VCE Specialist 3/4 Question Thread!

[keltingmeith]

why is there an asymptope at y = g(x)

Here's a good way of thinking about it:

Consider the equation . Now, just for kicks, let's figure out when this graph is equal to g(x) by finding when it intercepts the graph y=g(x):



However, this doesn't make sense, as this implies that , however we've decided that a doesn't equal zero (as when you make this equation, you'd make a some other number - say, 7). This means that the graph must never equal g(x), and so g(x) must be our asymptote.

[Zues]

also, why do we use the 1/x type graphs when dong addition of ordinates, like previous, it is addition of the asymptopes y values at a partciular x and hence using this means these grapsh are asymptopes?

[keltingmeith]

also, why do we use the 1/x type graphs when dong addition of ordinates, like previous, it is addition of the asymptopes y values at a partciular x and hence using this means these grapsh are asymptopes?

A little confused with what you mean, but let's consider the function (the function above).

Well, we could split this up into a sum of two separate functions - say, and . This means that , and so when sketching h(x), we may also choose to sketch g(x) and f(x) so that we may use addition of ordinates to sketch h(x) (as they did above).

[Zues]

• That isn't the asymptote, it's the graph of .
  
  Edit: I guess it gives you an idea of the addition of ordinates
  

  
  why did they graph this if it is not an asymptope? i am not using these values for addition of ordinates am i?

• also, why do we use the 1/x type graphs when dong addition of ordinates, like previous, it is addition of the asymptopes y values at a partciular x and hence using this means these grapsh are asymptopes?
  

  
  what i meant here is, i am not using the hyperbola graphs when i do addition of ordinates? then what is the point of sketching it on?

thanks

[Zues]

so my asymptotes are y = x+3and x=-1. i did the dotted lines for both, now if we look at the bottom graph, why have they also sketched 2/(x+1) .  using addition of ordinates when x=-2 i get a y value of 0 (x=-1) and 1 (y=x+3) so this means it has to cross the x axis twice, why has it not done so here

[Zues]

same here why have they drawn the other hyperbola and not just used y = x-4 and x = 3 ?

[keltingmeith]

so my asymptotes are y = x+3and x=-1. i did the dotted lines for both, now if we look at the bottom graph, why have they also sketched 2/(x+1) .  using addition of ordinates when x=-2 i get a y value of 0 (x=-1) and 1 (y=x+3) so this means it has to cross the x axis twice, why has it not done so here

Because you've done something wrong, I'm assuming? Because if you check the quadratic in the original numerator, it has a discriminant less than 0, and so no solutions can exist.

Also, they've drawn the other hyperbola so that they can visualise the addition of ordinates. See here for a nice video explanation on it.

[Zues]

Answer gets same asymptopes as me..?

[keltingmeith]

Answer gets same asymptopes as me..?

It should return the same asymptotes?

I'm getting just as confused as you are right now, maybe try to explain your issue again, or walk me through everything you're doing. If I still can't help after that, I'll just give way for the next person to try.

[Zues]

alright look at my screenshot, and you can see the asymptotes are x = 3 and y = x - 4

now using addition of oordinates with these two doesnt yield that graph (i may be wrong). I assume that they have also graphed the hyperbola and used that as a substitute to x=3 when using addition of ordinates since how can you use x = 3 for this ? does this make better sense? hopefully haha 

[keltingmeith]

alright look at my screenshot, and you can see the asymptotes are x = 3 and y = x - 4

now using addition of oordinates with these two doesnt yield that graph (i may be wrong). I assume that they have also graphed the hyperbola and used that as a substitute to x=3 when using addition of ordinates since how can you use x = 3 for this ? does this make better sense? hopefully haha 

Yes, I think I see your issue.

So, addition of ordinates is a tool used when you want to draw a function which is a sum of two other functions. Sometimes, this sum of functions has asymptotes - BUT you do not use the asymptotes to draw the function.

For example, take f(x)=x+2. We can consider this as a sum of two functions - f1(x)=x and f2(x)=2. So, we can draw y=x and y=2 on the same cartesian plane, and then use addition of ordinates to get an estimate on the shape of y=x+2.

Now, let's consider the harder example, . We can consider this as a sum of the functions and .

Now, to assist with us drawing, we would dot in the two functions and (as they've done). We've these two functions, we can now use addition of ordinates to get an estimate of the shape of the graph, remembering that these two functions aren't necessarily asymptotes.

After graphing it, you will notice that we do see asymptopic behaviour on the lines and , and so we would then dot those lines in to label them as asymptotes.

However, something with these graphs you can do is consider them in the general form , where g(x) is some arbitrary function.

Now, imagine drawing the graph . Next, what you do is sort of move that graph so that instead of it having asymptotic behaviour at y=0, it instead starts to asymptote towards g(x). (it's a little hard to explain, and really requires the help of a gif - but, I don't know how to make gifs, so we're at a loss there)

If you're still confused about why they drew the hyperbola, just say and I'll let someone else takeover. If you still don't understand, that's a statement about how I teach vs. how you learn, not how smart or dumb you are. (because really, we're all smart. )

[Zues]

yeap got it! thanks Euler!

also, the general form/basic form os a hyperbola is x/a^2 - y/b^2 = 1

thus, why is the translated version e.g. (x-h)^2.../ ...   have a squared on the numerator?

same with other graphs such as x^2/25 - y^2/16 = 1 ?

thanks

[Sanguinne]

yeap got it! thanks Euler!

also, the general form/basic form os a hyperbola is x/a^2 - y/b^2 = 1

thus, why is the translated version e.g. (x-h)^2.../ ...   have a squared on the numerator?

same with other graphs such as x^2/25 - y^2/16 = 1 ?

thanks

The general form of a hyperbola has both the x and y squared.
like this

What you may be getting confused with is the a and b. In the example, provided "x^2/25 - y^2/16 = 1", the a is actually 5 and b is 4. Thus when you square them you get 25 and 16.

[Zues]

The general form of a hyperbola has both the x and y squared.
like this

What you may be getting confused with is the a and b. In the example, provided "x^2/25 - y^2/16 = 1", the a is actually 5 and b is 4. Thus when you square them you get 25 and 16.

sorry, my notes had it printed out wrong and had x not x^2 (School notes)

also, with this partial fraction, is there a reason why we wouldnt keep going further e.g. a(x-2) + b = ....

and splitting the (x-2)^2?

[Sanguinne]

sorry, my notes had it printed out wrong and had x not x^2 (School notes)

also, with this partial fraction, is there a reason why we wouldnt keep going further e.g. a(x-2) + b = ....

and splitting the (x-2)^2?

well the thing is you cant split it up any further as the numerator is 1. generally in specialist you turn expressions into partial fractions when the numerator is of a higher degree than the denominator. This technique is commonly used so integration techniques can be applied (calculus).