VCE Methods Question Thread!

[Zues]

Zues, ln x^2 is undefined at x=0. So is (ln x)^2

ok, so why is there like a reflection in the y axis with these type of graphs, and dont go to -y values. it has something to do with absolute values but im not sure.

eg. sketch (logx)^2

[keltingmeith]

The mistake is that expands to , not . Eulerfan's method will work, but he made a slight error (forgot some of the coefficients in the binomial expansion).

I did fix this in the original, but apparently I was beaten to that...

ok, so why is there like a reflection in the y axis with these type of graphs, and dont go to -y values. it has something to do with absolute values but im not sure.

eg. sketch (logx)^2

[log(x)]^2 shouldn't have a reflection in the y-axis... But you're right in it being positive.
log(x^2), however, should have negative values - but it does have a reflection in the y-axis.

Here's a thought - what happens when you square a negative number? Is it still negative?

[knightrider]

Firstly, a distinction - in the form y=a(x-h)^3+k, (h, k) is the STATIONARY point of inflection. ALL cubics have a point of inflection, but the TP cubic is special in that it's the only one with a STATIONARY point of inflection.

Now, you must remember that when you differentiate a function, you get an x-value, not a y-value. So, by differentiating and finding an x-value, you can then use this to find your k. For example, let's use my function as earlier.

1) Differentiate, dy/dx=9x^2-36x+36

2) Solve for zero, x^2-4x+4=0 ===> (x-2)^2=0 ===> x=2

3) Substitute this into the cubic:

f(x)=3x^3-18x^2+36x-21
f(2)=3(2)^3-18(2)^2+36(2)-21=24-72+72-21=3

So, our cubic is of the form f(x)=a(x-2)^3+3. An expansion of this would quickly show that the x^3 has no coefficient, which means a 3 must have been factorised out of the original, so we can put it as f(x)=3(x-2)^3+3, which is exactly what I found earlier with inspection.

Thanks eulerfan 101 

[Zues]

I did fix this in the original, but apparently I was beaten to that...

[log(x)]^2 shouldn't have a reflection in the y-axis... But you're right in it being positive.
log(x^2), however, should have negative values - but it does have a reflection in the y-axis.

Here's a thought - what happens when you square a negative number? Is it still negative?

is the graph defined at 0? since 0^2 = 0 which is undefined e.g. loge(0).
is it the same as saving loge(|x|)^2, idk where im going...

is there any asymptote? help me out haha...

[Orb]

I know this is a basic question, but if you look at the pic attached.

We find the exact value of sin 150.
Isnt sin 150 already the angle created by positive x-axis? Why do we need to minus 150 from 180, if 150 degrees is the angle already from the x-axis?

With relevance to the unit circle, the sine is simply the point on the y-axis.
If you look at sin 30 and sin 150, the points there are the same height, they just have different x-values.

Hence, they're equal

[shivaji]

Yeah i know that, but it doesnt really answer my question of why we have to minus 150 from 180? I know Sin 150 and Sin 30 are the same, and that sin is the point on the y-axis. But when they say find the exact value of Sin150, what does that really mean?

Thanks

you have to work out the smallest angle b/w the whatever and the x-axis, hence sin150 is 150 degrees from the positive x-axis, 30 degrees from negative x-axis, so if you take the smaller of the two angles, it will be sin30

[keltingmeith]

Yeah i know that, but it doesnt really answer my question of why we have to minus 150 from 180? I know Sin 150 and Sin 30 are the same, and that sin is the point on the y-axis. But when they say find the exact value of Sin150, what does that really mean?

Thanks

This can all go back to our unit circle definition of the circular functions. Remember back in year 9, when you were introduced to the circular functions as a ratio of two sides of a right angled triangle? In this case, we said that , where O was the opposite side of the angle, and H the hypotenuse.

Well, we run into a slight problem when theta is not an acute angle - if theta is greater than 90 degrees, all of a sudden our right angled triangle doesn't exist. I encourage you to try and draw a triangle where one of the inner angles is 90 degrees, and the other is obtuse - you will never succeed! ;P

So, this begs a question - how do we calculate the sine of an angle if it isn't acute? Well, by using the unit circle, we all of a sudden can draw triangles if the angle is obtuse, reflex, and anything else. Consider this picture:



By doing some creative thinking, we now have a triangle created by an obtuse angle - and so we can use this new triangle to find its sine, cosine, tangent or any other trig function. However, to use this triangle, we must consider the angle in the triangle. Whilst this triangle corresponds to the obtuse angle (in your case, 150 degrees), we aim to consider the angle inscribed inside the triangle to actually create our value. This is what we call the "symmetry properties" of circular functions.

