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.