This week in Pre Calc 11 we learned how to Solve Absolute Value Equations both Graphically and Algebraically. In order to solve an abosulte value equation graphically, we must first seperate the equation into two parts. In the linear equation example: the seperated equations would be: and Next, we would graph both equations in order to visually determine how many possible solutions the absolute value equation might have and the vaules of the possible solutions. Linear absolute value equations can have 0, 1 or 2 solutions. We must also remember that since the equation has abosulte value bars around it, the equation can not be graphed in the negative zone of the graph and will instead reflect back up into the positive zone, making a V shape.

Once we have graphed both equations, we are now able to see that the equations intercept at and This also shows us that this absolute value equation has two solutions.

It is also possible to have a quadratic abosulte value equation, which follows similiar steps to that of a linear abosulte value equation. In the example: we must seperate the equations to be able to graph them. The sperated equations are: and Next we graph both equations to determine their points of intersection. Quatratic absolute value equations can have anywhere from 0 to 4 solutions.

Once we have graphed both equations, we are now able to see that the equations intercept at and This also shows us that this absoulte vlue equation has three solutions.

This week we also learned how to solve absolute value equations algebraically.  In the example: we must first start by getting rid of the absolute value bars by putting this equation into two different equations using piecewise notation. This means that our two new equations will be and Once we have our new equations we must solve to determine the value for which will tell us how many times the eqautions will intercept. In the first equation:  we must first make the equation equal to zero by moving to the other side which gives us from there you continue to isolate and in the end you will determine that In the second equation:  we must follow similar steps, however we must first distribute the negative into the brakets before we can make the equation equal to zero. Once you isolate for you will determine that This tells us that the equation  will have two solutions. Once you have determined the values for you must verify that they are not extraneous answers by plugging them back into the original equation.

This week in Pre Calc 11 we learned how to graph linear inequalities in two variables. We recalled that a linear inequality is different from a quadratic inequality because linear means that it only has a degree of one. We also learned how to identify the y-intercept and the slope of a linear inequality. For example in the inequality: the y-intercept is the second term which is positive and the slope is the first term which has a rise of and a run of We also learned how to determine whether the boundry line of the linear inequality will be broken or solid. If the linear inequality is greater than, less than or equal to (includes), then the boundry line will be solid, however if it is only greater than or less than (excludes), then it will be broken. For example in the inequality: the inequality sign tells us that the value for must be greater than to make this inequality a true statement, this means that the boundry line for this inequality will be broken.

Once we learned how to determine the y-intercept, the slope and whether the boundry line will be broken or solid, we began to graph. In the first example we used, we know that the slope is the y-intercept is and that the boundry line will be broken. If we were to draw this linear inequality on a graph it would look like this:

Once we have graphed the linear inequality, we must determine the possible solutions that will satisfy the inequality by shading the one side of the graph. To determine which region will satisfy the inequality we choose points (called the test points) on either side of the boundry line and input them into the linear inequality. If the inequality sign is true to the numbers then shade in the region of those points, if not shade in the opposite region of the graph.

I have included an example below that shows the detailed steps I would take to determine which region on a graph to shade to find True possible solutions.

This week in Pre Calc 11, we spent most of the week studying and reviewing for the upcoming midterm. I decided to focus my studying on the first unit we learned this year which was Arithmetic and Geometric Sequences and Series. I found that I needed to review the differences between them and also how to calcualte the value of the last number in an Arithmetic Sequence when the number of terms is given, as I have found these things hard to remember.

While studying, I have recalled that the difference between an Arithmetic and a Geometric Sequence is that an Arithmetic Sequence, increases or decreases by a common difference. Where as a Geometric Sequence increases or decreases by a common ratio.

I also reviewed how to find the value of the last number in an Arithmetic Sequence, when the number of terms is given. For example, in the sequence:  7+11+15+19… we need to determine the value for term 12. We already know that is our first term and that this sequence has a common difference of which means that in order to determine tha value of term 12 we will need to input the information that we already know into the equation = as I have shown in the example below.

After solving the equation, you now know how to calculate any term in the sequence that might be asked. This week I also reviewed the differences of how you can tell if a Geometric Series is diverging or converging. To determine if it is diverging or converging, you look at the value of the common ration of the Geometric Series. If it is diverging, it means that or that If it is converging, it means that or that If the Infinite Geometric Series is diverging it means that it will have no sum, however if it is converging than you are able to caculate the sum.

Each of the things that I have mentioned have all been things that I focused my review on over the past two weeks. They were all units that I needed to review as I had forgotten the majority of it.

This week in Pre Calc 11 we learned about equivalent forms of Quadratic Equations. Each Quadratic Function can be written in three different ways: Vertex Form, Factored Form and General Form and each form reveals different information that you may need to be able to graph a function.

A quadratic function in Vertex form tells us the scale factor and the coordinates of the vertex It looks like this:

A quadratic function in Factored form tells us the scale factor and the x-intercepts. It looks like this:

A quadratic function in General form tells us the scale factor the y-intercept and it also tells us if the parabola opens up or down. It looks like this:

We also learned how to find the vertex of a Quadratic function when it is in General Form. To do this you must first complete the square to solve for the vertex, because there are two variables in the function it will not be possible to solve completely, however it will give you the coordinates of the vertex in order to later graph it. Ex. In this example you must first get rid of the coefficient by dividing the first two terms by three. Ex. Then you must find the zero pairs by dividing the middle term by two and then squaring it. Ex.  Then you must simplify the first three terms and multiply the coefficient by to remove it from the brackets. Ex.  You are then able to combine like terms. Ex.  Once you have changed the equation into Vertex Form, you will be able to find the vertex needed to graph it. Vertex

If you wanted to find the x-intercepts, you would have to change the function from General Form to Factored Form by expanding the function and then factoring it.