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# Week 17 in Pre-Calc 11

This week, we have learned trigonometry and I will discuss one of the things I learned: the Sine Law.

It’s a very useful to solve any triangle as long as you have:

• an angle that has a side corresponding it.
• two angles / two sides / one angle and one side.

Here’s how it works: every angle needs to be written in CAPITAL LETTERS and every side needs to be written in LOWER CASE LETTERS. angle A corresponds with side a or side BC.

angle B corresponds with side b or side AC.

angle C corresponds with side c or side AB.

what’s the formula?

if you’re finding an angle… $\frac{sinA}{a} = \frac{sinB}{b} = \frac{sinC}{c}$

and if you’re finding a side… $\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$   Posted on

# Week 15 in PreCalc 11

One thing I’ve learned this week is about multiplying rational expressions.

It doesn’t really need a lot of explanation as it’s just your simple multiplying of fractions, however with variables this time.

Let’s have a recap: $(\frac{2}{5}) (\frac{10}{8}) = \frac{20}{40}$

We just multiply straight across, but remember we need to simplify our answer.

20/40 = 1/2

But we can also do it this way: It’s like prime factorization method. So, you take out all of the prime factors, then just cancel out. Well, that’s exactly what we’re going to do! But let’s add variables to the fun. So we have this not-so-pretty looking fractions and we might be thinking, how do we solve that?!

Well, I got good news! They’re pretty easy to solve as long as you’re good at factoring!

So first, factor everything. Now that you’ve factored everything, just cancel out same factors, just like what we did earlier in that simple fraction. Next – don’t forget about this! Remember the rule that you cannot have a zero on your denominator? This also applies to this! So, after everything’s factored out, calculate the values of (or other variables) that will result the denominator to equal zero. Now you’re done!

If you’re wondering ‘couldn’t we have just cancel numbers out before we factor’? Take note that if numbers have +/- between them, they’re called terms and you cannot cancel them. So we factor them so we can get factors, which has multiplication sign between them.

YOU CAN NOT CANCEL TERMS

YOU CAN ONLY CANCEL FACTORS!

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# Week 14 in Pre-Calc 11

One thing I’ve learned this week is about reciprocal functions. We’ve always use that word when we divided fractions. We ‘reciprocate’ the fraction we’re dividing with.

Anyways, let’s recall what a reciprocal is:

• the reciprocal ofis $\frac {1}{x}$
• Basically, the reciprocal of a number is just the numerator and denominator switched.
• Remember that all whole numbers and variables are over one. They have a denominator of one, basically.

What we need to remember about this topic are asymptotes, one’s called vertical asymptote (x value), one’s horizontal asymptote (y value).  An asymptote is a line that corresponds to the zeroes of your equation.

So, let’s say we have $y = \frac{1}{3x + 6}$

We must know that we cannot have a zero on the denominator, so we’ll first find out which value of x would result to have a zero denominator.

3x + 6 = 0

3x = -6

x = -2

Like was stated earlier, asymptotes are the zeroes to the function. So basically, since we just found out the zero of the denominator, we also found out one of the asymptotes. Remember, the vertical asymptote is a value of x, so our vertical asymptote is (x=-2)

Right now, though, we don’t need to bother with the horizontal asymptote. Take note that if the function’s denominator is 1, our horizontal asymptote will always be zero. (y=0)

So this is how you graph it:

• put a dashed line on where x= -2 is.
• put a dashed line on where y = 0 is.
• If the slope of your original line (3x + 6) is positive, then you will draw the hyperbolas on quadrant I and III, and if it’s negative draw it on quadrant II and IV.
• (https://www.varsitytutors.com/hotmath/hotmath_help/topics/quadrants)
• The hyperbolas must never touch the asymptotes, since they are your non-permissible values.
• The graph should look like this.:
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# Week 13 in Pre-Calc 11

This week, I’ve learned about absolute values function on graphing.

And in this blog, I’m going to talk about how to graph an absolute value function of linear equations.

Just a recap, but we do know that for absolute values, we must remember that:

• the sum inside the absolute value symbols will always be positive !

Taking the main idea that we can never get any negative answer in absolute values, since we have an equation of y=|mx+b| then if we graph our linear equation, we must never have any lines within the negative y side of the graph.

So how exactly do we graph it?

First of all, we must know what the parent function is of the graph equation we are given.

For example, we have y = |3x + 6|

If we take the absolute value out, we get y = 3x + 6

Secondly, we must figure out what the x-intercept is, doing it by graphing or algebraically doesn’t matter as long as we figure out what the x-intercept is accurately.

And in our equation y = 3x + 6, we want the y to equal zero so we could figure out our x-intercept.

In this case, it’s

0 = 3x + 6

-6 = 3x

-2 = x

Now that we’ve figure out what the x-intercept is, we can start graphing by taking the y-intercept, which is (0, 6) in our equation. After that, make a vertical dashed line to where the x-intercept is. Then, graph the equation without the absolute value. Then, what you’re going to do is to make the graph “reflect” from the x-axis from where the x-intercept is. Remember that the reflected graph HAS THE NEGATIVE SLOPE OF YOUR ORIGINAL LINE!!! If you’re stuck, just remember the idea of absolute value not having a negative as an answer!!

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# Week 11 in PreCalc 11

Last week in Pre-calculus was the 11th week. Sorry for the one-week delay.

One thing that I’ve learned is solving linear inequalities in one variable.

Overall, this topic should be quite straightforward. You just need to be careful of your signs (+/-) and the inequality that you need to use.

First of all, since we’re only solving in one variable, we don’t need to use the x-y graph, just a number line would be fine.

These are all the inequality signs that we’re going to use: Keep in mind that if you graph them on the number line, there’s a slight difference between greater/less than and greater/less than or equal

If you graph an inequality with greater than or less than signs, the number that it’s based on is NOT shaded or part of the solutions.

On the other hand, if you graph it with greater than or equal or less than or equal signs, the number it’s based on IS shaded or part of the solutions…

(greater/less than – left) (greater/less than or equal – right) NOTE: keep in mind that if you DIVIDE or MULTIPLY by a NEGATIVE, the inequality sign must be SWITCHEDcheck example 2!!

Examples:  Posted on

# 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.  Posted on

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

• Simplify if possible.
• $\sqrt[n]{x}$
• n is index, and x is radicand.

Examples: $\sqrt{3} + 2\sqrt{3} = 3\sqrt{3}$ $4\sqrt{5} - \sqrt{5}$

is simplified to as… $4\sqrt{5} - \sqrt{5}$

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

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# 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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# 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.