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Week 12 in PreCalc 11

We dealt with a lot of things this week, especially about systems. And for this blog, I’m going to talk about linear-quadratic system.

 

Let’s have a quick recap:

  • Linear equation deals with a straight line on the graph. Basically, an equation of a line.
    • The equation being known as y = mx + b, where is the slope, and is the y-intercept.
  • Quadratic equation deals with a parabola on a graph, with the equation having at least one squared variable.
    • The standard equation being known as y = a (x – p)2 + qFind out more on my blog post about quadratic equations!
  • Together, they form a relation called System of Linear and Quadratic Equation. 

 

Systems (where they intersect) of these two equations can be find out or solved:

  • graphically.
  • algebraically.
    • substitution.
    • elimination.

However, in this blog I’m only going to talk about solving it graphically.

(At this point, you should be able to graph both of them!)

 

NOTE: There are three possible cases that can happen if you finished graphing them!

(graphs used are in courtesy of desmos!)

  • Two solutions – if the line passes on two points of the parabola
  • One solution – if the line passes only on one point of the parabola.
  • No solution – if the line and the parabola doesn’t cross at all.
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Week 10 in PreCalc 11

As our math midterm is getting near, this week has been a review week which has been a great refresher for my memory. It was good and bad to know that I’ve forgotten a lot of things. But it’s a good thing to reflect to when studying!

Not much was learned, but rather a lot of refreshers. Here are some of them:

I wouldn’t go into much details about them all as I’ve posted them already in my edublog, but I would do some examples. Here are the links to the blog posts:

 

Absolute Values:

  • how far away a number from zero is in a number line.
  • \sqrt{x}^2
  • The answer will always be positive inside the absolute value sign ” | | “
  • Can act like a parenthesis.

Examples:

| 5 – 7 |

= | -2 |

= 2

 

-|-4|

= – | 4

= -4

 

Arithmetic Sequence:

  • A sequence of numbers that are added or subtracted by the same value.
  • Addition of a number sequence. E.g. 1 + 2 + 3 + 4 + 5 + 6….
  • S_n = \frac{n}{2}(a + t_n)
  • S_n = \frac{n}{2}(2a + (n-1)d)
  • n = the number/amount of terms you’re calculating
  • a = the first term
  • tn = last term
  • d = common difference

Example:

2 + 5 + 8 + 11…

n = 20

a = 2

 

Since we don’t have tn, we’ll be using the second formula.

S_n = \frac{n}{2}(2a + (n-1)d) S_{20} = \frac{20}{2}(2(2) + (20-1)3) S_{20} = 10 (4 + 57) S_{20} = 10 (61) S_{20} =610

 

Adding and Subtracting Radicals

  • add/subtract radicals with the same index and radicand.
  • Simplify if possible.
  • \sqrt[n]{x}
  • n is index, and x is radicand.
  • Never add radicand and index. Just add the number outside of the radical.

Examples:

\sqrt{3} + 2\sqrt{3} = 3\sqrt{3}

can be added because they have the same index and radicand…

 

4\sqrt[4]{5} - \sqrt{5}

is simplified to as…

4\sqrt[4]{5} - \sqrt{5}

can’t be subtracted because although they have the same radicand, they don’t have the same index..

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WWI Artifact: ‘The Revolution’ at Camp Ruhleben

(Source: http://ww1.canada.com/memory-project – ‘The Revolution’ at Camp Ruhleben, Germany in November, 1918 – Eileen Campbell, Thelma Johnson, Gail Campbell and Joan Chittick)

 

What does this artifact tell us about the First World War?
It tells us that not only the war is happening outside the countries, it’s also happening inside. Because of the declaration of wars, people who were the from the enemy’s country were considered as prisoners, like how British civilians became prisoned in German. Camp Ruhleben is an example, they imprisoned British people as the war started and they were treated horribly. This shows the effect of the war had to the people of their own country.
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Week 9 in PreCalc 11

Something I’ve learned this week is about modeling. No – not the modeling one where you pose for a picture – what I’m talking about is creating a quadratic equation based on a word problem, like regarding revenue, money, finding numbers, projectile motions, etc.

