Bazza's 3/4 Question Thread

[dc302]

Yeah domf refers to the function f, the only thing is that it is not necessary to put the x in because the x is just an element whereas you're considering the whole function. It shouldn't be marked down if you put it in though, and as TT said, it's just a notation convention.

[WhoTookMyUsername]

thanks all


what's the answer to the q (in attached file)

the book says D but i thought i would be B/C (which are equal unless i'm wrong)

The book seems to suggest that because the coordinate (a) is negative we must take this into account and do (x+a)
but doesn't this produce -a not a?

explanation please

[dc302]

Yeah I believe the book is wrong.

[WhoTookMyUsername]

1) so the answer is either B or C? is there a "best answer"?

2) What's the rule for taking out negative (-1) in is it if n is odd number and if n is even it's ? (i'm pretty sure its that, just confirming

[Shenz0r]

This is what the solutions for the book says:

8. Negative Cubic A or D
Point of inflection is (a,b)
a<0 so
y = -(x+a)^3 + b
The answer is D.

[TrueTears]

1) so the answer is either B or C? is there a "best answer"?

2) What's the rule for taking out negative (-1) in is it if n is odd number and if n is even it's ? (i'm pretty sure its that, just confirming

2) (x-a)^n if n is odd then (-(a-x))^n = (-1)^n(a-x)^n = -(a-x)^n since (-1)^n = -1 for all odd n.

[enwiabe]

This is what the solutions for the book says:

8. Negative Cubic A or D
Point of inflection is (a,b)
a<0 so
y = -(x+a)^3 + b
The answer is D.

Nah, book is wrong, the answer is B or C (both correct). If it was , then the shift would be a units to the left, meaning the co-ordinates of the inflexion would be (-a, b). Because the point given is (a, b), it should be and the translatation is a units to the right. Now if a is a negative number, then translating right by a negative number means you in fact go left. All the way to the negative side of the x-axis.

[paulsterio]

The answer is either B or C.





is the right answer.

Thus B, although C is equally right.

[WhoTookMyUsername]

Can someone explain addition / subtraction or ordinates (sum etc. functions?). I really don't get it how to decide the shape of the resultant graph. Also do you actively take steps to ensure both graphs are to scale?
thanks

i've searched for a good video / explanation through google but i couldn't find one :S (link anyone?)

[pi]

Can someone explain addition / subtraction or ordinates (sum etc. functions?). I really don't get it how to decide the shape of the resultant graph. Also do you actively take steps to ensure both graphs are to scale?
thanks

I find its best to do these questions visually (scale is important, but it doesn't have to be exact, just realistic). The steps I take (tech-free): For h(x) = f(x) + g(x)
1) Sketch each graph f(x) and g(x) lightly (dotted lined maybe), try and make sure they look realistic relative to each other
2) Keep any vertical asymptotes where they are!
3) Plot a few points, some you may have from calculations (t.ps), other you can just make up visually by literally adding the graphs together at certain points
4) From there, you should be able to see the shape of the graph, so you can sketch it in
5) Rub-out any f(x) and g(x) that aren't oblique/curved asymptotes

[Aurelian]

Can someone explain addition / subtraction or ordinates (sum etc. functions?). I really don't get it how to decide the shape of the resultant graph. Also do you actively take steps to ensure both graphs are to scale?
thanks

I find its best to do these questions visually (scale is important, but it doesn't have to be exact, just realistic). The steps I take (tech-free): For h(x) = f(x) + g(x)
1) Sketch each graph f(x) and g(x) lightly (dotted lined maybe), try and make sure they look realistic relative to each other
2) Keep any vertical asymptotes where they are!
3) Plot a few points, some you may have from calculations (t.ps), other you can just make up visually by literally adding the graphs together at certain points
4) From there, you should be able to see the shape of the graph, so you can sketch it in
5) Rub-out any f(x) and g(x) that aren't oblique/curved asymptotes

Just adding to that, I found there are a number of 'important' point when doing addition of ordinates. For example, where the two component graphs touch, you no the resultant graph has double the y-value. Similarly, when one graph crosses the x-axis, you know that the y co-ordinate of the resultant graph is simple the y co-ordinate of the other component graph etc

Plotting these 'certain' points helps a lot in trying to figure out the overall resultant graph shape as well as aid with scale and specificity.

[pi]

Can someone explain addition / subtraction or ordinates (sum etc. functions?). I really don't get it how to decide the shape of the resultant graph. Also do you actively take steps to ensure both graphs are to scale?
thanks

I find its best to do these questions visually (scale is important, but it doesn't have to be exact, just realistic). The steps I take (tech-free): For h(x) = f(x) + g(x)
1) Sketch each graph f(x) and g(x) lightly (dotted lined maybe), try and make sure they look realistic relative to each other
2) Keep any vertical asymptotes where they are!
3) Plot a few points, some you may have from calculations (t.ps), other you can just make up visually by literally adding the graphs together at certain points
4) From there, you should be able to see the shape of the graph, so you can sketch it in
5) Rub-out any f(x) and g(x) that aren't oblique/curved asymptotes

Just adding to that, I found there are a number of 'important' point when doing addition of ordinates. For example, where the two component graphs touch, you no the resultant graph has double the y-value. Similarly, when one graph crosses the x-axis, you know that the y co-ordinate of the resultant graph is simple the y co-ordinate of the other component graph etc

Plotting these 'certain' points helps a lot in trying to figure out the overall resultant graph shape as well as aid with scale and specificity.

+1, also there may be a "cancellation" of ordinates, resulting in an x-int of the resultant graph

[WhoTookMyUsername]

As always thanks for the help

Why cant a log to the base of a negative work?

If you have doesnt x = 2? why cant you rearrange it and solve using the log function in the cas?

[Panicmode]

Log functions can't exist from negative bases. Think about it this way;

It's because this function only works with powers that are rational numbers with odd integers.

Imagine the graph of f(x) = 2^x

It is a nice smooth, continuous curve.


Now imagine the graph of f(x) = (-2)^x

Every time a fractional power with an even denominator is obtained, the graph is undefined. [For example (-2)^(1/2) = sqrt(-2)]

If x is an even integer, the graph is positive.

If x is an odd integer, the graph is negative.


This doesn't really amount to a function but rather a jumble of points and dots that can be defined in one instant and undefined in the next depending on how zoomed in you want to go. Therefore, thinking logically, this graph doesn't exist and so its inverse doesn't exist either.

[dc302]

Yeah there are a few reasons, another one being that the log laws don't hold up.

For example, if log(-2)4 = 2, then by log laws, we should have log(-2)2 + log(-2)2 = 2 as well, but log(-2)2 has no real solution.

Another thing is that log(-2)4 can and is expressed as: log[4]/log[-2], and this of course also does not have a real solution.