Translations... (?) (1 Viewer)

[Aerlinn]

Ok, this completely puzzled me.
For any function with form something like this: y=a(x-b)^n+c
... basically, hyperbolas, truncus, linear, quadratics, cubics, quartics (and higher powers) all apply.
So, for any function with this form, apparently if there is a negative in front of the x, and you raise this sign to the power outside the brackets (I think), if the answer is positive, there is no reflection in the y-axis, and if answer is negative, there is reflection in the y-axis.

Can someone tell me whether that's true or not/ correct it if it's wrong. I've never heard it, but then, there are lots of tricks with these things that confuse you.



:wave:

 

[bos1234]

could you please give an example. I didnt understand the question. " and you raise this sign to the power outside the brackets (I think), if the answer is positive"

But you could try and graph both curves using the old method (x and y intercepts etc)
one curve is this : y=a(x-b)^n+c with certain values for a b c
and another curve with the question you were asking

and then you could compare the 2 graphs

i dont know if this has answered ur qn..... but i had a crack :rofl:

 

[Aerlinn]

I wasnt talking about odd and even functions, but thanks for the explanations all the same ^_^ You know how if you replace x with -x you get a reflection in the y-axis. I'm trying to work out whether trickier graphs are reflected in y, because it seems, just because there is -x doesn't MEAN it's reflected.

I don't suppose I can explain my question any clearer. It means just what is says. Maybe I'll just give an example...

This is the theory, true or not, I don't know:
So, for any function with this form, apparently if there is a negative in front of the x, and you raise this sign to the power outside the brackets (I think), if the answer is positive, there is no reflection in the y-axis, and if answer is negative, there is reflection in the y-axis.

Eg. A truncus, f(x)= -3/(2-x)²+1
Trying to work out whether the graph reflects in the y-axis.
According to this theory... (-)² = +. No reflection?
I dont have have a graphics calc here so I dunno if that proves the theory right/wrong. Can someone tell me whether it is?

 

[jyu]

I wasnt talking about odd and even functions, but thanks for the explanations all the same ^_^ You know how if you replace x with -x you get a reflection in the y-axis. I'm trying to work out whether trickier graphs are reflected in y, because it seems, just because there is -x doesn't MEAN it's reflected.

I don't suppose I can explain my question any clearer. It means just what is says. Maybe I'll just give an example...

This is the theory, true or not, I don't know:
So, for any function with this form, apparently if there is a negative in front of the x, and you raise this sign to the power outside the brackets (I think), if the answer is positive, there is no reflection in the y-axis, and if answer is negative, there is reflection in the y-axis.

Eg. A truncus, f(x)= -3/(2-x)²+1
Trying to work out whether the graph reflects in the y-axis.
According to this theory... (-)² = +. No reflection?
I dont have have a graphics calc here so I dunno if that proves the theory right/wrong. Can someone tell me whether it is?

In general, for f(x) = a(b - x)^(+/-n) + c, the negative in front of x indicates reflection in the y-axis.

If n is even, a(- x)^(+/-n) is the reflection of a(x)^(+/-n) in the y-axis before the translations. But because a(- x)^(+/-n) = a(x)^(+/-n), the reflection is the same as the original, so you can say there is no reflection.

If n is odd, a(- x)^(+/-n) is the reflection of a(x)^(+/-n) in the y-axis before the translations. This time a(- x)^(+/-n) = -a(x)^(+/-n) =/= a(x)^(+/-n). The reflection is different from the original, it is the same as the reflection in the x-axis.

Note that reflection is done before translations.

:wave:

 

[bos1234]

its like a parabola under the x-axis with an intercept between 0 and -1 and an asymptote at y=0

did u get something like that??

 

[Aerlinn]

jyu, I'm not sure I follow what you're saying. Can you explain it clearer, eg. examples might be a good idea, or explain it differently? =S

hmm... what makes a(- x)^(+/-n) = a(x)^(+/-n) ??
what does this =/= mean?

 

[stephenchow]

A curve is only reflected in the y - axis if the whole thing is put to the negative such as y = x^2 + 6 is reflected in the y axis if and only if y = -(x^2 + 6) and not if "y" equals something like y = -x^2 +6

 

[jyu]

jyu, I'm not sure I follow what you're saying. Can you explain it clearer, eg. examples might be a good idea, or explain it differently? =S

hmm... what makes a(- x)^(+/-n) = a(x)^(+/-n) ??
what does this =/= mean?

For any function f(x), f(-x) is always the reflection in the y-axis.

However, some functions are symmetrical about the y-axis (i.e. show symmetry under reflection in the y-axis), the reflection f(-x) is the same as the original function f(x), so f(-x) = f(x) and you can say there is no reflection.

Example 1

f(x) = 3x^2 + 1
f(-x) = 3(-x)^2 + 1 = 3x^2 + 1 = f(x)

In this case f(x) is symmetrical about the y-axis, the reflection f(-x) is the same as the original function f(x), so f(-x) = f(x) and you can say there is no reflection.

Example 2

f(x) = 3(x-1)^2 + 1
f(-x) = 3[-x-1]^2 + 1 = 3[-(x+1)]^2 + 1 = 3(x+1)^2 + 1

In this case f(x) is not symmetrical about the y-axis, the reflection f(-x) is not the same as the original function f(x), i.e. f(-x) =/= f(x).

Note that the negative taken out of the brackets to the power of 2 became positive and there is reflection in the y-axis.

Quoting 'for any function with this form, apparently if there is a negative in front of the x, and you raise this sign to the power outside the brackets (I think), if the answer is positive, there is no reflection in the y-axis, and if answer is negative, there is reflection in the y-axis.'

The part in italics is not true.

The way to determine whether there is reflection in the y-axis or not is to find out whether or not the function is symmetrical about the y-axis.

:wave:

 

[Aerlinn]

The part in italics is not true.

What then do you think is the true statement of the 'theory'?

The way to determine whether there is reflection in the y-axis or not is to find out whether or not the function is symmetrical about the y-axis.

So you're saying here is, if you know a function is symmetrical about the y axis, you would know that regardless of whether there was -x or not, there would always be no reflection/appear to be no reflection, right?

I don't completely understand, however...
eg. an example like this:
f(x)= 2(x-5)^2+1
f(-x)=2(-x-5)^2+1

Here, the graph is symmetrical in the y-axis before translation, but there is still a reflection in the y-axis...

Can you offer some explanation?