 

There’s actually no need for me to explain what it is as it’s pretty straightforward. All you have to do is analyze the word problem carefully and try and make an equation (of course, with a graph!) based on the problem.

 

Now, let’s take a look at this problem.

 

Find two integers that has a sum of 16, and the greatest possible product.

 

two numbers that has a sum of 16. In other words, it can be written like this algebraically,

x + y = 16

which then, we can arrange into…

y = 16 – x

 

Now, let’s say that

x = 1^{st} number

and…

$latex 16 – x = 2^{nd} number$

 

Now, if we add those two, we should get a sum of 16.

Why? Well, since our first number is x, and we arranged the equation to become y=16-x, which then means that y has the same value as 16-x, then if you add

x and 16-x, you should get 16 as in the first equation you’re adding x and y.

 

Anyways, since we also need to find the greatest possible product, we’re gonna multiply them instead of adding, which we’ll get the equation…

x (16 - x) = y

 

Right now, it’s in the factored form, so it means we already have our x-intercepts or roots, which are:

x = 0 ; and x = 16

Now what? Well, we need to find the value of the vertex, because the x-value is the value of each two integers that we need to find and the y-value is the product that we also need to find.

 

we can find the x-value of the vertex in two ways: graphing and calculating.

 

For calculating, we just need to find the average, which is pretty easy.

0 + 16 = 16

16 / 2 = 8

And x = 0 is your vertex’ x-value.

 

Now, for the graphing, we just need to sketch it. but since we already have the x-value let’s do a sketch…

 

Now, how do we figure out the product? Well, just plug the x-value of vertex in your formula.

 

f(8) = 8 (16-8)

f(8) = 8 (8)

f(8) = 64

 

Well, there you have it!

 

Two integers that has a sum of 16 is 8, and 8 with the greatest possible product of 64!

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Perceptions vs Reality: WWI

Explain the biggest differences between the perceptions of 1914 and the reality of war in World War One.

 

People, especially the poor ones, thought that enlisting to the army would give them a great opportunity to gain wealth, reputation, to show off, etc. Some were even excited to go to the war. However, little did they know that the life they would have was worse than the life they had back to their homes. They thought that after the war, they would live better than before, but the reality was very different. They had only thought that it would last for a year or two, but it lasted double than that. They weren’t prepared, so they didn’t get enough food later especially when the army doubled in size, some got depression due to the environment they live in, some inflicted wounds upon themselves, some committed suicide, and some even went insane. They didn’t get any medical treatment as technology wasn’t very advanced back then, so they got infected and most of them died. Not only did they not get the result they wanted, but it made their lives worse.

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Newton’s Three Laws of Motion

Hello! In this blog post, I will talk about and analyze Newton’s Three Laws of Motion and apply them to real-life situations.

This post will be divided into three parts: First Law, Second Law, and Third Law.

 

Section I: Newton’s First Law of Motion

Newton’s first law of motion states that a body or an object at rest will remain at rest, and a body or an object in motion will remain in motion, unless acted upon an outside force.

This law is typically called the law of inertia. So what is inertia?

It is the tendency for all the objects to resist change in their states of motion. It is dependent on mass.

 

So what does this all mean?

It means that any object has a tendency to not change what their state it: whether they are moving or at rest. All objects resist a change in their motion. If they are in rest, they will remain unmoving, and it they are moving, then they will not stop moving. Unless, they are overcame by an outside force.

 

Let’s take a look at this example to understand it a little better: a pushed, moving cart.

Let’s recall what inertia is: the tendency of an object to resist change in motion.

As we can see in the video, the cart was in rest, however it was pushed by an outside force and moved. After that, it went back to rest (not moving) again.

So, how exactly does the law apply to this situation?

Remember that every object resists change in motion. Even though the cart was moving, but due to the cart’s natural state (at rest), it will continue doing that. And when the cart was pushed, then it will move, and it will keep continue moving.

