Pre-Calculus 11 Week 10 – Midterm Review!

Our midterm is next week! So for this post, I’ll review one of the units that I had more trouble with. Absolute Values and Radicals. More specifically, rationalising fractions that have binomials as a denominator. Such as…

\frac{3\sqrt{8}+\sqrt2{5}}{\sqrt{2}+\sqrt{20}}

To rationalise a fraction means to make sure that there is no radical in the denominator. In the case that there’s a binomial as the denominator, you still multiply by the conjugate. The conjugate in this case will be a binomial, \frac{\sqrt{2}-\sqrt{20}}{\sqrt{2}-\sqrt{20}}. Multiplying the denominator by it’s conjugate eliminates the “middle term” that it usually creates. This creates a denominator with no radicals! Proof is below… (Excuse my handwriting)

Pre-Calculus 11 Week 9 – Different Forms for Quadratic Equations

Quadratic equations can take many different forms. There are three that we learned in this unit. Standard (Sometimes named Vertex) form, General form, and Factored form.

For this unit, the Standard form is generally (Heh.) the most useful, as it can show the most amount of information. The scale, whether it’s minimum or maximum, left/right/up/down translations, and the location of the vertex. That’s a lot. The Standard form is y = a(x - p)^2 + q. I go into more detail regarding the Standard form in my last week post here.

From the standard form, you can go back into the general form by distributing and then simplifying the equation. Treat it like a normal question and eventually, you’ll reach the General form. (NOTE: You cannot go from the Standard form to Factored form. To get from Standard to Factored, you must go Standard->General->Factored.)

The General form is one of the more familiar forms you’re probably used to prior to this unit. The general form can tell us the scale, whether it’s minimum or maximum, and the y-intercept. is the scale and whether the parabola opens up or down. c represents the y-intercept. However, the middle term, b is almost useless to us. b is a combination of the parabola translating left/right and up/down. It’s hard for us to do anything with it, so it’s disregarded for this unit. The General form is ax^2 + bx + c, just to refresh your memory.

From the General form, you can go to the Standard form by completing the square. To go to factored form, simply factor the equation.

The Factored form is unique, as it is the only form to tell us our roots, or x-intercepts. It can also tell us our scale, and whether the parabola opens up or down. x1 and x2 represents our roots/x-intercepts. The Factored form is a(x - x1)(x - x2).

i don’t know how to do subscript please forgive me.

Pre-Calculus 11 Week 8 – Graphing Quadratic Functions

How do you graph a quadratic function? It’s the same way you would graph a normal equation. But instead of it looking like a line, it looks like this: 

Like, literally, that’s it. That’s all. Just make a table of values, x and y on either side, plug some x values in, calculate for the y values. That’s all. But how bendy and twisty the line is isn’t random. You can spot certain patterns in quadratic functions. Take y = z(x - p)^2 + q for example.

z represents the scale factor (More on that in a bit), and whether or not the parabola will open up or down. If it opens up, z is positive. If it’s negative, the parabola will open down.

p represents if the parabola moves to the left or to the right of the y-axis. If p is +5, then the parabola’s axis of symmetry, or the middle, will be x = 5. However, this means that in the equation, it’ll appear as y = z(x - 5)^2 + q. Because if we remember, a positive and a negative make a negative, so in y = z(x - 5)^2 + q, it’s actually telling us to translate five units to the right. If our equation was y = z(x + 5)^2 + q (Notice the +), then the parabola would be translated 5 units to the left. p also represents the location of the vertex on the x-axis.

q simply shows us where the vertex lies on the y-axis.

