Maths help please? (2 Viewers)

[MiseryParade]

1) Factorise: a^2 - b^2 - 3a -3b

I keep getting stuck at (a+b)(a-b)-3(a+b), since on the answer sheet, it tells me that from that step, the final answer is (a+b)(a-b)(a-b-3). I don't understand. ;_; Help please? It's a two mark question.

2) If you have 18^(4-n), for example, can you break this up into 2^(4-n) * 3^(8-2n)? The actual question was 6^(2n-1)/18^(1-4n). In the solutions, they broke 6 and 18 into its factors, which I haven't seen before - so can you always break large numbers into its components, all with the same power...? (I don't think I'm making sense D: )

3) Sketch y = (x-1)^2 + 3. State a largest possible domain for which the function is one to one.

What does "one to one" mean...? D:

4) If you have an asymptote, do you put arrow signs on either side of it, or just do a broken line?


Thanks guys. My 2U test is tomorrow, and I keep finding out just how much I don't know about all this ): Functions in particular will be the death of me - does anyone have any "procedures" in finding out the domain and range of a function? I don't even know what I'm doing half the time, and I can never get those D/R questions right the first time. ;_;

 

[kazemagic]

For question 1, you're just missing a step. Just check the above image to see what are u missing. I started it off with your answer

For question 2 think of 18^(4-n) as (2 x 9)^(4-n). Now lets take xy^(ab) as an example, to expand that, x^ab * y^ab. Now you can see how you can expand (2 x 9)^(4-n) to 2^(4-n) x 3^2(4-n), which then equals to 2(4-n) x 3(8-2n)

Asymptote can be just a broken line

 

[theind1996]

1) Factorise: a^2 - b^2 - 3a -3b

I keep getting stuck at (a+b)(a-b)-3(a+b), since on the answer sheet, it tells me that from that step, the final answer is (a+b)(a-b)(a-b-3). I don't understand. ;_; Help please? It's a two mark question.

2) If you have 18^(4-n), for example, can you break this up into 2^(4-n) * 3^(8-2n)? The actual question was 6^(2n-1)/18^(1-4n). In the solutions, they broke 6 and 18 into its factors, which I haven't seen before - so can you always break large numbers into its components, all with the same power...? (I don't think I'm making sense D: )

3) Sketch y = (x-1)^2 + 3. State a largest possible domain for which the function is one to one.

What does "one to one" mean...? D:

4) If you have an asymptote, do you put arrow signs on either side of it, or just do a broken line?


Thanks guys. My 2U test is tomorrow, and I keep finding out just how much I don't know about all this ): Functions in particular will be the death of me - does anyone have any "procedures" in finding out the domain and range of a function? I don't even know what I'm doing half the time, and I can never get those D/R questions right the first time. ;_;

1.





3. For a normal function, you can use the vertical line test to determine if the graph is a function or not. For a 1-1 function, use a vertical AND a horizontal line test, and if it passes both, then it is a 1-1 function.

4. Yes, an asymptote should be a broken line to distinguish it from a graph of another function.

 

[planino]

bump

 

[planino]

These questions have my brain tied in a knot, but I THINK the answer to the second one is 2⁷ⁿ*3^6n-2

Edit: I'm hopeless at maths, ignore this

 

[Timske]

One to one function is where for every x value there is only one y value

 

[MiseryParade]

Thanks guys. I wrote a reply to first post but my net derped on me. ;_;

(xy)^(a+2). Can this be split into x^(a+2) * y^(a+2)?

Edit: for every x value there's only one y value - doesn't that define all functions though?

 

[theind1996]

1) Factorise: a^2 - b^2 - 3a -3b

I keep getting stuck at (a+b)(a-b)-3(a+b), since on the answer sheet, it tells me that from that step, the final answer is (a+b)(a-b)(a-b-3). I don't understand. ;_; Help please? It's a two mark question.

2) If you have 18^(4-n), for example, can you break this up into 2^(4-n) * 3^(8-2n)? The actual question was 6^(2n-1)/18^(1-4n). In the solutions, they broke 6 and 18 into its factors, which I haven't seen before - so can you always break large numbers into its components, all with the same power...? (I don't think I'm making sense D: )

3) Sketch y = (x-1)^2 + 3. State a largest possible domain for which the function is one to one.

What does "one to one" mean...? D:

4) If you have an asymptote, do you put arrow signs on either side of it, or just do a broken line?


Thanks guys. My 2U test is tomorrow, and I keep finding out just how much I don't know about all this ): Functions in particular will be the death of me - does anyone have any "procedures" in finding out the domain and range of a function? I don't even know what I'm doing half the time, and I can never get those D/R questions right the first time. ;_;

BTW, your school's 2 unit tests have those questions?..?

 

[kazemagic]

Thanks guys. I wrote a reply to first post but my net derped on me. ;_;

(xy)^(a+2). Can this be split into x^(a+2) * y^(a+2)?

Edit: for every x value there's only one y value - doesn't that define all functions though?

Yes it can

 

[MiseryParade]

BTW, your school's 2 unit tests have those questions?..?

Yeah, these are from past 2U papers

Yes it can

Okie, thanks bro

 

[planino]

you might wanna post this thread in the HSC forum since you can find most of the (no offence, more experienced and on average, smarter) y12's on there who can help you more

 

[RivalryofTroll]

1) Factorise: a^2 - b^2 - 3a -3b

I keep getting stuck at (a+b)(a-b)-3(a+b), since on the answer sheet, it tells me that from that step, the final answer is (a+b)(a-b)(a-b-3). I don't understand. ;_; Help please? It's a two mark question.