But wait, didn’t the law state that if an object is moving, then it will continue to move? Then, why did it stop moving?
Well, this law works the best when there is no friction. This is one of the exceptions for this law. The object can’t keep moving on forever on a surface with friction, because the friction will oppose its movement. Like what we saw in the video, it stopped moving exactly because there was a force opposing it. Hence the law, “the object will continue what its doing unless an unbalanced force acts upon it.”

 

Section II: Newton’s Second Law of Motion

Newton’s second law of motion states that if an unbalanced force acts on a body, it will accelerate.

This law should be easy to comprehend, because we already know this in our heads unconsciously. We know that we should not kick a big rock like we kick a soccer ball, right?

Why? Because it’s heavy.

 

So what does this mean? Well, the greater the mass of the object, the greater the force needed to accelerate it.

 

Mathematically, it can also be expressed by this:

-acceleration is directly dependent on the unbalanced(net) force and it’s inversely dependent on the mass of the object.

**Don’t forget, Fnet is the sum of all forces acting upon an object!**

So basically, a is acceleration; m is mass; Fnet is the unbalanced force.

a = \frac{Fnet}{m}

we can also say:

Fnet = ma

For a deeper understanding, let’s look at this video.

As we can see here, we have two objects: a mouse and a popsicle stick.
We all know that the mouse is a lot more heavier than the stick. And as said earlier, more mass means more force needed to accelerate it.

Now, as seen on the video, when both of them was flicked, the popsicle stick moved away faster and farther than the mouse.

So how does the law apply to this? Well, since the popsicle stick had less mass than the mouse, it accelerated faster with less net force needed. On the other hand, the mouse needed more net force to accelerate it.

This law won’t always work on something that is moving uniformly (constant speed), because it has no acceleration. Why? Well, since acceleration is directly proportional to the net force, if there’s no acceleration, then there are still forces acting upon it, but the forces are balanced with one another.

Another perspective is looking at the formulas. Since Fnet = ma, if your acceleration is 0, then there’s no Fnet.

 

Section III: Newton’s Third Law of Motion

Newton’s third law of motion states that for every action, there is an equal and opposite reaction.

The law itself is pretty much self-explanatory. So, for every movement or action, there will be an action and a reaction.

Now, let’s take a look at this video of a ball dropping and bouncing off the floor.

Assuming that the ball was dropped, the ball pushes against the floor, whilst the floor pushes against the ball with an equal amount of force. And that’s the first part of the law, “equal reaction.”

 

The opposites reactions would be the same, but in different meanings.

Action: The ball pushing down against the floor.

Reaction: The floor pushing up against the ball, making the ball bounce, which makes an opposite reaction.

 

 

Thank you! 🙂

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Week 8 in PreCalc 11

***All graphs used are taken from desmos.***

 

Another week in Pre-Calc 11, another blog post!

Something that I’ve learned this week is about the equation y = a (x - p)^2 + q

 

So, what exactly is this equation and what does this do?

 

Now, let’s do some basics. For example, y = x^2

If you graph it, it will be like this:

Let’s analyze some parts…

The vertex, which is the lowest or the highest point of the graph, depending on where it is. Where, as you can see in the graph, is in (0,0)

We can also find the domain (x values) and the range (y values) with our vertex.

Always remember that our x-values are always an element of real numbers, or x ∈ R. Why? Because take note that our graph widens horizontally, and take note that if it goes on forever, then our x values can be anything.

Not the same thing goes for our range, though. As you can see, our graph has either a maximum or a minimum, depending on the equation (which I’ll go through later at Section III). And in this case, our minimum is y=0. So it means that y must be greater than or equal to zero, or y ≥ 0.

There are many points to consider such as x-intercepts, which we know as where the line crosses the x-axis or if the equation is y=0. And the y-intercept, which we know as where the line crosses the y-axis or if the equation is x=0.

 

And we can see that in the graph, both y and x-intercepts are (0,0).

We also need to take note of the line of symmetry.

