Yes, you're right; a negative answer there simply doesn't make sense in the context of the problem. Nice catch!

Here's the link to the entire playlist. Please keep in mind this was made as a project for graduate courses in 2015, so the GeoGebra platform has changed quite a bit, though these activities are still possible in the newer online format of GeoGebra. Link: kzbin.info/aero/PLHjriFxuDbV6MP6bcfp7y5lB7ioHe2VOT

I'm so glad you found them helpful!

Good question. We can only remove a factor the radical when the factor is a perfect square. So something like √12 would become √(4*3) which would become √4*√3 and then 2√3 since the factor of 4 is a perfect square. The √6 doesn't have a perfect square factor, so that radical doesn't simplify.

You're totally right! Nice work!

Oh wow nice catch! You're totally right, that should be a negative answer.

Good question! The only time you need the +/- is when you *take* an even root of both sides. In this case, the 4th root was already present from the beginning, meaning we already knew we were only working with the positive 4th root for this problem.

@@curtaincommander@curtaincommander Oh, I get it now. Thank you so much.

You're right! Although it shouldn't be +/- 9 for the answer... I should have had +/- 4 after I'd taken the 4th root of each side. So we actually get two solutions: x=5+4 and x=5-4... x=1 and x=9 for final solutions.

Thank you! I really just post these for my students, but figured they'd be helpful to others as well, so I just made them public. Glad you're finding them helpful!

Thanks for your reply! By the way, are all the lessons and homework from the Bigideasmath textbook?

Yeah, I believe so. I haven't taught Algebra 2 in years, and we've gotten a new book since then. It should be noted that although their chapters and sections were used, these notes videos are often accompanied by my own explanations and wording.

Great, thanks for letting me know!

I looked it over again, and I'm pretty sure it's correct. What are you seeing as the issue with it?

Wow this is a big question... First, you use dy/dx when you are trying to find the slope of a parametrically-defined curve at a particular point in the parameter. For instance, finding dy/dx for the curve x=sin(t); y=cos(t) would allow you to plug in a t-value to find the slope of the curve at that point. Points of discontinuity or where the derivatives (either dx/dt or dy/dt) don't exist could be vertical or horizontal tangents, depending on how they're behaving as they approach their discontinuities, and how they contribute to the larger dy/dx. In general, if dy/dt=0, that's the numerator of dy/dx overall, and gives a horizontal tangent. Similarly, if dx/dt=0, then you have a denominator of dy/dx becoming 0, and thus a vertical tangent. I hope this helps!

@@curtaincommander it did, thanks!

Not the way that this program is written, no. However, if you look up my RAM program, you could add an output that takes LRAM+RRAM/2, which would allow an odd number of trapezoids to be used.

Same 💀💀💀💀

I wish I had a straightforward answer to that, but unfortunately, all I can say is.... sort of. Personally, I like the second form you use better, as it more readily shows the degrees of numerator and denominator. However, given our knowledge of transformations, that first form can be much faster in finding asymptotes. There's nothing wrong about teaching that transformational format... it's just more for the student to remember. From a teaching standpoint, I like to teach students to convert to the second format, just to keep information and analysis consistent. I could see students easily forgetting about the k at the end and just evaluating the a/(x-h) part, which would be catastrophic.

⁠got it ! Thank you Mr Hartz.

It takes a little foresight. If I'm working with a function using a square root, I want to choose inputs that will make a perfect square under that square root. That way, they come out cleanly. It's not necessary to do this, but it does produce easier points to plot. If you're allowed a calculator, you could just type any inputs in or look at the table.

Good question! When you have a coefficient that isn't 1 on a trinomial, like the 3 in this example, this factoring technique has the student divide out that coefficient from the two 'magic numbers.' On the one magic number, the dividing by 3 simplified -15 to a -5, but not with the other magic number of 4. If that's the case, the denominator of that fraction becomes the coefficient on that factor, and the numerator is left as the constant. So our divided magic number of 4/3 becomes a factor of (3r+4). If our divided magic number were something like -5/2, the factor would be (2x-5). It's a wonderful pattern to follow when factoring tricky trinomials!

@@curtaincommander thank you so much for the reply and again for your videos. this makes sense thankfully 🙏 I didn’t know if you would reply or not, it’s good to see you’re still here 💪 also I apologize for the late reply, I did complete my math exam just the other day

Thank you! In situations like 1d and 3, I like to use ln personally, because it's a button easily accessible on any scientific calculator. A common log (base 10) can be used also, or any base for that matter. With newer calculators able to handle inputs for logs with different bases, you could technically choose whatever base you want to arrive at the answer. ln is just used for simplicity and consistency.

Second

Third

Lmao this guy be carrying me

Same

This is a great question! Let's use your f(x)=(x+6)^2+3. If you are applying those transformations, the order does matter. To do "a reflection over the y-axis" followed by "a shift right 1", you would first carry out the reflection only. This means replacing the any x variables with -x. So now, our function would be h(x)=((-x)+6)^2+3.... I don't want to call in g(x) yet since that name is reserved for our final answer... Then, we'll do the shift right 1, which replaces any x with an (x-1). Now, we have g(x)=((-(x-1)+6)^2+3. We can clean up those parentheses by distributing and combining like terms to arrive at g(x)=(-x+7)^2+3 for a final answer.

Oh ok! I see. Thank you so much for the detailed explanation. I finally got it! Thanks again.😊

You're right! Whenever taking an even root as a step in solving, both positive and negative solutions are possible. Thanks for pointing that out!

thank you for the quick response! have a great day!