This week in pre-calc, I had some trouble simplifying expressions with three factors. So, instead of explainning how to do it step by step, I will use this picture to show it so it it less confusing. Here is how to solve an expression with three factors.

This week in Pre-Calc, I had some trouble understanding how to solve absolute values in piecewise notation, and how to solve reciprocal questions. So, today I will explain the steps on how to solve these.

To solve for piecewise notation, first, you will put the equation in absoulute value form. Next, you will will find the zero of the equation (isolate for “x” if it is a linear equation, or factor if it is a quadratic equation). After you have done this, you will put the number that you got from solving for “x” on a number line. You will choose two (if there is one “x” value) or three (if there are two “x” values) test points on your number line. Once youo have tested the points (by putting them back into the original equation), and you have determined which points are negative and positive, you can write the equation in piecewise form. To do this, you write “y” and then “{” and multiply in a negative for the negative value, writing whether it is < or > where you “x” value or values are on the number line. And you will leave the positive points as the original equation, and write < or > to represent where on the numberline “x” is positive.

To solve a reciprocal function, first you will put it into a fraction. Next, you will take the original equation (not in a fraction), and find the NVP. To do this, you will find the zero (the equation cannot be equal to zero) of the equation, and do this by isolating “x” (linear) or by factoring (quadratic). Once you have found what “x” cannot equal, this will be your asymptote. After, you go one to the left from your asymptote and up or down (positive, up or negative, down) and go one to the right from the asymptote and one up or down. You will then draw the shapes (two if it is linear, or three, one is a parabola, if it is quadratic).

This week in Pre-Calc, I had some trouble understanding how to put equations into piecewise form, and how to solve absolute value equations. So, today I will explain how you do them.

When you are putting an equation into piecewise form, you first take the equation out of the absolute value signs. Once you have done this, you make the equation equal to zero. After this, you put all the numbers (without variables) on the other side of the equation. Next, you isolate “x” by diving by the number that is next to it (coefficient) on both sides, and then you will have your answer for what “x’ equals. Once you have solved for “x”, you put the point or points on a number line, and choose two (one point) or three (two points) on the number line. Based on which point is negative, you will then write the formula in piecewise form. To do this, you will write the original formula as it is, and then write “x” is < or > depending on where your point is positive and negative for all your points (one or two). Next, you will write out the original equation again, this time for the negative point, and since it is a negative, you will have to multiply in a negitive sign in the front of the original equaiton to make it positive. Once you have writtent that, you will then write “x” is < or > the point or points that are negative on your number line.

To solve an equation, you seperate it into two different equations. One will be the original equation (without the absolute value symbols) and the other will be the equation with the negative symbole multiplying in (to make the negative part of the number line positive). To solve your first side (the original one), you will move any numbers without variables to the other side. Once you have done this, you divide both sides by the number in front of “x” (coefficient), and this will give you your answer. For the second side (the original one with the negative multiplied in), you will start by multiplying the negative symbole in. Once you have done this, you move all the numbers without variables to the other side. Next, you will divide both sides by the number in front of “x” (coefficient), and this will give you the other answer.

This week in Pre-Calc, I didn’t understand how to solve quadratic systems of equations algebraically. As I was going through the notes trying to understand how to do it, I practiced a question, following all the directions, and got the wrong answer. I tried every possible correction to try and get it right, but I just couldln’t seem to do it. I re-read over the question and through my steps, checking the steps in the notes to make sure that the order was right, only to realise, that I forgot to carry down a negative symbole on one of the numbers in the equation. I would like to take a moment to remind myself how important it is to revise your work as you are solving an equation, to be sure that you did not make any mistakes that could be fixed, like this one, and to pay attention to the little details and to what you are doing in an equation. Today, I will be explainning how to solve a linear equation algebraically.

First, you put the first equation into the second equation (where the “y” is in the second equation, since the first equation is equal to “y”). After you have done this, you simplify the equation, first by foiling the equation, and then by making it equal to zero. After you have done this, you factor the equation. Once you have the factor of the equation, you are able to put those points into the the first equation where “x” was. After you have put both points into the first equation (seperatly), you now have the second points for both numbers. Your first point will be the first number you got from factoring the equation, and the first nunber you got from putting that number as “x” into the first equation. The same will be for your second point of the equation.

This week in Pre-Calc, I have been trying very hard to understand the conecpts we have been learning. However, there are a few different things that I will explain.

First, when you are solving a linear equation, you want to isolate for “x”, and when it is a quadratic equation you want to isolate for “y”. To solve a linear equation, the first step is to seperate the numbers from the numbers with variables (“x”). After you have put both numbers on the other side of the less than or greater than sign, you find what the answer is to both of the numbers on the other side. Once you have done this, you divide both sides of the equation by the number in front of the variable. Once you have done this, you have the point or points that you will use. After, you put the point or points on a number line, and choose two ( one point) or three (two points) test points to determine which way works for the equation. If you are solving a quadratic equation, you could first find the factors for the equation, and write (x  + 3) (x – 2). Once you have done this, you take the first number from the original equation, and would divide both numbers in the brackets by this number. Once you have done this, you take the answer that you get from dividing and flip the symbols of both numbers and use these as the points for the number line. Once you have put them on a number line, you choose three test points, and you can determine this way which parts of the number line work for the equation.

