Week 18 – My TOP 5 things from PC 11

This is my last blog post for Precal 11 and I will be going over my top 5 things from Precalc 11.

  1. Factoring Trinomials. This was one of my favourite parts of PC11 because I found it super easy to understand the concept, especially since I can think of factors quickly.
  2. Sine Law. This topic was extremely easy to understand because it only needed 3 pieces of info out of 4 and it works with more complex triangles that are not 90-degree triangles.
  3. Quadratic Formula. The quadratic formula was really annoying to remember in the beginning but after you have it memorized you can find answers that factoring can’t. The discriminant will also tell you how many solutions the equation has.
  4. Vertex Form. This formula was really easy to use and remember. Once you understand what each letter/number means. It made it really easy to graph it out. Once you learn how the stretch value works, it makes graphing so easy.
  5. Multiplying/Dividing Fractions with Nomials. This topic was easy to understand and helped a lot especially when they are trinomials that can be factored and cancelled out.

Week 16 in Precalc 11 – Reference Angles and CAST Rule

This week in Precalc 11, I learned about reference angles and the CAST Rule.

I learned that for reference angles, they are angles that start from the x-axis and are less than 90 degrees, here is an example…

The angle of the rotation is 170 degrees, so I take 180 minus 170 which will leave me with a reference angle of 10 degrees.

Here is another example, the angle of rotation is 200 degrees so I take 200 and minus 180 which will leave me with a reference angle of 20 degrees.

In Quadrant one, the reference angle is the rotation angle, in Quadrant 2, it is 180 degrees minus the rotation angle, in quadrant 3, it is rotation angle minus 180 degrees, In Quadrant 4, the reference angle is the rotation angle minus 270 degrees. Now that we know how to calculate reference angle, we can now move onto the CAST rule.

When we are using triganomic functions in each quadrant, it is the easiest for us to have a positive numerator and denominator, so we need a certain trig function to help us achieve that. Here is where the CAST rule helps us. In Quadrant 4, we have C. This means that in order to have a positive answer, we must use Cos in this quadrant. In Quadrant 1, we have A, meaning that all trig functions in this quadrant will result in a positive answer. In Quadrant 2, we have S, meaning that we need to use Sin to in this quadrant to get a positive answer. And last, in Quadrant 4, we have T which requires us to use tan in order to get a positive answer.

Here is an example of finding an angle in quadrant 4. As we can see, using a trig function other than cos in this quadrant will result in a negative answer.

Week 14 in Precalc 11 – Multiplying Fractions with Variables

This week in precalc 11, I learned how to multiply fractions with variables. Let’s get started

So there are 4 easy steps, first factor if possible, multiply the numerator, multiply the denominator, then simplify across.

Here is an example.

First we should see if there is anything that is factorable and it looks like there isn’t anything. Then we just multiply across (Just Do It).

Let’s look at our next example. We can see that 2x+4 can be factored so let’s do that. After we factor out the 2, we see that we end up with 2(x+2) and we also see that there is the same expression across the equation, so we can just cancel those out.

Since we have 2 of the same expressions across from each other and is not tied to any addition or subtraction, we can just cancel them out. So we end up with this.

Week 13 in Precalc 11 – Fractions with Variables

This week in Precalc 11, I learned more about fractions with variables.

We learned that we can have unfactored expressions on the numerator and denominator.

Here is an example.

As we can see, both top and bottom can be factored so let’s do so.

From here since we have 2 of the same factors, we can just cancel them out since it is equivalent to 1 over 1. Now there is an important rule. We can only do this because the factors have a multiplication side between them. If it had a + or -, we CANNOT cancel them out. At this point we can figure out the restrictions, we need to know that the denominator cannot equal 0 so our restrictions would be +1 and -3

So here is what we end up with and this is our final answer as it cannot be factored out anymore.

Back to what to what the rule was. If I had a fraction like this, I CANNOT cancel any factors. This would be the final answer.

We also learned about adding fractions with different variable denominators.

We first make sure we can’t factor. Once we made sure, we can just multiply each side by the other denominator. The left side would be multiplied by 3y and the right side would be multiplied by 5x. We will end up with this. We can easily tell that our x restrictions would be cannot equal 0 and that our y restriction would be cannot equal 0 as well.

 

Week 12 in Precalc 11 – Parabolic Inequalities

This week in Precalc 11, I learned about parabolic inequalities. It’s very similar to my last post about linear inequalities but this time, instead of being a linear line, it will be a parabola and we will be using vertex form.

Let’s get started.

In my last post I explained the different meanings of each symbol and those will apply here as well. I will quickly explain that the area shaded is where the solution of y can be. So if y is bigger than an equation, then the shaded area will be above the line. Another way to figure out if the area you shaded is correct is by taking a coordinate and placing it into the equation, if the equation is true then the area that you shaded is correct.

So first let’s start with an equation

y>(x-2)^2+2

When we put this into desmos we get this…

We can see that our line is dotted since y does not include the coordinates of the parabola. We can also see that the shaded area matches up with our equation since the value of y is bigger than the coordinates of the parabola. We can double check by taking a coordinate from inside the shaded area and putting it into the equation. Let’s use (2,6)

6>(2-2)^2+2. We end up with 6>2 which is true so that means our shaded area is correct.

The same rules will apply to the less than (<) equations as seen below. We can double check by using the coordinates (0,0) since it is in our shaded area.

0<(0-2)^2+2. We end up with 0<6 which is true so our shaded area is correct.

The same rules apply for this but the line is solid since our symbol is including the coordinates of the parabola.

