Author: Sybil Shaver

We began with a diagnostic designed to check your understanding of basic fraction information and vocabulary, and tell you what you need to work on. You can see the diagnostic at this link  (it starts on the second page) and you can also read the reasoning behind the choice of answers, if you are interested. Please respect the (rather strict) copyright which the authors assert, and do not reproduce this.

Then on to the new topic, Rational Expressions: reducing them to lowest terms. We will be looking at the basic operations addition, subtraction, multiplication, and division. They work in the same way as the operations on fractions (rational numbers) with a few warnings we must pay attention to.

For review of how we build up the Real number system starting with the Natural Numbers, see the links in this post

It is very important that you become familiar with the various sets of numbers: their names, and what they are (and also where they come from). For example, why do I say that the number 0 (which is indeed a number!) is a very important invention – specifically, why is it an invention?

Rational Expressions are just algebraic expressions which are formed by ratios of polynomials, in the same way that rational numbers are formed by ratios of integers. (The links are to the Website Math Is Fun, which is a wonderful resource for reviewing basic concepts.)

The trouble we can have with rational expressions comes from the fact that, if a variable appears in the denominator, then it is possible for the denominator to become 0 for certain values of that variable. For example, the simple rational expression $\frac{1}{x}$ is undefined when $x = 0$. We must be careful to be aware of this potential problem.

Two warnings that come from this fact:

• When we reduce a rational expression by canceling a common factor which contains a variable, the reduced form is only valid as long as that factor is not equal to 0. Simple example: $\frac{3x}{x} = 3$ only if $x \neq 0$

• When solving an equation, it is important not to divide both sides of the equation by a variable or any expression containing a variable (because you might be dividing by 0.) For example, some people try to solve $x^{2} = 3x$ by dividing both sides by x, which gives $x=3$. But that is not the only solution of the equation! Find the other one by “guess and check” or by solving it properly.

We will also see that the problem of zero denominators causes difficulties when we solve equation which contain rational expressions.

Important note: for many reasons, we usually prefer to see rational expressions in the form where their numerator and denominator are completely factored. So when reducing a rational expression, leave your answer in this factored form.

Information about the assignments and the order and timing of them:

If you feel that you could benefit from practice with operations on fractions (and it really does help), I recommend the Math Is Fun pages on reducing, addition and subtraction, multiplication and division, which are linked from the Fractions menu there. Each page has a number of questions at the bottom that you can use for practice or learning more, and the website checks your answers and gives more information as you work them!

There are several open assignments on WeBWorK. You should start with the assignment on Reducing Rational Expressions. I would recommend then starting the assignment on Adding Rational Expressions, which is not too hard if you remember how to add fractions that have the same denominator. I plan to work on the Multiplying and Diving Rational Expressions tomorrow in class, so hold off on that unless you are very ambitious and/or bored.

The related chapter in the textbook is Lesson 2 (because we are not going in the same order as the book does!), and there are also video resources.

Question to think about: Why is it that division by 0 is not allowed? After all,  when we could not take the square root of $-1$ in the Real Numbers, we just invented a new number for it. It turns out that there is no way to invent a number that would equal $\frac{3}{0}$, for example, but why?

To understand the reason, you will need to think mathematically about what division is. Resist the urge to use clever wordplay like, “well, you can’t divide by nothing” – I could convince you that anything at all is impossible by that kind of vague semantic trickery.

In mathematics, in the branch that is called algebra, there are two basic operations (addition and subtraction) and the other operations are defined in terms of those. This means that division is defined in terms of multiplication: $12 \divide 3 = 4$ because $4 \cdot 3 = 12$. (In the clumsy way I talk about square roots, I could similarly say that “12 divided by 3 is the number that, when you multiply it by 3, you get 12.”) That’s the definition of division.

When we say that division by 0 is undefined, what that means is that division by 0 cannot be defined in terms of multiplication the way I did it in the previous paragraph, and furthermore there is no way to define division by 0 so that it “works” inside our number system.

To see a little bit of why this is true, try to definen 3 divided by 0 using the definition in terms of multiplication. (This is anyway how you check a division problem.) With some thought you should be able to see what goes wrong.

It’s less easy to see why there is no alternative way to define 3 divided by 0, but I’ll give you a writeup or a link next time.

And it turns out that 0 divided by 0 is even worse! We say that $\frac{0}{0}$ is indeterminate. See if you can figure out why.

That’s the news. Check back later for more information.

Here are notes and the classwork from Thursday the 10th, including advice on the homework. Please read!

If you want to use Desmos as I did in class, here is the webpage. There is also an app which works very well. See the desmos.com homepage for links. Also see below!

Here is a page with some fun and informative stuff that you can enjoy reading or viewing (I hope) – including some pretty amazing graphs made using Desmos.