It’s basically the line that divides the graph equally, which in this case is also x = 0.

 

Moving on to the main topic, what exactly is y = a (x-p)^2 + q

In this blog post, we will divide them first into three sections to understand them more.

the three sections are:

I.     y = x² + q

II.    y = (x – p)²

III.   y = ax²

 

Section I: y = x² + q

Let’s go briefly over this.

So what ‘q’ does in our equation is translating (or sliding, shifting,…) our graph vertically.

Just note that -q slides downwards and +q slides upwards.

i.e. comparing q = 3 and q = -3

y = x² + 3

y = x² – 3

As you can see, if our equation is y=x²+3, our graph slides upwards by 3, and if our equation is y=x²-3, our graph slides downwards by 3.

 

Section II: y = (x – p)²

What ‘p’ does to our graph is slightly the same as ‘q’, however it translates or shifts/slides the graph horizontally. Another thing is that this equation is very confusing, because as we are familiar to, we move to the right if it’s positive, and we move to the left if it’s negative. However, this is the exact opposite. If it’s (x-p)², then it slides to the right or positive side of the graph. And if it’s (x+p)², then it slides to the left or negative side of the graph.

i.e. y = (x – 2)² vs. y = (x + 2)²

 

y = (x – 2)²

y = (x + 2)²

As you can see, if our equation is y=(x-2)², our graph would translate to the right, and if our graph is y=(x+2)², our graph would translate to the left.

 

Section III: y = ax²

Now, this part of the equation is a lot more different the other two.

What a does in the equation is changing the wideness of the graph, and where the graph opens up, or the reflection depending on its sign.

To make sense out of it better, let’s analyze these graphs.

y = 1/2 x² (TOP)

y = -1/2 x² (BOTTOM)

 

y = 2x² (TOP)

y = -2x² (BOTTOM)

 

What we can conclude with that is if the sign (-/+) of a is positive, then it opens up, or the graph has a minimum value or “lowest point.” And if the sign is negative, then it’s opening downwards, or it reflects, or you can say that the graph has a maximum value or “highest point.”

Also, what we can conclude from those graphs is the wideness of graph. We can see that if it’s greater than zero but less than one (0 < a < 1), then our graph will be wider. And if it’s greater than one (a > 1), then the graph will be narrower.

 

Now, let’s try applying it to the equation y = a (x – p)² + q

So basically, the summary of what we did is:

a = changes the wideness and the direction the graph opens up. (negative reflects down, positive opens up) (0 < a < 1, wider graph) (a > 1, narrower graph)

p = translates the graph horizontally. (negative translates to the right, positive translates to the left)

q = translates the graph vertically. (positive translates upwards, negative translates downwards)

 

Example: y = 2 (x – 3)² – 4

Applying our skills, we know that a is 2, which means our graph will be narrow and opens up.

We know that our p value is – 3, which means it will translate to the right by 3.

We know that our q value is 4, which means it will translate downwards by 4.

 

That’s everything! Thank you.

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Week 7 in PreCalc 11

Late post, but in this blog, I will talk about what I’ve learned last week (week 7) in Precalc 11!

 

To briefly go over it, one of the things I’ve learned is about the discriminant. The discriminant is a part of the quadratic formula used to find how many “roots,” “solutions,” “zeros,” or what we usually call as “x’s” in our formula.

 

Before we go over that, let’s talk about the quadratic formula quickly.

 this is the formula used to find the “x’s” of a quadratic equation. This is usually used when you can’t “factor” your equation.

 

Anyways, the “discriminant” that I was talking about is the root part of the equation or “b2 – 4ac”

 

What does it do?

Like I’ve said, it’s used to find if there are any roots or solutions in your equation and how many there is(are).

 

So how exactly do we determine how many “roots” there is?

