Every parabola has the base parent function of y = x² , but when a parabola changes shape and location, here’s what’s happening to that base function:

The function y = x² + 3 will change our parabola by moving the vertex 3 units upwards.

The function y = x² – 4 will change our parabola by moving the vertex 4 units downwards.

The function y = (x – 1)2 will change our parabola by moving it 1 units to the right.

The function y = (x + 9)2 will change our parabola by moving it 9 units to the left.

The function y = -x² will flip (reflect) our parabola upside down

The function y = 4x² will make the stretch of the parabola much thinner

The function y = 1/4x² will make the stretch of the parabola much wider

When each part takes effect, they can all wombo-combo together to give us something a lot more complicated, but still pretty straight forward. Once you understand what each part does individually, the whole is really easy to understand and plot out.

The product-sum technique, also known as the AC method or the decomposition method, is a strategy used to factor quadratic trinomials. It involves finding two numbers that multiply to he product of the leading coefficient and the constant term and add up to the coefficient of the linear term.

Here’s how it works:

1. Identify the Coefficients:
   Identify the values of a, b, and c
2. Find the Product and Sum: Calculate the product of a and $c,$and then find two numbers that multiply to ac and add up to b
3. Rewrite the Middle Term: Rewrite the middle term of the quadratic trinomial using the two numbers found in the previous step. This effectively splits the middle term into two terms.
4. Factor by Grouping: Group the first two terms together and the last two terms together. Factor out the greatest common factor from each group.
5. Factor Further: If possible, factor the expression obtained in the previous step.
6. Check: Verify that your factored expression matches the original quadratic trinomial.

This method is particularly useful when the quadratic trinomial cannot be factored easily by other techniques like trial and error or factoring by grouping alone. It’s a pretty systematic approach that can help efficiently factor most easy quadratic expressions.

Factoring trinomials is pretty easy, and there are a few different approaches possible. The main two today are simple factoring and difference of squares. Always make sure to check if it’ll Factor 1, 2, 3.

1. Scan through all terms in the trinomial to identify if there’s a common factor.
   If one exists, factor it out.
2. Identify the Type of Trinomial:
   Determine whether the trinomial fits the criteria of a quadratic trinomial, a perfect square trinomial, or another type.
3. Quadratic Trinomials (ax² + bx + c):
   For quadratic trinomials, attempt factoring using either the AC method or grouping:
   Find two numbers whose product equals the product of the leading coefficient (a) and the constant term (c), and whose sum equals the middle coefficient (b).
4. Express the middle term using these two numbers.
5. Difference of Squares:
   For expressions in the form of a difference of squares (a² – b²), factor them into (a + b)(a – b).

Overall, there are quite a few different options available for us to use, and as long as you get to the right answer, it doesn’t make to big of a difference.

1. One of the most memorable things I learned this year from precalc11 was the importance of going to class. I attended the course every day, except for a four day period at the beginning while I was on vacation. Those four missed days really put the emphasis on showing up, in part because the course moves pretty quickly, and so even missing one or two lessons can keep you behind.
2. The previous one ties into this one as well, as I found that equal in importance to showing up is asking questions. When unsure, ASK QUESTIONS. Asking questions in class is important because it helps clarify doubts, deepens understanding, makes sure you’re actively learning, and can bring up class discussions that benefit everyone. And also, unless the question is something the teacher has already explained, it really can’t hurt to ask a question.
3. Third is the quadratic formula, I mean it’s just so versatile and useful. The quadratic formula is useful because it gives a straightforward way to solve any quadratic equation, regardless of whether it can be factored easily. It works universally, handling cases where other methods like factoring or completing the square may be annoyingly complicated or I guess even impossible. This method is way more comprehensive, and also simplifies the entirety of solving quadratics. Overall, the quadratic formula is a powerful tool in algebra, making it really important for us to learn, as I’m sure we’ll hear all about again next year in math.
4. A personal favourite of mine this year was algebraic fractional expressions and equations. Something about all the numbers and variables just made sense in my head. I found that for the equations, cross multiplication was really helpful. I also found the idea of finding a common denominator really straightforward, and it clicked in my head pretty early one so I didn’t have to much trouble with this unit which was really nice.
5. Finally, one that I can’t say was a personal favourite, but was still INCREDIBLY important was trig. I couldn’t leave trig out in part because it’s just so important. Trigonometry introduces you to trigonometric functions, which are critical for understanding more complex mathematical concepts and their applications. Trig will come back to haunt you no matter what if you decide to move forward in math. Trig will be everywhere and so understanding it is entirely crucial to your success.

How I Use the Quadratic Formula: A Simple Guide

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form ax² + bx + c = 0  While there are other methods to solve these equations, such as factoring and completing the square, the quadratic formula works for any quadratic equation, and if you’re not sure, it can’t hurt to throw it into the formula. Here’s how I use the quadratic formula step-by-step.

Step 1: Identify the Coefficients

The first thing I do is identify the coefficients a, b, and c in the quadratic equation ax² + bx + c = 0. For example, in the equation 2x² + 3x – 2 = 0, the coefficients are: a = 2,  b = 3,  c =−2

Step 2: Write Down the Quadratic Formula

Write it all out with the values for a, b, and c not yet all subbed in.

Step 3: Calculate the Discriminant

The discriminant is the part of the formula under the square root: (b² – 4ac). It tells me about the nature of the roots. I calculate it first by plugging in the values. The discriminant is interesting in that it actually tells us how many solutions there will be. If it’s negative, there will be no answers, if it’s positive there will be two, and if it equals zero, than there will only be one solution.

Step 4: Plug the Values into the Formula

Next, I plug the rest of the values of b, a, and the discriminant into that quadratic formula.​​

Step 5: Simplify to Find the Roots

I then just simplify the expression to find the roots. This will give me up to two answers.

Overall the steps are pretty simple, and when followed, you shouldn’t be having any issues with quadratics. I kind of like to stick to this method, even when I’m the slightest bit unsure about my work with another method.

Multiplying Radicals:

Multiplying radicals in math can seem tricky at first, but once you understand the process, it becomes much easier. I find it easiest to think of it like typical multiplication with an extra set of numbers.

Step 1: Multiply the Coefficients

The first thing I do is multiply the coefficients (the numbers in front of the radicals). For example, if I have 3√2 and 4√5, I start by multiplying 3 and 4, which gives me 12.

Step 2: Multiply the Radicands

Next, I multiply the radicands (the numbers under the radical sign). Using the same example, I multiply 2 and 5, which gives me 10. So, 3√2 × 4√5 becomes 12√10.

Step 3: Simplify

After multiplying, I simplify the result if possible. In the example above, 12√10 is already in its simplest form. However, if the radicand can be simplified, I do so. For instance, if I end up with √18, I simplify it to 3√2 (since √18 = √(9×2) = 3√2). Both ways are equivalent, but it’s good to simplify as far as logically possible.

Example:

1. Example 1: Simple Multiplication2​√3 x 4√5

   First, I multiply the coefficients:

   2×4=8
   Then, I multiply the radicands:

   √3 x √5 = √15
   ​Which means,

   2√3 x 4√5​ = 8√15

When it comes to trig ratios, in previous math courses we’ve used “SOH CAH TOA” to understand how to orient the different values within the ratios.

meaning:

SOH – Sin = opposite over hypotenuse

CAH – Cos = adjacent over hypotenuse

TOA – Tan = opposite over adjacent

However, we’ve now moved on to using terms that work much more smoothly with graphing. Now, instead of “SOH CAH TOA”, we now have:

Sin = y/r

Cos = x/r

Tan = y/x

In the short term, memorizing a new set of ratios might feel a bit rough, especially since we already have “SOH CAH TOA” memorized, but swapping to these new ratios will be crucial for our learning. It’s also important to note that these “new” ratios are actually exactly the same as the ones we’ve been using before. In this setup…

x = Adjacent

y = Opposite

r = Hypotenuse

The best way to understand this is just by viewing every one of our triangles like they’re on a graph (because they are). While at first, this concept might take a bit of time getting used to, once you’ve got it down, everything will go smoother.

When you have to solve a math word problem, the most important thing you can do is read the problem carefully. This first step sets you up for everything else. This is why it’s so important and some steps to follow when reading one…

Because word problems tell a story, you need to understand them to know what the problem is about. Knowing the story helps you see the problem more clearly and figure out what math you actually need to use, which is one of the most important steps.

Because word problems have specific details and values, it’s really important to keep track of all the details. Missing these can mess up your solution. It can help to write some notes on your first read.

On the first read it’s also really just important to get the Main Idea. It’s best to read the problem once to understand the general story and what’s being asked.

On the second read, that’s when we can start to focus on details. You can highlight or underline key information like numbers and single out what you’re actually looking for in the question.

The third read is where I finally start to write down the equation, and sometimes I read the question more times depending on how well I understand it.

Reading the problem several times helps you catch all the details and understand better.