Blog post week 14

This week I learned how to multiply and divide rational expressions by factoring to get the restrictions for only the denominator.

My example is how to divide rational expressions. The first step is to factor, the second step is to state the restrictions, the third step is to flip over the fraction (reciprocal) and multiply from across to bottom. Then scratch out what is common and whatever is left that is your answer. I find this really easy because I love working with fractions and it is not that complicated. I find multiplication easier than dividing because then you can factor the expressions and then state your restrictions and the last step to multiply and scratch out the common expressions. There is no reciprocal required. And in my example there is a difference of squares so the factored form will be conjugates which is really easy. With this unit factoring is the key component because it tells the most important part the restrictions for x or any variable.

My division example that I used and simplified.

Blog post week 13

This week I learned how to graph and solve reciprocal functions. So to solve the function you must factor it to get the roots, or also known as the x ints. So the graph has 2 solutions. And to know if a reciprocal function has no solutions that means that it doesn’t factor and has extraneous roots. And if you don’t have an equation but a graph instead you would know that it doesn’t have any solutions if the graph doesn’t cross the x ints. My example I solved had 2 solutions because it had 2 roots.

My example that I used and explained


Blog post week 12

This week, I have learned how to graph the absolute value of a linear function. And since you can’t have negatives the graph goes down and then comes up which makes it a shape of a v. My example that I used was y=3x+3. And then I graphed it and labeled the x int and the y int. Also the domain and range. And its a linear function because it does not have an x^2 so it won’t be a parabola. The domain is always going to be that XER and the range can be that y \ge 0. Or any number greater than zero in this case my range is that y is greater or equal to zero. I can’t wait to learn more of these functions and so far graphing them is easy. And its a lot easier to find the x ints when it is in vertex form because then you can easily plot the vertex on the graph and then convert it to general form to get the y int.



Blog Post Week 11

This week I learned how to translate word problems to equations. I found it pretty hard because some of the wording was confusing or I put it in the wrong order. I should do more problems to get the hang of it and to understand it more, and make it easier for me.

Step 1: Underline the key words ex: sum,product,minus, greater or equal to, less then and other important signs.

Step 2: Choose variables ex: x and y or other variables of your choice

Step 3: Then create your equation

After I created the equation I was able to figure out the y int and then graph it.

Here is my example and I underlined the key words that helped me to create the equation.




Blog Post Week 10 – Finding the Discriminant

Today I decided to do my blog post about how to find the discriminant in a quadratic equation. The discriminant is used to find if an equation has 2 solutions or no solutions. An equation can have no solution if the discriminant is negative and if the discriminant is positive it has 2 real solutions. And if the equation is not equal to zero then you have to rearrange it so it will equal to zero. I did my example with that kind of quadratic equation. Then when you rearrange it you substitute the variables with the numbers given and then you calculate your discriminant and see how many solutions. It was the easiest for me to do because it didn’t involve a lot of steps or any factoring.

My example that I chose to complete and show the steps how to do it.


Blog post week 9 Modeling Problems with Quadratic Functions

This week I learned how to model problems using quadratic functions. The first step you do is make an equation in factor form because it will be a lot easier to find the roots. With the factor form formula you are able to find the axis of symmetry which will be you x intercept.

(x-x1)(x-x2) This formula is the factor form formula. So after you find the value of x you just substitute it into your equation. With my example I first had to use the rise/run to find my slope and after I found my slope I used the slope point formula.

y-y1=m(x-x1) This is the slope point formula.

Then you just substitute your values to get your equation. After you use a formula that helps you find the revenue of a certain object that is for sale. After you got that equation substitute your values again and then find the axis of symmetry or rearrange the equation. Then to find the axis of symmetry take your x value and divide it by 2. Then you will get your price and answer. For my example it has 2 answers but I mostly got $5. In this chapter I found the rearranging confusing but after I did more questions I understand how to arrange from different forms to the forms that I need in order to solve the quadratic equation. And if you get stuck just use standard form to get a vertex and then draw it on your parabola because you will find the axis of symmetry a lot faster and your x intercepts. This is what I learned in Math 11 pre calculus in this week.


this week. My example is number 6 in the workbook.


Finding parabola characteristics week 8 Block A

Last week I learned how to find a vertex, the x intercepts, y intercetps, the domain and range, the axis of symmetry, minimum or the maximum and if the parabola opens up or down. By using the formula y=a(x-p)2         

With this formula I am able to find the vertex with the p+q values. The “a” tells me if the parabola compresses or stretches and the (x-p)2 tells me the horizontal translation. The formula at the top is in standard form which is the easiest to use. 

And with the standard form formula you can also use algebra to solve a missing variable so that your equation will be inm that form. In the example below I will show you how to find all of the characteristics of the following equations.

Here is my example and the equation is positive so my parabola will be opening up and if my equation was negative then it will be opening down. The a also tells me if the parabola opens up or down so the formula is easy to understand and use to find all of the characteristics. I also used desmos to see the y int which was 13. Desmos has been very helpful to me because then I can see how the parabola looks like and know the x and y intercepts. The standard form also tells me all of the information and their is enough to graph a lot of points. Remember when you are graphing a parabola make sure to have a lot of points because its a curve and not a straight line and if you are stuck make a table of values and it will make it easier to graph. That is what I learned in class last week and can’t wait to learn more about parabolas.


Solving Quadratics by Factoring week 6

This week I learned how to solve quadratic equations using the quadratic formula. The first step will have to be make sure your equation is equal to zero. In the example I have provided it is also equal to zero so all I need to do is substitute the values. Make sure to simplify the equation to get an answer. Some quadratic equations can have 2 solutions. In this case there are 2 solutions and some can have one solution. That is what I learned in math last week.

Multiplying and dividing radicals

This week I learned how to multiply and divide radicals using simple steps. When you divide radicals if the denominator has a square root then you have to rationalize the denominator which means that I multiply the bottom with the top and then I use the distributive property or foil to get like terms and then I collect them and get my answer. For some other dividing problems you have to use a conjugate which means that I multiply the radical at the bottom by the same one except I need to have different signs, if the expression is positive then you have to multiply it by the same numbers but instead of the plus it has to be a minus. Because then it will create one zero pair and you get the expression that you need to have at the bottom. After you get that expression you just multiply it by the top and then you get your denominator that has no radical and its simplified. In my example I used the conjugate one so their is a visual to show how you solve those expressions. And if the bottom square root can be simplified to a simpler radical then you should change it so then you can work with easy radicals. I worked smarter than harder and that ugly radical becomes an easy one.