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How do I algebraically express transformations of a sigmoidal?
The Next CEO of Stack OverflowWeibull Distribution Behaving Weirdly?Interpret the graph of $fracax+bcx+d$ as a transformation of $y=frac1x$Ordering angles using max and min functionsContour graphing algorithmrotating a sigmoidal curveWhat is the logic behind the method used to shift a graph horizontally?Calculating y = sec(x) functions given points?Express the following data in the form: $y-2=k(1+x)^n$Position $f(x) = x^3$ by inputting $2$ variablesHow to transform a rational function into a straight line (or viceversa)
$begingroup$
I am graphing a sigmoidal of the form
$$
y=d+fraca1+e^-(x-b)/c
$$
I am investigating how the shape of the graph changes when each of the parameters a, b, c, and d are altered.
I understand that d will shift the graph vertically, b will shift the graph horizontally, c will dilate the graph and a will change the size of the graph.
However, I am unsure how to express these parameters change the graph in algebraic form.
So far I have:
d=constant (k) and has no bearing on x, so it simply shifts the graph up or down by a value of d
x-b=0, so x=b. The horizontal shift of x is equal to the value of b.
Am I on the right track? Is there a more fluid way of expressing these values algebraically? How can I express a and c as well?
Thankyou
graphing-functions
edited Mar 18 at 11:09
Matti P.
2,2811514
asked Mar 18 at 10:53
AdspAdsp
152
$endgroup$
add a comment |
$begingroup$
I am graphing a sigmoidal of the form
$$
y=d+fraca1+e^-(x-b)/c
$$
I am investigating how the shape of the graph changes when each of the parameters a, b, c, and d are altered.
I understand that d will shift the graph vertically, b will shift the graph horizontally, c will dilate the graph and a will change the size of the graph.
However, I am unsure how to express these parameters change the graph in algebraic form.
So far I have:
d=constant (k) and has no bearing on x, so it simply shifts the graph up or down by a value of d
x-b=0, so x=b. The horizontal shift of x is equal to the value of b.
Am I on the right track? Is there a more fluid way of expressing these values algebraically? How can I express a and c as well?
Thankyou
graphing-functions
edited Mar 18 at 11:09
Matti P.
2,2811514
asked Mar 18 at 10:53
AdspAdsp
152
$endgroup$
2
$begingroup$
I'm not really sure what you want exactly (what do you mean by 'express $a$ and $c$ alebraically' ? ), but you can play around with this: desmos.com/calculator/kbegf3a3j2
$endgroup$
– Matti P.
Mar 18 at 11:12
$begingroup$
@MattiP. I want to show how each parameter (a,b,c and d) changes the curve. I have found how is it done graphically, but want to be able to show these transformations algebraically as opposed to graphically.
$endgroup$
– Adsp
Mar 18 at 11:27
$begingroup$
I think anyway you will get a lot of insight from playing around with the graph. For example, the parameter $a$ bears a relationship between the difference in height of the two "legs" of the graph, namely the value at $-infty$ and $+infty$ ...
$endgroup$
– Matti P.
Mar 18 at 11:35
add a comment |
$begingroup$
I am graphing a sigmoidal of the form
$$
y=d+fraca1+e^-(x-b)/c
$$
I am investigating how the shape of the graph changes when each of the parameters a, b, c, and d are altered.
I understand that d will shift the graph vertically, b will shift the graph horizontally, c will dilate the graph and a will change the size of the graph.
However, I am unsure how to express these parameters change the graph in algebraic form.
So far I have:
d=constant (k) and has no bearing on x, so it simply shifts the graph up or down by a value of d
x-b=0, so x=b. The horizontal shift of x is equal to the value of b.
Am I on the right track? Is there a more fluid way of expressing these values algebraically? How can I express a and c as well?
Thankyou
graphing-functions
edited Mar 18 at 11:09
Matti P.
2,2811514
asked Mar 18 at 10:53
AdspAdsp
152
$endgroup$
I am graphing a sigmoidal of the form
$$
y=d+fraca1+e^-(x-b)/c
$$
I am investigating how the shape of the graph changes when each of the parameters a, b, c, and d are altered.
I understand that d will shift the graph vertically, b will shift the graph horizontally, c will dilate the graph and a will change the size of the graph.
