This week we mostly did review because we have a midterm coming up. While I was looking through the practice test I noticed a question from one of our earlier units and I didn’t get it right first try. Then I realized that I had to use exponent laws:

At the beginning of this week in pre calc 11 we learned about Vertex/Standard form. This is personally my favourite form because it provides more information about us graphing the quadratic function than the quadratic or factored forms give us.

Vertex/Standard form looks like this:

$y=a(x-p)^2+q$

This form provides you with the following information about a parabola:

a:

1- tells us whether a parabola is opening up or down. If it’s a positive # than it’s opening up, and if it’s a negative # than it’s opening down or being “reflected”.

2- determines the size of our parabola, whether it is stretched or compressed. If a is greater than one then the parabola is stretched, if a is less than one then the parabola is compressed.

p:

1- tells us the horizontal translation of a parabola (whether the vertex is moving left or right).

2 – tells us what our line of symmetry is.

q:

1- tells us the vertical translation of a parabola (whether the vertex is moving up or down).

p & q:

1- the p and q are very important becuase they let us know our vertex *the vertex is the most important part of any parabola. The p tells us our x co-ordinate and the q tells us our y co-ordinate for when we are graphing a quadratic function.

This week in pre calc 11 we started chapter 4, and the first thing we learned was properties of quadratic functions. Now it’s important that before we start listing off the properties of each of these quadratic functions, that we know 100% that this is a quadratic equation. How can we tell if a function is linear, quadratic, or neither. In this blogpost I’m going to show you how you can tell the difference between a quadratic function and a linear function by looking at the table of values for any type of function.

Linear – look at the first differences, if they are equal to each other than it is a linear function.

Quadratic – Look at the second differences, if they are equal to each other than it is a quadratic function.

Neither – Than it just simply doesn’t follow any of the rules from above and is not a linear or quadratic function.

Ok so lets begin, suppose we have this table given to us:

The first step is to find the first difference, how you do this is to find how much space is between each of the #s.

As you can see, the first differences are equal to each other therefor this is a linear function.

Let’s look at another example:

Follow the same steps as you did for the function above:

As you can see, the first differences aren’t equal to each other so now we have to find the second differences:

The second differences are equal! Therefor this is a quadratic function.

Lets look at one last example:

First – find the first differences:

The first differences aren’t equal so we move on to step 2 – finding the second differences:

For this example both the first and second differences are not equal therefor this is neither a quadratic or linear function.

This week in pre – calc 11 we learned chapter 2.6 – solving equations algebraically. One thing that stood out to me was learning how to rationalize the denominator to solve an equation that looks like this one:

The first step is to multiply the $\sqrt{3}$ “rationalize” it. The $\sqrt{3}$ multiplies with everything and cancels out the $\sqrt{3}$ on the bottom of the equation.

The next step would be to further simplify but as we can see we no longer can simply this equation further, so we leave it like this

Here is an example of an equation which also requires to be rationialized, take a couple seconds and try to solve this using what you know about rationalization:

Below is the answer and me solving this equation using rationalization:

This week in Pre calc 11 we learned how to factor polynomials. One of the ways we can factor a polynomial is through using substitution.

Lets suppose you have something like –

Step 1: for substitution you can just substitute a variable of your choice to equal the terms in the brackets so that it can become factorable –

Step 2: once it becomes factorable you can start factoring, in this case I’m going to be using the box method –

Step 3: when we substituted we just made this polynomial factorable so that we can factor it and make it easier but it doesn’t mean that it’s the final answer. Now that we have the two multiples we can add them to the original (x-6) that we substituted a for –

This week in Precalc 11 we learned how to factor an “ugly trinomial” which is a trinomial that has a coefficient in front of the $x^2$. This is when we can’t use our factoring 1 – one thing in common, 2 – 2 terms (difference of squares), 3 – 3 terms pattern (product & sum). Factoring a trinomial is actually kind of easy once you have figured out the way you like to solve them. There’s two ways, the first way is just guessing and plugging in numbers which can be difficult because you are guessing until you find the right number and that can take forever. The second way is to use a visual piece that’s called the box method. I personally like it better this way because it’s more straight forward and efficient. So today I will be showing how to factor an ugly trinomial using the box method correctly.

This is the ugly trinomial –

Step 1 create a square and plug in the first and last terms in the trinomial *the first term goes in the top left hand corner and the last term goes in the bottom right corner –

Step 2 fill in the empty spots in the box. Do this by multiplying the first and last terms together. Then you find the multiples of the number until you have hit the two that add or subtract to the term in the middle which in this case is our -5mn. Once you have found your two multiples then you plug in those numbers into the box *it doesn’t matter which spot you place it in so long as it’s an empty one *if the leading coefficient sign has more positives than the negatives than the bigger number in the multiples is going to be positive –

Step 3 find what’s common and there you go. You find whats common across and then up and down *whatever sign the term is on the outside is going to determine the sign of the term that’s common –

To check if you did it right just use foil and see if you get the trinomial you had to start with –

Here’s another example –

A question that I came across this week that stumped me was:

Now for this question I distributed and used foil properly and when I was solving it, it was too difficult because this is what I was trying to break down all at once. Which…led to me making silly mistakes. A tip that I received from my teacher was to do them seperately and then combine them at the end once they are easier to manage.

So I drew a line in the middle so that I know not to combine them and redid the question, which then gave me the correct answer. Now you might be wondering what I’m talking about because even though I did them seperately, once I combined them together the question looked just as long as it was before. Well the size of it didn’t change but what did change is that we now have like terms that we can combine which takes only a few seconds as oppose to a few minutes trying to find where you made a mistake if you did it the non organized way.

This week in class we started unit 2 called Radical Operations and Equations , and the first thing we discussed was how to simplify radicals. We went from an entire radical to a mixed which we already knew how to do from the previous unit so I felt pretty good about it, and then we added a variable with an exponent to the radical like $\sqrt{24x^3}$  When I took a look at this I understood how to turn it into a mixed radical except I had no idea what to do with the variable with the exponent. So here’s the way that I deal with that variable and some tips that will make it easier for you to remember how to deal with that variable when it pops up in more complicated questions.

1st – Find the prime factorization, when you have a variable with an exponent you also find its prime factorization. * Realize that $\sqrt{24x^3}$ is a squared root therefor you are looking for pairs when finding the prime factorization. What I do right away is highlight my pairs right away so that way I won’t miss them.

2nd – once the prime factorization is found you will take the pairs and square root them and put them on the left side of the radical symbol taking it “out” or “removing it from the radicand area and then taking whats not highlighted multiplying them together and putting it on the inside of the radical symbol making it the radicand.

Station 1:

Station 2:

Station 3:

Station 4:

Station 5:

Station 6:

A big “ahhh” moment for me this week was when I had come across this question:

What came to mind first was to put it in a radical and then solve from there, but this is what happened.