VCE Methods Question Thread!

[saba.ay]

Could someone please help with the following:

Y=f[ (-(f^-1 (3x))]

Where f(x)= 2ln(x+1) and f^-1(x)= e^(x/2) -1

It has to be in the form (ax) / (bx + c)

Please and thank you and sorry bout all the messy brackets.

[Phy124]

Hi,

Could you explain how to sketch sin/cos graphs that have no y translations  by solving for x intercepts instead of transformations?

i.e. for graphs such as y=-1/2cos(2x-5pi/3) or y= -3sin(3x-7pi/2), etc.

Since my SAC doesn't cover trig equations, all sin/cos graphs cannot have translations in the y direction or else the x intercept cannot be worked out just be transformations alone. However, I find the transformation method of graphing tedious.. so could you please explain how to solve the x and y intercepts of these types of sin/cos graphs (that have no y translations) through trig equations instead of transformations?

Thanks!!

"spoilering" the answer due to length of post;

Spoiler

y-intercepts occur when





x-intercepts occur when





for









This can also be written as

Have a go at the second one yourself

Spoiler

y-intercepts occur when





x-intercepts occur when





for







This can also be written as

Also note that if you had something of the form you should see on inspection that this is the same as

[itsdanny]

Multiply the whole thing by -1, you just multiplied the square root, +2 will become -2.

So becomes

Wow this is some really good LaTex practice =)

Thank you but sorry, I think you misunderstood my question. What I was trying to say was that what is the difference between the initial points of the equation y=-(sqrt[x-3])+2) and y=(sqrt[x-3])+2), i.e. they both assume initial points at (3,2), although I ask why is this so? Because the graph without the reflection in the x-axis, would have a point at (3,2) but even with the reflection in the x-axis, the point is still at (3,2). What I thought was that since it's being reflected in the x-axis, which would multiply the y values by -1, therefore why isn't the graph of y=-(sqrt[x-3])+2) start at (3,-1[2]) or (3,-2)?

Thanks,

[abcdqd]

Thank you but sorry, I think you misunderstood my question. What I was trying to say was that what is the difference between the initial points of the equation y=-(sqrt[x-3])+2) and y=(sqrt[x-3])+2), i.e. they both assume initial points at (3,2), although I ask why is this so? Because the graph without the reflection in the x-axis, would have a point at (3,2) but even with the reflection in the x-axis, the point is still at (3,2). What I thought was that since it's being reflected in the x-axis, which would multiply the y values by -1, therefore why isn't the graph of y=-(sqrt[x-3])+2) start at (3,-1[2]) or (3,-2)?

Thanks,

the graph y=-(sqrt[x-3])+2 has been reflected in the x-axis first and then translated 2 units up. thus, the starting point is still (3,2)
if it was translated first, and then reflected, the equation would be what ovazealous said, and the starting point would be at (3,-2).

[itsdanny]

the graph y=-(sqrt[x-3])+2 has been reflected in the x-axis first and then translated 2 units up. thus, the starting point is still (3,2)
if it was translated first, and then reflected, the equation would be what ovazealous said, and the starting point would be at (3,-2).

Okay that definitely makes more sense, thanks abcdqd. So in the case of when reading any function rule, is to basically stick with conventions of reading the dilations or reflections and then lastly, the translations? Or have I misunderstood a part of your explanation?

[fleet street]

Hi, could someone please show me the algebraic steps of finding the intersection between the graphs e^-x + 3 and -loge(x-3) ?

I know that you would let e^-x +3 =x, to find the x-value, but how do you go from there?

thanks

I'm pretty sure that you don't need to solve this exactly for methods. There is an approximate solution of (3.047,3.047), however. (By the way, if you're interested, an exact solution can be found. Look up "Lambert W function").

[Zealous]

I'm pretty sure that you don't need to solve this exactly for methods. There is an approximate solution of (3.047,3.047), however. (By the way, if you're interested, an exact solution can be found. Look up "Lambert W function").

Haha I thought so, I was looking at it and I had no idea what to do (was just gonna let someone else answer that one =p). CAS calculator would work better for that in a Method's context.

[brightsky]

yeah that is an example of a transcendental equation. transcendental equations 'transcend' algebra, and thus cannot be solved using normal algebraic methods. an approximation can, however, always be found using something like newton's method or the lambert w function.

[fleet street]

Hi,

A function is mapped to the curve . Create a matrix equation that will map to .

I know that you could do it by writing out all of the transformations (e.g "Reflection in the x-axis, etc") and then constructing the matrices from that, but is there a quicker way (that can be written down in a SAC/Exam)?

Thanks!

[Zealous]

Hi,

A function is mapped to the curve . Create a matrix equation that will map to .

I know that you could do it by writing out all of the transformations (e.g "Reflection in the x-axis, etc") and then constructing the matrices from that, but is there a quicker way (that can be written down in a SAC/Exam)?

Thanks!

Maybe you could try just by observation seeing what has happened to the x values and the y values. I saw that for every x value in g(x), h(x) would replace it with 4(x+1) and then it was reflected in the x-axis and translated 3 units upwards. So the reflection and the translation upwards has to do with y values and the dilation and the translation sidewards had to do with x values.

+
That's the matrix I got, I think it will work...

How I got the matrices...

Spoiler

What has been mapped onto the x values is 4(x+1), if you know how to use matrices to map equations, you can imagine the x is x' and rearrange backwards.



=

=

Then convert that into matrix form.

Then the y, we know it has been reflected in the x axis then translated 3 units upwards:



Use y':





Then convert that into matrix form =)

Note these are my improvised methods of solving these questions, what I've done is just improvising since I haven't been taught how to do it by a teacher.

[JieSun92]

quick question. solve for x

[brightsky]

log9(x) = y
9^y = x
3^(2y) = x
2y = log3(x)
y = 1/2*log3(x)
so the equation becomes
log3(2x+1) - 1/2*log3(x) = 0
log3(2x+1) = log3(sqrt(x))
2x+1 = sqrt(x)
etc
etc

[JieSun92]

thanks for the help brightsky. is there any other way of doing it?

[fleet street]

quick question. solve for x

I got









Hence, no real solutions.

Although, still, this method seems cumbersome.

Maybe you could try just by observation seeing what has happened to the x values and the y values. I saw that for every x value in g(x), h(x) would replace it with 4(x+1) and then it was reflected in the x-axis and translated 3 units upwards. So the reflection and the translation upwards has to do with y values and the dilation and the translation sidewards had to do with x values.

+
That's the matrix I got, I think it will work...

How I got the matrices...

Spoiler

What has been mapped onto the x values is 4(x+1), if you know how to use matrices to map equations, you can imagine the x is x' and rearrange backwards.



=

=

Then convert that into matrix form.

Then the y, we know it has been reflected in the x axis then translated 3 units upwards:



Use y':





Then convert that into matrix form =)

Note these are my improvised methods of solving these questions, what I would do if I hadn't been taught a way of doing these specifically.

Thanks! I thought it would be something like that. However, would you please be able to explain the way of doing these specifically that you were taught?

[JieSun92]

Ahh it was supposed to be 2x-1 in the bracket but I got it now. Thanks fleeet