However, I am unsure how to express these parameters change the graph in algebraic form.
So far I have:
d=constant (k) and has no bearing on x, so it simply shifts the graph up or down by a value of d
x-b=0, so x=b. The horizontal shift of x is equal to the value of b.
Am I on the right track? Is there a more fluid way of expressing these values algebraically? How can I express a and c as well?
Thankyou
graphing-functions
graphing-functions
edited Mar 18 at 11:09
Matti P.
2,2811514
asked Mar 18 at 10:53
AdspAdsp
152
edited Mar 18 at 11:09
Matti P.
2,2811514
asked Mar 18 at 10:53
AdspAdsp
152
edited Mar 18 at 11:09
Matti P.
2,2811514
edited Mar 18 at 11:09
Matti P.
2,2811514
edited Mar 18 at 11:09
Matti P.
2,2811514
2,2811514
asked Mar 18 at 10:53
AdspAdsp
152
asked Mar 18 at 10:53
AdspAdsp
152
asked Mar 18 at 10:53
AdspAdsp
152
152
2
$begingroup$
I'm not really sure what you want exactly (what do you mean by 'express $a$ and $c$ alebraically' ? ), but you can play around with this: desmos.com/calculator/kbegf3a3j2
$endgroup$
– Matti P.
Mar 18 at 11:12
$begingroup$
@MattiP. I want to show how each parameter (a,b,c and d) changes the curve. I have found how is it done graphically, but want to be able to show these transformations algebraically as opposed to graphically.
$endgroup$
– Adsp
Mar 18 at 11:27
$begingroup$
I think anyway you will get a lot of insight from playing around with the graph. For example, the parameter $a$ bears a relationship between the difference in height of the two "legs" of the graph, namely the value at $-infty$ and $+infty$ ...
$endgroup$
– Matti P.
Mar 18 at 11:35
add a comment |
2
$begingroup$
I'm not really sure what you want exactly (what do you mean by 'express $a$ and $c$ alebraically' ? ), but you can play around with this: desmos.com/calculator/kbegf3a3j2
$endgroup$
– Matti P.
Mar 18 at 11:12
$begingroup$
@MattiP. I want to show how each parameter (a,b,c and d) changes the curve. I have found how is it done graphically, but want to be able to show these transformations algebraically as opposed to graphically.
$endgroup$
– Adsp
Mar 18 at 11:27
$begingroup$
I think anyway you will get a lot of insight from playing around with the graph. For example, the parameter $a$ bears a relationship between the difference in height of the two "legs" of the graph, namely the value at $-infty$ and $+infty$ ...
$endgroup$
– Matti P.
Mar 18 at 11:35
2
2
$begingroup$
I'm not really sure what you want exactly (what do you mean by 'express $a$ and $c$ alebraically' ? ), but you can play around with this: desmos.com/calculator/kbegf3a3j2
$endgroup$
– Matti P.
Mar 18 at 11:12
$begingroup$
I'm not really sure what you want exactly (what do you mean by 'express $a$ and $c$ alebraically' ? ), but you can play around with this: desmos.com/calculator/kbegf3a3j2
$endgroup$
– Matti P.
Mar 18 at 11:12
$begingroup$
@MattiP. I want to show how each parameter (a,b,c and d) changes the curve. I have found how is it done graphically, but want to be able to show these transformations algebraically as opposed to graphically.
$endgroup$
– Adsp
Mar 18 at 11:27
$begingroup$
@MattiP. I want to show how each parameter (a,b,c and d) changes the curve. I have found how is it done graphically, but want to be able to show these transformations algebraically as opposed to graphically.
$endgroup$
– Adsp
Mar 18 at 11:27
$begingroup$
I think anyway you will get a lot of insight from playing around with the graph. For example, the parameter $a$ bears a relationship between the difference in height of the two "legs" of the graph, namely the value at $-infty$ and $+infty$ ...
$endgroup$
– Matti P.
Mar 18 at 11:35
$begingroup$
I think anyway you will get a lot of insight from playing around with the graph. For example, the parameter $a$ bears a relationship between the difference in height of the two "legs" of the graph, namely the value at $-infty$ and $+infty$ ...
$endgroup$
– Matti P.