Here is a page with Quiz solutions (not quite up to date yet)

Here is a page with the (tentative) schedule of the Tests.

Here is a page with links to the tutoring schedules and also other resources

here is the classwork:

MAT1275coGraphingCircles

See Lesson 11 in the textbook for more notes on this.

Please complete the problems on finding the perpendicular bisector for next Wednesday. The answers are here:

There is an assignment on WebWorK about Nonlinear Systems, and you can take a look at it if you are ambitious: I would recommend looking at Problem 1 to start with. You can graph both of the equations in Desmos and see if you can figure out the solution or solutions that way! (In fact, you can try that with all of them.)

Please do work the CircleLab and try to get it correct 4 times, as directed. Since this problem uses Geogebra, you may want to work in the Computer Lab.

The nQuiz next time will be on putting the equations of circles into standard form by completing the square, and reading off the center and radius. (No graphing. But remember that you will have to draw the graphs on the Final Exam!)

This is purely a bribe to get you to read the OpenLab posts for today, but it is slightly related to mathematics.

If you are not already familiar with the song “My Shot” from the musical Hamilton (or you just want to hear it again), here it is:

The lyrics are also printed on that video in case you want them.

So here is the magical question:

When one of our students was at the board having worked out one of the “graphing parabolas” problems, another student said a short phrase (which I heard, of course) and it reminded me of this song, so I said of the student at the board, “He’s Alexander Hamilton”.

Which line in this song “my Shot” was I thinking of?

Post your best guess as a comment on this post, one per customer, and please only students enrolled in this course. The first person to post the correct answer will get a free 10/10 quiz grade. (And since I am offline until Wednesday night, just keep posting answers and I’ll look at them and announce the winner after that.)

P.S. The answer to the question “What’s my favorite song from Hamilton?” is always, for me, “the last one I listened to.”

Recall what we know already about parabolas:

• They are the graphs of functions of the form $y = ax^{2} +bx + c$, where $a$, $b$, and $c$ are real numbers and $a \neq 0$.
• There is a distinguished point on the graph, called the vertex. It is either the lowest or highest point on the graph, depending on whether the graph opens upward or downward.
• The graph is symmetric across a vertical line through the vertex.
• The coefficient $a$ determines whether the graph opens upward or downward; upward if $a > 0$, downward if $a<0$.
• Otherwise $a$ influences the steepness or flatness of the graph. If $|a| > 1$ the graph is steeper than the standard parabola $y = x^2$; if $|a| < 1$ the graph is shallower and flatter than the standard graph.
• From shifting parabolas, we found that if $a=1$, the equation of the function has the form $y = (x-h)^2 + k$, where the vertex is $(h, k)$
• In general, the formula for any parabola can be put into the form $y = a(x-h)^2 + k$, where the vertex is the point $9h, k)$ and $a$ influences the steepness or shallowness of the graph and whether it opens upward or downward.

Today we take an equation in the form $y = ax^2 + bx +c$ and rewrite it into the “vertex form” $y = a(x-h)^2 + k$. I showed you two ways to do this. One is the way I was taught (and the way it is usually shown in textbooks), and the second way is the way the WeBWorK does it. You can use either method, whichever you prefer, but WeBWorK walks you through the second method.

Here  is my slideshow with notes on the second method:

MAT1275parabolasCompleting Square-slideshow

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Finally, we saw a formula for finding the x-coordinate of the vertex, namely $x= -\frac{b}{2a}$ (which, I’m sure you will not be surprised to learn, comes from the Quadratic Formula).

Once we have found the x-coordinate, the y-coordinate of the vertex can be found by substituting into the formula for the parabola.

Example: for the parabola with formula $y = 3x^2 -4x +1$,

the x-coordinate of the vertex is given by $x = \frac{-(-4)}{2(3)} = \frac{4}{6} = \frac{2}{3}$

Find the y-coordinate: $y = 3\left(\frac{2}{3}\right)^2 – 4\left(\frac{2}{3}\right) + 1 = 3\left(\frac{4}{9}\right) -\frac{8}{3} + 1 = \frac{4}{3} – \frac{8}{3} + \frac{3}{3} = -\frac{1}{3}$

The vertex is the point $\left(\frac{2}{3}, -\frac{1}{3}\right)$

Sorry for the delay. I had to fix a bug in some of the new WeBWorK problems and it took far longer than I expected.

Here are the notes from Wednesday the 25th and some advice on the order of the homework.

I recommend that you do the WeBWorK “ParabolaParameters” before you do “ShiftingParabolas”.

If you want to use Desmos as I did in class, here is the webpage. There is also an app which works very well. See the desmos.com homepage for links. Also see below!

Here is a page with some fun and informative stuff that you can enjoy reading or viewing (I hope) – including some pretty amazing graphs made using Desmos.