1. Factoring: Knowing how to factor was the most important thing to know in this course. Coming from Pre Calculus 10, I didn’t really remember how to properly factor. I can say now that it is engrained in my brain after using it in almost every unit this semester. Especially as the equations were only going to become increasingly harder, we constantly reviewed and went over factoring.

Example: = (x+4)(x-2)

2. Sine Law: Learning the Sine Law made things much easier for me. The Sine Law is a formula/law that basically allows you to find any angle/side in the triangle as long as you know a matching side+angle, and the opposite side/angle to the one you are trying to find. Making sine law a very useful tool when it comes to trigonometry. Angle A/B/C and side a/b/c are opposite to each other.

3. Completing the square: Due to the course of PreCalc 11 having lots of graphing and quadratics, you are going to need to know how to solve in the semi-frequent situation of when you cannot factor the equation. In this case you complete the square. Completing the square is when you take the middle term in the quadratic (also known as the “b” term), divide it by 2 and then square it. Then you take the original last team (or “c” term) and move it outside of the original quadratic, and solve to find the zeros of the equation.

4.

This week we learnt about the Sine Law. The Sine Law is basically a formula that you can use to find a missing side/angle if you know a matching angle+side and the opposing angle/side to which you are trying to find.

The formula: 

Using the formula itself is pretty simple, you input the known information. Then remove the one of the a/b/c you do not have enough information to use, and solve for what you are looking for.

1. Since the denominators are all equivalent, the first step is to combine and simplify the numerators by combining all like terms.
2. You can then simplify further by eliminating x-3
3. This will leave you with 1/x+3

Simplify: 1/3a+2a/5a

The first step to adding these together is to find the common denominator of 3 and 5. This will be 15. You will have to multiply 1 and 3 by 5 and 2a and 5a by 3. This will result in 5/15a + 6a+15a=11/15a

Solving equations algebraically

9= -1(-2x+6)

9=2x-6

15=2x

x=7.5

Simplify 5/a+3/a

=8/a

1. Find common denominator (multiply denominator to do so)
2. add or subtract across like normal.

Simplify 4/a-2/3

12-2a/3a

An asymptote is a line that identifies the boundaries of a curve. As of pre caluclus 11, the y asymptote (horizontal)=0. The x asymptote (vertical) is a little bit more complicated. Looking at the y axis, identify where the point of line is at -1 and 1. In between these points is where the x asymptote will be.

Today we reviewed some key points on absolute values:

-The values are ALWAYS positive.

-When graphing, the linear or quadratic equation will never enter the negative area (y.) As shown below, the line reaches 0 on the y axis and reflects up.

This week, we learned how to read a linear equation.

y=2x-1

In this equation, the form is y=mx+b. The b in this situation is -1. This represents the y intercept. The mx in this situation is 2x. This represents the slope. The slope is 2/1 (2 right, 1 up.)

This week we worked on converting from standard for to vertex form by completing the square.

Ex: y=x²+4x-5

y= (x²+4x+4)-5-4

y=(x+2)²-9