Mar 18 at 11:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are in a particular case of the following general setting :
How to describe in a geometrical way the transformation of the graphical representation of $y=f(x)$ into the graphical representation of
$$y=d+a.fleft(fracx-bcright) ?tag1$$
Here are the successive actions, in this order :
1) $x$-axis translation $b$ units rightwards (this must be considered algebraically : if $b<0$, the translation is $|b|$ units on the left).
2) $x$-axis directional enlargment if $c<1$, shrinking if $c>1$ by a factor $c$.
3) $y$-axis directional enlargment if $a>1$, shrinking if $a<1$ by a factor $a$.
4) $y$-axis translation $d$ units upwards (considered algebraically as for 1)).
Important remark : there is an equivalent way to write down (1):
$$underbracefracy-da_Y=fleft(underbracefracx-bc_Xright) tag2$$
which is symmetrical in $x$ and $y$.
(2) can be written as well under the form :
$$Y=f(X) textwith begincasesx&=&cX+b\y&=&aY+dendcases (3)$$
(old coordinates expressed as - affine - functions of the new ones, as usual).
(3) provides a "dual view" : the new curve can be interpreted "statically" as the ancient curve "seen" with respect to a change of origin and scaling on both axes...
answered Mar 18 at 12:45
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
My function takes the form y=d+a/f(x−bc) Does the divide by a change any of your results or have you rearranged it?
$endgroup$
– Adsp
Mar 18 at 20:30
$begingroup$
I have considered that $f(x)=1/(1+e^-x)$. If one considers $f(x)=1+e^-x$, we need indeed transformation $f(x) to 1/f(x)$ which cannot be treated in a simple way (one could say that this transformation is a "violent" operation compared to others)
$endgroup$
– Jean Marie
Mar 18 at 20:36
add a comment |
$begingroup$
As $d$ and $b$ are vertical/horizontal shifts respectively, then $a$ and $c$ can be interpreted as vertical/horizontal expansion/compression respectively ($a,c>1$ imply on expansion while $a,c<1$ imply compression). A typical shape for $d=b=0$ and $a=1$ is as follows:
answered Mar 18 at 11:26
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are in a particular case of the following general setting :
How to describe in a geometrical way the transformation of the graphical representation of $y=f(x)$ into the graphical representation of
$$y=d+a.fleft(fracx-bcright) ?tag1$$
Here are the successive actions, in this order :
1) $x$-axis translation $b$ units rightwards (this must be considered algebraically : if $b<0$, the translation is $|b|$ units on the left).
2) $x$-axis directional enlargment if $c<1$, shrinking if $c>1$ by a factor $c$.
3) $y$-axis directional enlargment if $a>1$, shrinking if $a<1$ by a factor $a$.
4) $y$-axis translation $d$ units upwards (considered algebraically as for 1)).
Important remark : there is an equivalent way to write down (1):
$$underbracefracy-da_Y=fleft(underbracefracx-bc_Xright) tag2$$
which is symmetrical in $x$ and $y$.
(2) can be written as well under the form :
$$Y=f(X) textwith begincasesx&=&cX+b\y&=&aY+dendcases (3)$$
(old coordinates expressed as - affine - functions of the new ones, as usual).
(3) provides a "dual view" : the new curve can be interpreted "statically" as the ancient curve "seen" with respect to a change of origin and scaling on both axes...
answered Mar 18 at 12:45
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
My function takes the form y=d+a/f(x−bc) Does the divide by a change any of your results or have you rearranged it?
$endgroup$
– Adsp
Mar 18 at 20:30
$begingroup$
I have considered that $f(x)=1/(1+e^-x)$. If one considers $f(x)=1+e^-x$, we need indeed transformation $f(x) to 1/f(x)$ which cannot be treated in a simple way (one could say that this transformation is a "violent" operation compared to others)
$endgroup$
– Jean Marie
Mar 18 at 20:36
add a comment |
$begingroup$
You are in a particular case of the following general setting :
How to describe in a geometrical way the transformation of the graphical representation of $y=f(x)$ into the graphical representation of
$$y=d+a.fleft(fracx-bcright) ?tag1$$
Here are the successive actions, in this order :
1) $x$-axis translation $b$ units rightwards (this must be considered algebraically : if $b<0$, the translation is $|b|$ units on the left).