As for z, z is usually 1. If the scale factor is one, the parabola follows a 1, 3, 5 pattern. As in one up, one over. Three up, one over. Five up, one over. And so forth. If the scale factor is 2, then all of that is multiplied by two. Two up, one over. Six up, one over. Ten up, one over. And so forth. Scale factor is 3? Three up, one over. Nine up, one over. Fifteen up, one over. And so forth. So the higher the scale factor is, the thinner the parabola is going to be. The lower, the fatter.

http://amsi.org.au/teacher_modules/Quadratic_Function.html

Pre-Calculus 11 Week 7 – Using the Discriminate to Help Solve Quadratic Equations

This week, we learned how to learn if a quadratic equation had one solution, two solutions, or no solutions using the discriminate. What is the discriminate, you may ask? Look at your quadratic equation, see the expression inside the square root? That’s your discriminant. b^2 - 4ac

If you don’t remember, a represents the first term’s coefficient, %latex b$ represents the second/middle term’s coefficient, and %latex c$ represents the third/last term’s coefficient.

Say your quadratic equation is x^2 + 4x - 7 = 0, then the discriminate would be…

(4)^2 - 4(1)(-7) 16 + 28

The discriminant is 44

The discriminant can tell us whether the answer has two solutions, one solution, or no solutions. If the discriminant is positive, then there are two solutions to your quadratic equation. If the discriminant equals zero, there will be one solution. However, if the discriminant is negative, then your quadratic equation has no solutions.

A quadratic equation graphed, showing two solutions.

What’s a solution? Glad you asked, if you attempt to graph a quadratic equation, your line isn’t a line! It’s this little bendy thing, which is called a parabola. For a quadratic equation to have a solution, a part of it has to be touching the x-axis. So, if the para

bola touches two points on the x-axis, then it has two solutions. However, if the apex of the parabola touches the x-axis, then that means the equation only has one solution. And if the parabola doesn’t touch the x-axis? No solutions.

A quadratic equation graphed, showing no solutions.

A quadratic equation graphed, showing one solution. Notice only the very top, the apex of the parabola touches the x-axis.

Pre-Calculus 11 Week 6 – Solving Quadratic Equations Using the Quadratic Formula

This week, we’ve learned how to solve quadratic equations using the quadratic formula, which is different from simplifying When you attempt to solve a quadratic equation, there will be two possible answers. The quadratic formula looks like this (It’s not that scary, bear with me).

Now, in the quadratic formula, a represents the first term, b represents the middle term, and c represents the third term.

Say we’re given the question 3x^2 - 4x - 1 = 0. Here is what our quadratic equation, and our steps will look like:

You now have two answers for x, because you have a +or- sign.

Pre-Calculus 11 Week 5 – Factoring Quadratic Expressions

This week, we learned how to solve quadratic equations. A quadratic equation looks something like this: x^2 + 3x + 2

x^2 + x + n (n representing a number) Is what you’ll see most of the time. To solve these questions you factor it into two binomials. To check if your answer is correct, distribute the two binomials, and your answer should be the same as your quadratic equation.

The best way to find out what your two binomials are is to think about this: what are two numbers that multiply together to make your term, but also add together to made your middle term? For the above question, your answer is simple. The only numbers that multiply together to make 2 are 1 and 2. If you add 1 and 2 together, you will get your middle term, 3! What about a harder question?

Our question is x^2 + 7x + 12. I wrote down the multiples that could make 12. Then I find out which ones can add together to make our middle term, 7. The multiples 3 and 4 work! So our answer is (x+3)(x+4)!

 

 

 

Pre-Calculus 11 Week 4 – Adding and Subtracting Square Roots

This week, we discussed adding and subtracting numbers with square roots, such as 2\sqrt{2} + 3\sqrt{2}

The easiest way to learn how to add these numbers is simply looking at them the same way you would look at 2x + 3x in algebra. You can simplify this expression. 2x + 3x = 5x

But you wouldn’t be able to add, say, 2x + 2y together. The expression is already simplified. The same thing applies to roots!

You can add 2\sqrt{2} + 3\sqrt{2} together to make 5\sqrt{2}!

However, you are unable to simplify 4\sqrt{2} + 5\sqrt{3}, because the radicands aren’t the same.