This week in Pre-Calc, I struggled with understanding how to solve quadratic inequalities, so here is how to do them.

First, you factor the equation. Next, you take the two numbers that will be used to factor and write them in two separate brackets next to “x”. Next, you take the first number from the original equation, and use that as the bottom number of a fraction of each number next to “x” in each bracket. After, you simplify the fractions, and if the number is still in fraction form, then you take the bottom number and put it in front of the “x” in the bracket, because you cannot have it in fraction form. Next you take the numbers next to “x” and give them the opposite sign that they originally had (positive or negative) and then you use those numbers, however if one or both of the numbers was a fraction, then you would put it back into it’s simplified (if possible) fraction form, and still reverse the original sign that it had, as you did with the other number. After, you put both numbers on a number line. You would use a closed circle if it is not = to zero (greater or lesser), or an open circle if it is = to zero (greater or lesser). Once both numbers are placed on the number line, you choose one number in each section (left, middle and right) of the numbers, and test them by plugging them into the equation that you created, and then you are able to determine which sections work with the equation, and therefore determine which numbers are greater, lesser or equal to “x”.

Today I am going to explain how to word problems. This is something that I have had trouble with this week, so I am going to give some tips that I have learned to help.

First, every word problem wants you to create two equations from it’s poblem. These two equations will be what you use to solve the problem. From these equations, you will create a quadratic equation. Then, you will use that equation to solve the problem. Next, if the equation is asking you about people and money (or two different things) you would put people with people and money with money in your equation. Then, like the last one you will turn this equation into a quadratic equation by using FOIL and add all the like terms together to give you a quadratic equation that you will use to solve the question.

This week, I had trouble creating and knowing how to make the “x” and “y” table of values from an equation, and knowing how to graph it.

If you had y = 2x^2 + 8x + 6 you would begin by finding the vertex. You would do that by factoring out 2 from the equation (because that is the value of the first number). So the equation would now become y = 2 (x^2 + 4x + 3) The next step would be to take the 4 divide it by two and then square it. Then, you would take that number (4) and create a zero pair after the 4x in the equation. The equation would now becomre y = 2 (x^2 + 4x + 4 – 4 +3). Then you put brackets around the first three numbers of the equation, y = 2 (x^2 + 4x + 4)(-4 +3). After this, you would go back to the number that you had to divide by two and then square (4), and root it, which gives you two. So the equation would now become, 2 (x + 4) (x + 4 ) (-1), which would become 2 (x + 4)^2 (-1). This would give you your vertex of (-4, -1). Once you have found your vertex, you are able to put it into the middle of the table, and move up by two, and down by two. The top number on the “y” values of your table will be the same on the top and bottom. These will give you your points for your graph. Once you have these you will be able to plot them on the graph, which you can draw based on where you points on (move the graph higher or lower). That is how you find the vertex, points, and graph an equation.

For this chapter in math, I have struggled. However, I learned a trick that can help make long factoring questions faster and more clear.

For these types of questions, you can substitute what’s in the brackets on both sides for a variable. If they are both the same then you can make the the same variable (a) and if they are both different you can make them different variables (a and b). After, you re-write the equation with the variables in place of what was in the brackets. After this, you can factor it. In this equation, the root of 49 is 7, so you write 7 on both sides of the equation, since the variables are squared you put an “a” next to the 7. Next, you root 9, which is three. Since three is negative, one three on a side will be negative, and the other three will be positive in the other brackets. You would do that same as the “a” and put the “b” variable next to both threes. Once youo do this, you re-write the equation replacing what the variables where for what was originally in the brackets. After, you factor what your equation is. You would factor the number that is in front of the brackets to the brackets in front of it. After you so this, you are left with your answer.

This week in math I have had some trouble figuring out some of the different strategies for factoring. I will explain how to factor using the completing the square technique.

For this example, the first thing you do is add a zero pair at the end f the equation but before the equal sign, in this equation that was +16 and – 16. Then, you take take the first three terms as your equation to solve first. To do this, you take the second term, and half it, and then square it, in this equation that was 8. Next you take one of the answer you get from that, half all of the equation (only the first three terms), and write them out next to each other (which can be simplified by writing it once and squaring it), adding on the rest of the equation that we where ignoring before. Next, you want to isolate x. To do this, you first have to take away what is not in the brackets first. In this equation, that is -6 (you would add +6 to both sides of the equation). Next, you have to get rid of the square, and to do that youo would root both numbers of both sides of the equation. When you root the other side of the equation, it becomes +-. Since the square and brackets are now gone, you can isolate the x. In this equation, you moved the four to the other side of the equation by adding -4 on both sides. This, would give you the answer.