This is why we do the whole thing - you're actually skipping the step of drawing that triangle inside the unit circle, and going straight to the using symmetry properties, having derived them earlier in your education. (in truth, few schools seem to properly derive them, for any number of reasons - the essentials textbook has a good explanation contained within, though, and a few webpages should have a proper one, if you do want to see it)

you have to work out the smallest angle b/w the whatever and the x-axis, hence sin150 is 150 degrees from the positive x-axis, 30 degrees from negative x-axis, so if you take the smaller of the two angles, it will be sin30

This, however, begs the question as to why it has to be the smallest angle.

[IndefatigableLover]

Thank you so much guys for the help, I definitely understand it now (not 100%, but ill get there soon).

A question for the people who did both spesh and methods, i started doing the circular functions on spesh but i still havent done it in methods yet, should i do the methods chapter first (im guessing its the basics of the spesh one) and then continue with spesh circular functions? Thanks!

I'd definitely suggest going through the Methods Chapter first since after learning the techniques used for Methods, you'll apply them to different trigonometric graphs. Essentially in Methods you're only dealing with 3 types but in Specialist you'll be dealing with 9 types (add on inverse and reciprocal trigonometric functions).

Though if you feel confident in the step up in Specialist then you could go from there as well since it's a lot more fast-paced but easy to follow as things do get repetitive with circular functions (until you hit application/ER).

[keltingmeith]

A question for the people who did both spesh and methods, i started doing the circular functions on spesh but i still havent done it in methods yet, should i do the methods chapter first (im guessing its the basics of the spesh one) and then continue with spesh circular functions? Thanks!

It depends on how confident you are with circular function, really. If you're good with all the circ functions from 1/2 methods, namely symmetry properties and what the graphs look like, you should be fine with specialist.

There is definitely a problem in doing the two concurrent in that often you'll hit things in specialist which are "assumed" from methods, such as circular functions, and particularly differentiation (possibly even integration) in calculus. HOWEVER, this in no way means you're more disadvantaged than other students, so don't feel like students who did methods the year before you have a leg-up - both have pros and cons.

[keltingmeith]

Normally wouldn't double post, but I forgot about this, so just in-case Zues has already come through...

is the graph defined at 0? since 0^2 = 0 which is undefined e.g. loge(0).
is it the same as saving loge(|x|)^2, idk where im going...

is there any asymptote? help me out haha...

Nope, you're working this one out yourself.

Okay, so you've figured out that the graph isn't defined at x=0. (at this point, I'm only discussing log_e(x^2) - discussing both might get confusing)

Okay, so you're on the right track with log_e(|x|)^2, but that isn't the same as log_e(x^2). However, you have identified something important - when you square a number, it becomes positive, like a modulus. So, could log_e(x^2) maybe look similar in shape to log_e(|x|)?

You tell me. Sketch log_e(x) using only squared values, instead of your normal average line (i.e, plot y against x^2 instead of x). Does it look like there is still asymptotic behaviour?

[SE_JM]

Hello,
I just started addition of ordinates (it feels like i'm working backwards lol)
Anyway, I know the basic blah blah like the domain of f+g (x) is the common domain of the two functions, and I know to sketch the addition function, you have to add the y-value of the two functions and then plot it.

But I was wondering someone could help me sketching them more efficiently? Yeah, there is that tedious method of adding y-values constantly But it's tooo tiresome and time-consuming... Surely there is a faster way? I know you develop a better understanding of how these graphs will look like (without really adding stuff) when you practice a lot enough. But I was wondering if you (who have solved many of these than me lol ) could inform me of the little details and tricky bits. (if there is any) Or what knowledge you have formed as you worked these out (like, in what kind of graphs could you detect asymptotes? (without actually adding the y-values)) and other things..

Thank you once again. You guys help me so much

[pi]

Maybe this will help you? [GUIDE] Techniques for Sketching Nice-Looking Graphs

[SE_JM]

Hello
Thank you for directing me here. Could you explain a few things for me?

First, I don't understand the bit in the picture attached. IGNORE THE FIRST PICTURE Sorry weird thing happened there  How do you find the range of the addition functions? ( i know how to find the domain, but not sure about the range)

Also how would you know if your asymptote was vertical or oblique?

I'm also unsure about number 4, where it explains how to find the oblique or curved asymptotes.

Sorry, but could you please explain these in more detail? The implied domain bit (at the start ) was really good and easy to understand but this bit is a bit fuzzy. Probably because i'm not used to this concept right now.... Thanks!

[pi]

Firstly, the guide was written for spesh, so it might be a little above the Methods level (the part you have attached to your post regarding 'm' and 'n' is a restriction on the graphs they offer in questions in spesh, not relevant for you).

Well say we had the equation:

We can see that there is obviously a vertical asymptote at as can't exist at that point.

To see if there are other asymptotes, we say
Hence,
Hence, which is your other asymptote!

[SE_JM]

Thank you