.125 + …

Information given:

a = 8
r = $\frac{1}{4}$

Plug in the information given in the formula and solve.

Remember to flip the fraction (reciprocals) when doing division.

Week 9 – Quadratic Functions

Posted on April 14, 2018 by shannonf2015

Throughout the week, I have learned how to convert a quadratic function in general form to standard form and factored form to be able to find different information about how the graph looks like.

Let’s start with an example:

$y=2x^2-20x+32$

Right away, we can tell that the y-intercept is 32 because it is in general form.

We can also tell the direction of opening as it is positive, which means it opens up and has a minimum value.

Start by removing the greatest common factor which is 2.

$y=2(x^2-10x+16)$

Next, factor the three term in the bracket to find the x-intercepts (the roots)

$y=2(x-2)(x-8)$

x intercepts: 2, 8

Now we need to find the vertex, to do that we need to change the general form to standard form by completing the squares. We cannot change from factored form to standard form.

$y=2x^2-20x+32$

First take out what is common between the first two terms, leaving the 32. Leave two spaces for a + value and – value between middle term and last term to create zero pairs.