VCE Methods Question Thread!

[cosine]

Remember the order of DR. T: Dilations, Reflections and Translation.

Dilation by factor 2 from y axis
Reflection in x axis
Translation 6 units in negative direction of x axis
Translation 4 units in positive direction of y axis

1/2 from y-axis
3 in the neg direction

:3

[Floatzel98]

Remember the order of DR. T: Dilations, Reflections and Translation.

Dilation by factor 2 from y axis
Reflection in x axis
Translation 6 units in negative direction of x axis
Translation 4 units in positive direction of y axis

1/2 from y-axis
3 in the neg direction

:3

Thanks guys. I'll be sure to remember DRT.

[Floatzel98]

Just incase you do not know, if the question specifically states transformations in their own order, you must follow it. You only use DRT when a specific order is not prescribed

Applying the same transformation values in different orders can affect the graph differently then?

[lzxnl]

Just incase you do not know, if the question specifically states transformations in their own order, you must follow it. You only use DRT when a specific order is not prescribed

Again, DRT isn't the best.
If I tell you to transform y = 1/x to y = 1/(2x+1), DRT means you have to factor out the 2 in the denominator to get 2(x+1/2) and then apply transformations.
If you do the translation first, you can just translate one unit to the left and then dilate by factor 1/2 from the y axis. The numbers are neater if you don't use DRT.

Applying the same transformation values in different orders can affect the graph differently then?

Yes. Try dilating by factor 2 from the x axis and moving a function up 1, and then try reversing the order of those.

[dankfrank420]

1/2 from y-axis
3 in the neg direction

:3

Forgive me if I'm wrong, but I thought that translations remain the same if they are applied after the dilation? I know I messed up the dilation from y-axis thing, but if I dilated it first doesn't the translation still be 6 units and not 3?

[knightrider]

How would you do this question?

If , find x when a = 2 and y = 3?

[IndefatigableLover]

How would you do this question?

If , find x when a = 2 and y = 3?

However we know that 'x' cannot be negative so we take the positive answer for 'x'

[knightrider]

However we know that 'x' cannot be negative so we take the positive answer for 'x'

thank you so much   IndefatigableLover 

but why cant x be a negative number?

[IndefatigableLover]

thank you so much   IndefatigableLover 

but why cant x be a negative number?

The best way I can explain it is to look at the original function you've been given and then substitute your answers in. You know that logs cannot be negative and must be greater than zero (that's the implied domain of logarithmic functions).

When I moved the '2' to make 'x' become 'x^2', I knew that I was going to get two solutions however in reality there is only one solution of 'x' (I just so happened to expand it as such through arithmetic) and afterwards I cancelled one of the  answers out to leave me with one value of 'x' because in reality, there was always only one value of 'x' to satisfy the equation and not two (since it was never x^2 to begin with).

That's how I see it and my wording is probably wrong but one thing you should remember is to always check with your original function when finding solving since sometimes you may have more solutions than needed!

[lzxnl]

thank you so much   IndefatigableLover 

but why cant x be a negative number?

Well, let's see how you'd define a log.
There are two main ways of defining the natural logarithm (all other logs stem from this definition).
One is via integral:
This integral is clearly not defined for negative t as that crosses the asymptote (and the improper integral doesn't exist but shh)

The other is via the inverse of the exponential function e^x, which can in turn be defined by , the limit of a sequence where n is an integer.
For this definition, it's provable that the exponential function is always bigger than zero. If the exponential function is never negative, then its inverse cannot have a negative input.

[cosine]

Well, let's see how you'd define a log.
There are two main ways of defining the natural logarithm (all other logs stem from this definition).
One is via integral:
This integral is clearly not defined for negative t as that crosses the asymptote (and the improper integral doesn't exist but shh)

The other is via the inverse of the exponential function e^x, which can in turn be defined by , the limit of a sequence where n is an integer.
For this definition, it's provable that the exponential function is always bigger than zero. If the exponential function is never negative, then its inverse cannot have a negative input.

Are you even human? o.O

Isn't it because an exponential can never produce a negative number, despite how large/small/sign of the degree?

[keltingmeith]

Are you even human? o.O

Isn't it because an exponential can never produce a negative number, despite how large/small/sign of the degree?

Which is the second definition that lzxnl was talking about, where you define the logarithm as the inverse of the exponential.

[knightrider]

The best way I can explain it is to look at the original function you've been given and then substitute your answers in. You know that logs cannot be negative and must be greater than zero (that's the implied domain of logarithmic functions).

When I moved the '2' to make 'x' become 'x^2', I knew that I was going to get two solutions however in reality there is only one solution of 'x' (I just so happened to expand it as such through arithmetic) and afterwards I cancelled one of the  answers out to leave me with one value of 'x' because in reality, there was always only one value of 'x' to satisfy the equation and not two (since it was never x^2 to begin with).

That's how I see it and my wording is probably wrong but one thing you should remember is to always check with your original function when finding solving since sometimes you may have more solutions than needed!

Well, let's see how you'd define a log.
There are two main ways of defining the natural logarithm (all other logs stem from this definition).
One is via integral:
This integral is clearly not defined for negative t as that crosses the asymptote (and the improper integral doesn't exist but shh)

The other is via the inverse of the exponential function e^x, which can in turn be defined by , the limit of a sequence where n is an integer.
For this definition, it's provable that the exponential function is always bigger than zero. If the exponential function is never negative, then its inverse cannot have a negative input.

Thanks IndefatigableLover  and Lzxnl 

[kinslayer]

How would you do this question?

If , find x when a = 2 and y = 3?

When y = 3 and a = 2,

[faso]

5^n+1-5^n-1/5^n+5^n-2

Need help with this question