Week 11 in Precalc 11 – Linear Inequalities

This week in Precalc 11, I learned about linear inequalities with the formula y=mx+b

We know that “m” represents the slope of the line and that “b” represents the y-intercept.

So for the symbols of equalities we need to know that (>) is the symbol for greather than, (<) is the symbol for less than, (≥) is the symbol for greater or equal to, (≤) is the symbol for less than or equal to. Let’s take a look at some equations using the formula y=mx+b while using the symbols.

We’re going to use the same base equation so that we can analyze the differences easily

Let’s start with our base line which is y=2x+3

Slope of 2 and y-intercept of 3

As we can see the shaded area is our solutions for our y-intercept.The solutions(shaded area) is under the line because the solutions are less than the line (2x+3). We can see that the line is dotted since the solutions of y do not include the points on the line.

But when we change the symbol to less than or equal to, the line becomes solid because y can equal the coordinates of the line.

The same concept applies to the opposite symbols. We can see that for y is bigger, the area that is shaded is above the line. And again, the line is dotted as y cannot equal the coordinates of the line.

The same thing applies for this as well. The line becomes solid as the solutions can equal the coordinates of the line.

 

Week 10 in Precalc 11 – Analyzing the Quadratic Function

This week, in precalc 11, I learned how to read and analyze quadratic functions. The standard form that we will be discussing is…

a$latex(x-p)^2$ +q)

With this standard form, we will be going over each part of the formula and discussing how it will effect our parabola.

Let’s start with “a”. This value will determine if our parabola will open up or down. “a” will also effect the stretch of our parabola. The higher the value, the more the parbola will stretch.

Let’s see what happens when the value of “a” is set to a positive number. As we can see when our “a” is set to a positive number, our parabola will open upwards.

 

Now let’s see what happens when the value of “a” is set to a negative number. As we can see when our “a” is set to a positive number, our parabola will open downwards.

Next, let’s cover “p”. So “p” will effect the vertex’s x position, but here’s the tricky part. It will move it in the opposite direction. So if we have -2 as our “p” value, the vertex will be +2 on the x axis. Let’s try it out. We can see that our “p” is -2 but out vertex’s x value is +2. This applies for postive numbers as well

Let’s take a look what happens if we have +2 in our “p”. We can see that our vertex shifted to -2.

“q” is also very similar to “p” except it moves it’s vertical positioning instead of horizontally.

For “q” the value is the real value that it will move up or down, for example +2 will move it 2 up on the y axis and -2 will move it down 2 on the y axis. As we can see when we put the “q” value as +2 it really moves it up by 2.

And when we set “q” as -2, it moves it down by 2.

Week 8 in Precalc 11 – Quadratic Formula

In week 8 of Precalc 11, I learned how to use the quadratic formula for equations that are hard to factor.

Let’s get started with the formula. The formula is below.

So let’s start with an equation that isnt easy to be solved at the top of your head…

We then think of these equations we look like at it like ax^2 + bx + c = 0 and take ‘a’, ‘b’, and ‘c’ numbers.

So we pull each of the numbers and we should get…

A=1        B=6          C=3

We then plot these numbers into our quadratic formula

In the photo above I went through each step basically simplifying and solving the little equations within the equation. We then have 2 answers since we have either a + or -.

 

Week 7 in Precalc 11 – Solving Perfect Square Trinomials

This week in Precalculus 11, I learned how to factor Trinomials with perfect squares.

Let’s get started. So first, what is a trinomial with perfect squares?

It is when the first and third numbers are perfect squares (Sometimes the second number may be a perfect square.)

So when we look at this equation, we see that a is a perfect square and 36 is as well. When we are solving these equations, the formula is square root of ax^2 multiplied by the square root of c, then multiplied by 2. This formula ONLY works when you have a perfect square for ax^2 and c.

Here is an example we can try to solve for bx. When we are solving these equations, the formula is square root of ax^2 multiplied by the square root of c, then multiplied by 2.

So we square root both sides, multiply those together then multiply by 2 as seen below.

After we solve bx, we then can factor it.

x^2 + 14x + 49

 

From my previous blog post we know that Product of c, Sum of b. Just by looking, I know that 7×7 is 49 and 7+7 is 14 which matches up with our b value.

So when factored correctly, it should look like this.

(x+7)(x+7)

and we can verify it as well.

Week 6 in Precalc 11 – [Solving Quadratic Equations Using Factoring]

This week in Precalculus 11, I learned how to factor quadratic equations and the differences between quadratic and linear equations.

So first of all, for linear equations, x will never have a power. Quadratic equations will always have a power.

For example, $latex x^2+5x+6=0 is a quadratic equation and 2x+6=0 is a linear equation.

Let’s get into factoring the quadratic equations.

Let’s start with a simple equation that has already been factored. In the example below, we want the left side of the = to be the same as the right side by finding the value of x. With some simple calculations, we know that if x is the conjugate of one of the number in one of the brackets, the bracket will equal 0, and the brackets are multiplied by each other so if one bracket is equal to 0 we know that the left side of the equals sign would be 0.

In the example below, we have one more step before solving which is to factor out the equation. As seen in one of my previous, we want the product of c, sum of b so we take the factors of 15 which are 1,5 and 3,5, our b value is 8 so 1 and 5 won’t work since it would equal 6 but 3,5 will work since it adds up to our B value of 8. We should also know that the variables will be in the beginning of our brackets and our original equation is x^2 and we know that x multiplied by x equals latex x^2. So the values in our brackets will be (x+5)(x+3). We can verify this by foiling the equation and seeing if it matches up with out original equation as seen in the image which it does. We then repeat the steps in the previous example.