This past week in Pre Calculus 11 we continued Chapter 6: Trigonometry. Something I learned was special triangles. There are two special triangles, one that is an equilateral triangle that was split down the middle and the other is a rectangle that was split diagonally in half. Below is both the triangles with their dimensions:

Questions would potentially ask sin315 and in the quadrants listed above you find 315 in one of the quadrants which happens to be in quadrant 4 which is in the rectangle special triangle. Sine is opposite/hypotenuse so from angle 45 degrees you find the opposite which is 1 and the hypotenuse which is root 2, at this point the answer is .

One other part you have to consider is it negative or positive:

315 degrees in highlighted and it is found in quadrant 4, since cosine is the only one that is positive in this quadrant then the  becomes negative –> .

This past week in Pre Calculus 11 we finished Chapter 7: Rational Expressions and Equations. Something I learned this week was how and when to cross multiply. The rules when cross multiplying are it has to be two fractions with an equal sign between them. Here is an example:

 

Multiply 3(q+4) and 5(q-2)

= 3q+12=5q-10

Isolate q: 11=q

 

An example of what an equation that would not be able to cross multiply is because it has 3 fractions and not 2.

This past week in Pre Calculus 11 we started unit 7: Rational Expressions and Equations. Something I learned this week was how to simplify an expression. If the equation was both (x+4) on the top and bottom cancel out leaving . A non-permissible value means if that number was substituted in for a variable it would make the denominator a 0 which is not what we want so it can not be these values. Looking at the original equation (x+4) and (x+6) are on the bottom so their non-permissible value would be -4 and -6.

This past week in Pre Calculus 11 we finished Chapter 8: Absolute Value and Reciprocal Functions. I learned how to do piecewise notation. Piecewise notation is used to show the domain that each part of the function is in.

Here is an example of an absolute value parabola shown in piecewise notation:

Here is an example of an linear absolute value shown in piecewise notation:

This past week in Pre Calculus 11 we finished unit 5: Graphing Inequalities and Systems of Equations and started unit 8: Absolute Value and Reciprocal Functions. Something I learned this week was piecewise notation and what it means. There is 3 sections to a parabola and it is divided whenever the line touches the x-axis. To find piecewise notation for a parabola we write out the original equation without the absolute value for whichever section(s) are above the x-axis and write the opposite for the sections below the x-axis. Here is an example of a parabola writing in piecewise notation (writing inside yellow box):

Piecewise notation essentially shows the different sections of the line on the graph and each sections domain. Here is an example of instead of a parabola it is a linear equation:

This past week in Pre Calculus 11 we worked on unit 5: Graphing Inequalities and Systems of Equations. This week I learned how to graph inequalities and find the area that the possible answers could be that would make the equation true.

Any point in the shaded part would make this a true statement.

*Key points to remember:

When the sign is greater than or less than the line on the graph is dashed and if the sign is greater than or equal to or less than or equal to the line on the graph is solid.

Any point on the graph could be chosen for step 3 as the test point as long as it is not on the parabola.

This past week we started unit 5: Graphing Inequalities and Systems of Equations. Something I learned this week is how to graph inequalities. Here is an example of graphing an inequality on a graph:

1. The parabola is facing up because 4x^2 is positive
2. The equation is > 0 so if its larger it is talking about the part of the parabola that is above the x-axis (shaded in dark on drawing)

This past week in Pre Calculus 11 I learned that when the a value is anything other than x^2 I take the value that it would have been in x^2 and multiply it by the a value I am trying to find. An example would be if a=2 then the first x value would be 1 so 1x3x2=6 so it is going up 6 instead of going up by 3. Shown on a chart for visualization it looks like this:

Here is the points on a graph:

This week in Pre Calculus I learned how to find the vertex of a quadratic formula if the equation is in standard form. Standard form looks like:

The x and y variables can represent any point besides the vertex on a graph. The p represents the x value of the vertex and the q represents the y value of the vertex. If the equation was:

The -7 is the x value in the vertex and the -1 is the y value in the vertex, the vertex of this equation would be (7,-1). There could be some confusion that the vertex should be (-7,-1) but this is not the case. The sign in ONLY the x value gets flipped essentially making it a positive in this case. This is what the parabola would look like on a graph: