Month: May 2018

Pre-Calculus 11 Week 15 – Solving Rational Equations

alexr2015

This week, we solved rational equations

Note that this is different from last week, where we only simplified rational expressions.

Simplifying looks like this:

\frac {x^2+5x+6}{x+3}

 

\frac {(x+2)(x+3)}{x+3}

 

x+2

And solving looks like this:

\frac {x^2+5x+6}{x+3}=

 

\frac {(x+2)(x+3)}{x+3}

 

x+2

 

x = -2

 

The major difference here is the equal sign that is present when solving vs no equal sign when simplifying.

A common issue with solving ration expressions is how to add together two or more fractions when they have different denominators. Simply remove the denominator like so

\frac {x-1}{x-3}= \frac{x+1}{x-4}

 

\frac {(x-1)(x-3)(x-4)}{x-3} = \frac {(x+1)(x-3)(x-4)}{x-4}

 

x^2-4x-3 = x^2-2x-3

 

= 0=3x-7

 

3x=7

 

x = \frac {3}{7}

My Time at the Microsoft Garage and the BCTech Summit

alexr2015

Recently, I got invited to join a pilot program started by Microsoft. It involved working with my peers from my school, and knowledgable mentors at Microsoft. The goal of the program was for the four of us (Me, Sara, Marcus, and Paige) to create a product that we could present at the BCTech Summit, all under our mentor, Stacy Mulcahy. It started off with us brainstorming a number of ideas. We got to play with the tech that they had at Microsoft, and that included 3D printers, Hololenses, and Mixed Reality/Virtual Reality headsets.

While the promise of developing something for virtual reality seemed enticing (And dear god, it really was), we also wanted to create a product that could help people who used it. After taking a day to brainstorm, we wanted to make something that could improve a user’s mental health. We decided to create a smart mirror that would detect the emotion that the user is feeling, and then react and give help on how to deal with that emotion.

 

 

 

 

Over the course of a month and a bit, we came to the Microsoft Garage in our own free time to work on developing the software to detect the users emotion, and the hardware to house everything. I worked on the software with our mentor Stacy Mulcahy. She taught me basic coding with JavaScript, how to add APIs, and how to use Visual Studio and GitHub. Marcus and Paige worked on creating a box to house the mirror. Stacy also ordered some 2-way acrylic panels for us.

The prototype was a taken apart monitor that was connected to a raspberry pi, which is all stored inside of the box. We have a keyboard and mouse connected to the prototype, but later, we would like it to be touch/voice activated. The monitor/pi is stored inside the box, and the two-way acrylic covers the monitor to create the smart mirror functionality. There is also a camera attached to the top of the mirror that scans the person sitting in front of it and sends the face to the API to get it analyzed. Once it’s analyzed, the information gets sent back to the mirror, which is displayed in the ultimate form of millennial communication: an emoji.

We’re super proud of what we created. Being able to proudly walk into the showroom, and talk to all these amazing people in the tech industry about what we built was an amazing feeling.

Pre-Calculus 11 Week 14 – Simplifying Rational Expressions

alexr2015

This week, we started a new unit. This unit is on rational expressions.

A rational expression you might find in grade 11 is \frac{(x+2)(x+3}{(x+2)(x-2)}

To simplify this expression, we find anything common between the numerator and denominator. In this case, the binomial (x+2) exists in the numerator and denominator. So we can cross it out, and our final answer is \frac{(x+3)}{(x-2)}

What if you had quadratics as your numerator and denominator like \frac{x^2+x-6}{(x-2)}

Well, we can factor the quadratic into (x+3) and (x-2). What do you see now? Now our rational expression can be simplified!

\frac{(x+3)(x-2)}{(x-2}

 

(x+3)

 

However, if you have a rational expression like \frac {x+3}{x+2}, remember that this cannot be simplified any further. This is it’s final form! You cannot take away the x from the numerator and denominator, as tempting as it is. If you follow through with it, your answers will be wrong. So beware!

 

 

Pre-Calculus 11 Week 12 – Graphing Absolute Values as a Linear Function

alexr2015

This week was absolute values! An absolute value describes how far a number is from zero. When a question asks for an absolute value, it looks like this: |6|

Not that hard. How far is the number six from zero? Well, six. What about |-5|?

|-5| = 5 because -5 is five away from zero. That’s all there is to it.

Now, how do we graph an absolute value? A rule that should be remembered is that there will never be a part of your graph below the x-axis. If your graph is below the x-axis, that means your number is negative, which is wrong. Because absolute values are never negative.

Below is a table and a graph for the function y = x + 1 and y = |x + 1|.

See how the line practically bounces off of the x-axis? This is how an absolute value looks when it is graphed. The red line represents y = |x + 1| and the black line shows what the line would look like if we weren’t looking for the absolute value. The black line is y = x + 1.

Pre-Calculus 11 Week 13 – Graphing Reciprocal Functions

alexr2015

Aight, this is really crazy, you’re gonna wanna strap in.

Imagine this function. It’s an easy one. f(x) = x

Nothing you haven’t seen before. The y-intercept is even zero. Easy peasy. Now this graph  is probably rustling your jimmies, and that abomination is what we call a reciprocal function. In fact, this is the reciprocal of f(x) = x. That

one in particular looks like this: f(x) = \frac{1}{x}

Quick reminder that when we’re asked to reciprocate something, it means the numerator and denominator switch places. In every whole number, there’s always a number 1 as it’s denominator. It’s just invisible. So when we’re asked to reciprocate 3, it becomes 1/3. In the case when we’re asked to reciprocate a function, the whole thing becomes a denominator. So to reciprocate f(x) = x+3, it looks like f(x) = \frac{1}{x+3}

Also good to know is that those bendy lines that are defying the laws of everything you know are called hyperbolas.

So you’re probably wondering: how the hell am I supposed to graph that thing? Well, good question! First, we’re gonna pretend that it’s not a reciprocal function. Let’s attempt to graph f(x) = x Just a diagonal line, nothing special. Now, we’re gonna label some special points.

We need the asymptote, which is an invisible line/fence/great wall that the graph can never cross over. There’s gonna be at least one for x and one for y. There can be more than one on the x-axis, but that’s for another blog post. The asymptote is going to be wherever the line crosses the x-axis. In this case, the asymptote is going to be x = 0. For the y-axis, the asymptote is always going to be y = 0. Don’t fight it, it just is. It’ll change in grade 12, but don’t worry about it. The x-axis asymptote is the red vertical line.

Now the last thing we need to find are the other special points the invariant points. To find the invariant points, take y = 1 and y = -1 and find wherever it would be on your line. I’ve marked them in green on the graph. In the case of this line, our invariant points are (1,1) and (-1,-1).

Now from here, draw your hyperbola, make sure it intersects with your cool dude points, and WHATEVER YOU DO, DON’T LET THE CURVE CROSS THE ASYMPTOTE. Now, I love accuracy as much as the next pre-calc nerd, but as long as your teacher sees a curve, and that you haven’t crossed the asymptote, your graph is golden. Draw a t-chart if it makes you feel any better, but generally, this is what most reciprocal graphs are gonna look like. They’re translate to the left and to the right if their b value, or the y-intercept is higher or lower, and the hyperbolas can be farther or closer to each other depending on the slope.