2) If you have 18^(4-n), for example, can you break this up into 2^(4-n) * 3^(8-2n)? The actual question was 6^(2n-1)/18^(1-4n). In the solutions, they broke 6 and 18 into its factors, which I haven't seen before - so can you always break large numbers into its components, all with the same power...? (I don't think I'm making sense D: )

3) Sketch y = (x-1)^2 + 3. State a largest possible domain for which the function is one to one.

What does "one to one" mean...? D:

4) If you have an asymptote, do you put arrow signs on either side of it, or just do a broken line?


Thanks guys. My 2U test is tomorrow, and I keep finding out just how much I don't know about all this ): Functions in particular will be the death of me - does anyone have any "procedures" in finding out the domain and range of a function? I don't even know what I'm doing half the time, and I can never get those D/R questions right the first time. ;_;

1. Yeah, everyone has basically answered it. You're just missing one more step.
a^2 - b^2 - 3a - 3b
(a-b)(a+b) -3(a+b)
(a-b-3)(a+b)

2. Yes, like for example, 18^5 is basically like (6x3)^5 = 6^5 x 3^5
6^2n-1/18^1-4n = (2x3)^2n-1/(2x9)^1-4n
2^2n-1 x 3^2n-1/2^1-4n x 3^2(1-4n)
And from there, you know what to do.

3. Never heard of the term 'one to one' to be honest. Which textbook is this? (Timske should be right)

4. Just a broken line will do.

For domain, it's usually quite easy.
For range, especially for certain [harder] discontinuous/continuous functions, drawing the graph will allow you to see the 'range' better.

Sorry if there's any silly mistakes

 

[RivalryofTroll]

BTW, your school's 2 unit tests have those questions?..?

Yeah, I thought question 3 would be in an Ext 1 test? (is it?)

 

[MiseryParade]

1. Yeah, everyone has basically answered it. You're just missing one more step.
a^2 - b^2 - 3a - 3b
(a-b)(a+b) -3(a+b)
(a-b-3)(a+b)

2. Yes, like for example, 18^5 is basically like (6x3)^5 = 6^5 x 3^5
6^2n-1/18^1-4n = (2x3)^2n-1/(2x9)^1-4n
2^2n-1 x 3^2n-1/2^1-4n x 3^2(1-4n)
And from there, you know what to do.

3. Never heard of the term 'one to one' to be honest. Which textbook is this? (Timske should be right)

4. Just a broken line will do.

For domain, it's usually quite easy.
For range, especially for certain [harder] discontinuous/continuous functions, drawing the graph will allow you to see the 'range' better.

Sorry if there's any silly mistakes

They're all from 2U papers at my school, I picked out ones I couldn't do.

When you say "harder dis/continuous functions", which ones are you exactly on about? I keep losing one, two marks every time a d/r question pops up. ;_;

 

[Timske]

Thanks guys. I wrote a reply to first post but my net derped on me. ;_;

(xy)^(a+2). Can this be split into x^(a+2) * y^(a+2)?

Edit: for every x value there's only one y value - doesn't that define all functions though?

Nope think of a parabola, thats why you restrict its domain to obtain a one to one function

EDIT: Passes both vertical and horizontal line test

 

[RivalryofTroll]

They're all from 2U papers at my school, I picked out ones I couldn't do.

When you say "harder dis/continuous functions", which ones are you exactly on about? I keep losing one, two marks every time a d/r question pops up. ;_;

Let's look at the range of y = 1/(x^2+1) [da bell curve]
y > 0 (it never reaches y = 0), y<= 1 (sub x = 0 to know this)
Therefore R = 0<y<=1

 

[MiseryParade]

Nope think of a parabola, thats why you restrict its domain to obtain a one to one function

Holy shit you're amazing. In a parabola, there are two arms - if it tells you to state the "largest" domain, does the "largest" refer to the size of x? (i.e. 2<x<4 and 4<6<x - the latter would be "bigger", number-wise?) I can't imagine they'd ask for the bigger arm itself, since the magnitude of the arms itself should be equal, yeah..? So lost. ;x

Let's look at the range of y = 1/(x^2+1) [da bell curve]
y > 0 (it never reaches y = 0), y<= 1 (sub x = 0 to know this)
Therefore R = 0<y<=1

y>0 simply because it's a hyperbola? Where did y <= 1 suddenly pop up from - wai not y=1 as the y int?

 

[RivalryofTroll]

If you ever are stuck on a D & R question, I suggest you quickly make a rough tiny sketch somewhere.

Doing this will allow you to see both the D & R quite clearly. Yes, time is wasted, but marks are not wasted.

 

[theind1996]

They're all from 2U papers at my school, I picked out ones I couldn't do.

When you say "harder dis/continuous functions", which ones are you exactly on about? I keep losing one, two marks every time a d/r question pops up. ;_;

Discontinuity refers to where there are no x-values for a function. So a 0 in the denominator for hyperbolas (which cause the asymptotes to appear) results in discontinuity, Piece-meal functions, A semi-inverse parabola (y= sqrt x) where the x inside the square root cannot be negative etc.

 

[RivalryofTroll]

Discontinuity refers to where there are no x-values for a function. So a 0 in the denominator for hyperbolas (which cause the asymptotes to appear) results in discontinuity, Piece-meal functions, A semi-inverse parabola (y= sqrt x) where the x inside the square root cannot be negative etc.

This.

In simple terms, continuous - you never need to lift your pen off the page to draw it (e.g. parabola)
discontinuous - you need to lift your pen off the page due to stuff like asymptotes. (e.g. hyperbola)

lul