If our square root is really big, like \sqrt {128} + 3\sqrt{2}, we can still add them together! We just need to simplify the square root first. In this case, \sqrt {128} simplifies into 8\sqrt {2}. Our steps are as follows:

\sqrt {128} + 3\sqrt{2}

 

\sqrt {8*8*2} + 3\sqrt{2}

 

8\sqrt{2} + 3\sqrt{2}

 

11\sqrt{2}

 

All these rules apply to subtraction as well.

7\sqrt {3} - 3\sqrt{3} = 4\sqrt{3}

and

\sqrt {243} - 3\sqrt{3}

 

9\sqrt {3} - 3\sqrt{3}

 

6\sqrt{3}

Pre-Calculus 11 Week 3 – Absolute Value of a Real Number

This week, we looked at what absolute values were. Absolute values are the distance that the number is from zero. In class, we used the analogy of attempting to throw crumpled paper balls into a garbage bin. One person made the shot, others missed and landed a certain distance from the trash can. While some paper balls will have landed in different positions, their distance from the trash can could be the same as another paper ball. Absolute values are written like so:

|-5| = 5

The answer is 5 because -5 is 5 away from 0.

|5| = 5

The answer is also 5 because 5 is 5 away from -5.

Numbers beside these terms act as coefficients. Like so:

3|-5|

3×5

=15

Or if there is an expression within the lines, you do those first before completing the equation:

3|-3+6|

3|3|

3*3

= 9

You can almost think of the lines like brackets.

 

Capital Punishment within the short story, “The Two Fishermen”

Capital Punishment within the short story, “The Two Fishermen”

Capital punishment is the legally authorized killing of someone as punishment for a crime. This was used for anyone who committed murder, rape, and treason. Canada used to have capital punishment for a while, up until 1976, where it was removed from the Canadian Criminal Code. The death penalty was life imprisonment, along with no chance of parole for 25 years for first-degree murders. Capital punishment remained within the Canadian National Defence Act, only for the most serious military offences until 1998.

If this story took place in Canada, the crime would have been committed between 1865-1961. This was the time when crimes of murder, rape, and treason all carried the death penalty. Then the events that happened in the story would be allowed to take place in Canada, and Thomas Delaney would be handed capital punishment for murdering the man molesting his wife.

Thomas Delaney should not have been hanged. Capital punishment was not appropriate for the crime he committed. He may have killed someone, but it was a) in the heat of the moment, b) to protect his wife, and c) the crime was really voluntary manslaughter, not murder. There was a fight between the two men, that was started by Matthew Rinehart  This resulted in one of them dying. This for sure warrants jail time, along with recuperating with therapy sessions or something. Either way, death was not the proper punishment for the crime he committed.

Gallows cartoons, Gallows cartoon, funny, Gallows picture, Gallows pictures, Gallows image, Gallows images, Gallows illustration, Gallows illustrations

Sources:

https://www.thoughtco.com/history-of-capital-punishment-in-canada-508141

https://www.cartoonstock.com/directory/g/gallows.asp

 

 

Pre-Calculus 11 Week 2 – Geometric Sequences

This week, I focused on geometric sequences.

Geometric sequences share similar characteristics to arithmetic sequences. For one, geometric sequences are patterns, just like our arithmetic sequences from last week! Arithmetic sequences have common differences, such as this one: “5, 10, 15, 20, 25, 30…” You can tell the pattern here is that 5 is being added for every new term!

Geometric sequences, however, have common ratios. Like so: “2, 4, 8, 16, 32, 64…” You should be able to see a pattern here. What’s the pattern? The terms are being multiplied by 2 each time! This is very different from our arithmetic sequence, where numbers were being added as the pattern was continuing. The ratio is determined like so: tn/(tn-1)=r.

The ratio is a very important number. It’s as important to geometric sequences as the common difference, d, is to arithmetic sequences, so get used to finding it!

There are a number of ways that you can utilize the common ratio, r. Say, you wanna find the 9th term in our geometric sequence here. Looks something like this:

t1*r^(n-1) = t9

2*2^(9-1) = t9

2*2^8 = t9

2*256 = t9

512 = t9

The general formula for geometric sequences is t1*r(n-1)=tn