2) $x$-axis directional enlargment if $c<1$, shrinking if $c>1$ by a factor $c$.
3) $y$-axis directional enlargment if $a>1$, shrinking if $a<1$ by a factor $a$.
4) $y$-axis translation $d$ units upwards (considered algebraically as for 1)).
Important remark : there is an equivalent way to write down (1):
$$underbracefracy-da_Y=fleft(underbracefracx-bc_Xright) tag2$$
which is symmetrical in $x$ and $y$.
(2) can be written as well under the form :
$$Y=f(X) textwith begincasesx&=&cX+b\y&=&aY+dendcases (3)$$
(old coordinates expressed as - affine - functions of the new ones, as usual).
(3) provides a "dual view" : the new curve can be interpreted "statically" as the ancient curve "seen" with respect to a change of origin and scaling on both axes...
answered Mar 18 at 12:45
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
My function takes the form y=d+a/f(x−bc) Does the divide by a change any of your results or have you rearranged it?
$endgroup$
– Adsp
Mar 18 at 20:30
$begingroup$
I have considered that $f(x)=1/(1+e^-x)$. If one considers $f(x)=1+e^-x$, we need indeed transformation $f(x) to 1/f(x)$ which cannot be treated in a simple way (one could say that this transformation is a "violent" operation compared to others)
$endgroup$
– Jean Marie
Mar 18 at 20:36
add a comment |
$begingroup$
You are in a particular case of the following general setting :
How to describe in a geometrical way the transformation of the graphical representation of $y=f(x)$ into the graphical representation of
$$y=d+a.fleft(fracx-bcright) ?tag1$$
Here are the successive actions, in this order :
1) $x$-axis translation $b$ units rightwards (this must be considered algebraically : if $b<0$, the translation is $|b|$ units on the left).
2) $x$-axis directional enlargment if $c<1$, shrinking if $c>1$ by a factor $c$.
3) $y$-axis directional enlargment if $a>1$, shrinking if $a<1$ by a factor $a$.
4) $y$-axis translation $d$ units upwards (considered algebraically as for 1)).
Important remark : there is an equivalent way to write down (1):
$$underbracefracy-da_Y=fleft(underbracefracx-bc_Xright) tag2$$
which is symmetrical in $x$ and $y$.
(2) can be written as well under the form :
$$Y=f(X) textwith begincasesx&=&cX+b\y&=&aY+dendcases (3)$$
(old coordinates expressed as - affine - functions of the new ones, as usual).
(3) provides a "dual view" : the new curve can be interpreted "statically" as the ancient curve "seen" with respect to a change of origin and scaling on both axes...
answered Mar 18 at 12:45
Jean MarieJean Marie
31k42255
$endgroup$
You are in a particular case of the following general setting :
How to describe in a geometrical way the transformation of the graphical representation of $y=f(x)$ into the graphical representation of
$$y=d+a.fleft(fracx-bcright) ?tag1$$
Here are the successive actions, in this order :
1) $x$-axis translation $b$ units rightwards (this must be considered algebraically : if $b<0$, the translation is $|b|$ units on the left).
2) $x$-axis directional enlargment if $c<1$, shrinking if $c>1$ by a factor $c$.
3) $y$-axis directional enlargment if $a>1$, shrinking if $a<1$ by a factor $a$.
4) $y$-axis translation $d$ units upwards (considered algebraically as for 1)).
Important remark : there is an equivalent way to write down (1):
$$underbracefracy-da_Y=fleft(underbracefracx-bc_Xright) tag2$$
which is symmetrical in $x$ and $y$.
(2) can be written as well under the form :
$$Y=f(X) textwith begincasesx&=&cX+b\y&=&aY+dendcases (3)$$
(old coordinates expressed as - affine - functions of the new ones, as usual).
(3) provides a "dual view" : the new curve can be interpreted "statically" as the ancient curve "seen" with respect to a change of origin and scaling on both axes...
answered Mar 18 at 12:45
Jean MarieJean Marie
31k42255
edited Mar 19 at 10:34
answered Mar 18 at 12:45
Jean MarieJean Marie
31k42255
answered Mar 18 at 12:45
Jean MarieJean Marie
31k42255
answered Mar 18 at 12:45
Jean MarieJean Marie
31k42255
31k42255
$begingroup$
My function takes the form y=d+a/f(x−bc) Does the divide by a change any of your results or have you rearranged it?