 

*NOTE: Examples used are easier to factor just for easier explanation, but typically you use the quadratic formula if there is no way of factoring it.*

 

Situation one: b2 – 4ac > 0

When the discriminant is more than zero or positive, then equation ax2 + bx + c = 0 (quadratic) has two real, unequal roots.

i.e. x2 + 5x + 4 = 0

a = 1; b = 5; c = 4

b2 – 4ac > 0

52 – 4(1)(4) > 0

25 – 16 > 0

9 > 0

So this means that x2 + 5x + 4 = 0 has two unequal, and real roots.

Notice that if you’d factor it, it will be (x + 4) (x + 1), in which your roots are

x = -4 and x = -1, that’s why you’d have two real, unequal roots.

 

Situation two: b2 – 4ac = 0

When the discriminant is zero, then there will be two real and equal roots. Technically, there is only “one” solution because both x’s have the same value. This usually occurs in perfect square trinomials.

i.e. x2 + 4x + 4 = 0

a = 1; b = 4; c = 4

b2 – 4ac = 0

42 – 4(1)(4) = 0

16 – 16 = 0

0 = 0

This means that the equation x2 + 4x + 4 = 0 only has one real root. But notice how you’d know it immediately if you factor a perfect square trinomial.

x2 + 4x + 4 = 0

(x + 2) (x + 2) = 0

x = -2; x= – 2. They just have the same value for both solutions, so technically, even though there are two solutions, we can say that it only has one.

 

Situation three: b2 – 4ac < 0

When the discriminant is less than zero or negative, then the equation has no roots or solutions.

Why? Well, think about it. The quadratic formula is like this: r

And we use the discriminant part, which is under a square root sign to figure out the roots.

So, if your discriminant is negative, then it will be an error because you can’t calculate the square root of a negative number.

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Week 6 in PreCalc 11

Another week, another blog post for Pre-Calc 11!

 

There were many things that I’ve learned the past week, but one thing we’re going to go over is “completing the square.”

 

First of all, we have to make sure that our QUADRATIC EQUATION (as explained in the previous blog) is EQUAL to ZERO (0).

Now, let’s do an example here. Let’s say that we have…

 

(Unfortunately, latex coding isn’t working well this time…)

^ = to the power of.

 

x^2 + 6x + 8 = 0

Now, what we’re going to do is take the first two terms and make it a perfect square trinomial.

(x^2 + 6x + __) + 8 – __ = 0

So, for this to work, as seen above, we’re going to technically “isolate” the first two terms from the 3rd term and make it a perfect square trinomial. But remember, you can’t just add it like that. Notice that the equation is equal to zero. So this means that you need to have a “zero pair.”

Which means that if you add (let v = ) v from your first two terms, then you must subtract v from the last term.

(x^2 + 6x + 9) + 8 – 9 = 0

So, how did I figure the value for the last term is 9? Here’s the trick.

Divide your middle term by half and square it. So basically if our equation is ax^2 + bx + c, then it will be: (b/2)^2

(x+3)^2 – 1 = 0

So how did we get this? We factor the trinomial (go to my prev blog post for more info) and combined the last two terms.

The next steps should be really straightforward…

(x+3)^2 = 1

Then, just square root both sides. It’ll be:

x+3 = ±√1

x = -3 ±√1

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Week 5 in Precalculus 11

What I learned this week is about quadratic trinomials and how to factor them.

 

A quadratic trinomial looks like this:

ax^2 + bx+ c

Remember that a quadratic trinomial always have a degree of 2.

 

Basically, what you need to do for factoring is find two numbers that will multiply to “c”, and also they must add together to have a sum that’s equal to “b”.

We must not forget signs (- / +) and how they work because they’re also necessary for this unit!

 

E.g.

x^2 + 5x + 6

Find factors of 6:

1 x 6 ; -1 x -6

2 x 3 ; -2 x -3

 

Now, what will add to “5”?

1 + 6 = 7 (NO!)

2 + 3 = 5 (YES!)

 

So now, we’re using 2 and 3 as factors.

x^2 + 5x + 6 =(x + 2) (x + 3)

 

As you’ve noticed, what you did was like doing the FOIL method backwards. For information about FOIL METHOD, check out my blog post about FOIL.