$endgroup$
– Adsp
Mar 18 at 20:30
$begingroup$
I have considered that $f(x)=1/(1+e^-x)$. If one considers $f(x)=1+e^-x$, we need indeed transformation $f(x) to 1/f(x)$ which cannot be treated in a simple way (one could say that this transformation is a "violent" operation compared to others)
$endgroup$
– Jean Marie
Mar 18 at 20:36
add a comment |
$begingroup$
My function takes the form y=d+a/f(x−bc) Does the divide by a change any of your results or have you rearranged it?
$endgroup$
– Adsp
Mar 18 at 20:30
$begingroup$
I have considered that $f(x)=1/(1+e^-x)$. If one considers $f(x)=1+e^-x$, we need indeed transformation $f(x) to 1/f(x)$ which cannot be treated in a simple way (one could say that this transformation is a "violent" operation compared to others)
$endgroup$
– Jean Marie
Mar 18 at 20:36
$begingroup$
My function takes the form y=d+a/f(x−bc) Does the divide by a change any of your results or have you rearranged it?
$endgroup$
– Adsp
Mar 18 at 20:30
$begingroup$
My function takes the form y=d+a/f(x−bc) Does the divide by a change any of your results or have you rearranged it?
$endgroup$
– Adsp
Mar 18 at 20:30
$begingroup$
I have considered that $f(x)=1/(1+e^-x)$. If one considers $f(x)=1+e^-x$, we need indeed transformation $f(x) to 1/f(x)$ which cannot be treated in a simple way (one could say that this transformation is a "violent" operation compared to others)
$endgroup$
– Jean Marie
Mar 18 at 20:36
$begingroup$
I have considered that $f(x)=1/(1+e^-x)$. If one considers $f(x)=1+e^-x$, we need indeed transformation $f(x) to 1/f(x)$ which cannot be treated in a simple way (one could say that this transformation is a "violent" operation compared to others)
$endgroup$
– Jean Marie
Mar 18 at 20:36
add a comment |
$begingroup$
As $d$ and $b$ are vertical/horizontal shifts respectively, then $a$ and $c$ can be interpreted as vertical/horizontal expansion/compression respectively ($a,c>1$ imply on expansion while $a,c<1$ imply compression). A typical shape for $d=b=0$ and $a=1$ is as follows:
answered Mar 18 at 11:26
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
As $d$ and $b$ are vertical/horizontal shifts respectively, then $a$ and $c$ can be interpreted as vertical/horizontal expansion/compression respectively ($a,c>1$ imply on expansion while $a,c<1$ imply compression). A typical shape for $d=b=0$ and $a=1$ is as follows:
answered Mar 18 at 11:26
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
As $d$ and $b$ are vertical/horizontal shifts respectively, then $a$ and $c$ can be interpreted as vertical/horizontal expansion/compression respectively ($a,c>1$ imply on expansion while $a,c<1$ imply compression). A typical shape for $d=b=0$ and $a=1$ is as follows:
answered Mar 18 at 11:26
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
As $d$ and $b$ are vertical/horizontal shifts respectively, then $a$ and $c$ can be interpreted as vertical/horizontal expansion/compression respectively ($a,c>1$ imply on expansion while $a,c<1$ imply compression). A typical shape for $d=b=0$ and $a=1$ is as follows:
answered Mar 18 at 11:26
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 11:26
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 11:26
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 11:26
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
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2
$begingroup$
I'm not really sure what you want exactly (what do you mean by 'express $a$ and $c$ alebraically' ? ), but you can play around with this: desmos.com/calculator/kbegf3a3j2
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– Matti P.
Mar 18 at 11:12
$begingroup$
@MattiP. I want to show how each parameter (a,b,c and d) changes the curve. I have found how is it done graphically, but want to be able to show these transformations algebraically as opposed to graphically.
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– Adsp
Mar 18 at 11:27
$begingroup$
I think anyway you will get a lot of insight from playing around with the graph. For example, the parameter $a$ bears a relationship between the difference in height of the two "legs" of the graph, namely the value at $-infty$ and $+infty$ ...
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– Matti P.
Mar 18 at 11:35