Graphing functions with mapping vector vs transformation vector












0












$begingroup$


Teacher 1 method to graph $y=2sqrt[3]{2(x-1)}+4$ says:




  1. Create the transformation vector of $<1,4>$ and plot this point as an ordered pair.

  2. Create a table of values for the parent graph of $y=sqrt[3]x$ using easy, reasonable x values $(-8, -1, 0, 1, 8)$.

  3. To this parent graph table, create 2 new columns and modify the x and y values with the stretch and compression values.

  4. From the plotted ordered pair of $(1,4)$ move over and up according to the new modified values.
    This works and creates correct ordered pairs.


Teacher 2 method:
1. Create a mapping vector of $<1/2 x+1, 2y+4>$




  1. Simply apply these to the parent graph ordered pairs.


teacher 2 method seems simpler. But then I ran into a function that doesn't seem to work with teacher 1 as follows:



$y=2^left(x+1right)-3$.



When I try to create the transformation vector of $<-1,-3>$ I see that this ordered pair is NOT on the graph. So applying the parent graph values to a point NOT on the graph clearly produce erroneous results.

When I use the method of creating $<x-1, y-3$ this works properly.



Why doesn't Teacher method 1 work for this simple exponential?






linear-algebra algebra-precalculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Isn't this the same as math.stackexchange.com/questions/3027869/… ?
    $endgroup$
    – Gerry Myerson
    Dec 6 '18 at 1:50






  • 1




    $begingroup$
    I deleted the duplicate.
    $endgroup$
    – user163862
    Dec 6 '18 at 6:51
















0












$begingroup$


Teacher 1 method to graph $y=2sqrt[3]{2(x-1)}+4$ says:




  1. Create the transformation vector of $<1,4>$ and plot this point as an ordered pair.

  2. Create a table of values for the parent graph of $y=sqrt[3]x$ using easy, reasonable x values $(-8, -1, 0, 1, 8)$.

  3. To this parent graph table, create 2 new columns and modify the x and y values with the stretch and compression values.

  4. From the plotted ordered pair of $(1,4)$ move over and up according to the new modified values.
    This works and creates correct ordered pairs.


Teacher 2 method:
1. Create a mapping vector of $<1/2 x+1, 2y+4>$




  1. Simply apply these to the parent graph ordered pairs.


teacher 2 method seems simpler. But then I ran into a function that doesn't seem to work with teacher 1 as follows:



$y=2^left(x+1right)-3$.



When I try to create the transformation vector of $<-1,-3>$ I see that this ordered pair is NOT on the graph. So applying the parent graph values to a point NOT on the graph clearly produce erroneous results.

When I use the method of creating $<x-1, y-3$ this works properly.



Why doesn't Teacher method 1 work for this simple exponential?






linear-algebra algebra-precalculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Isn't this the same as math.stackexchange.com/questions/3027869/… ?
    $endgroup$
    – Gerry Myerson
    Dec 6 '18 at 1:50






  • 1




    $begingroup$
    I deleted the duplicate.
    $endgroup$
    – user163862
    Dec 6 '18 at 6:51














0












0








0





$begingroup$


Teacher 1 method to graph $y=2sqrt[3]{2(x-1)}+4$ says:




  1. Create the transformation vector of $<1,4>$ and plot this point as an ordered pair.

  2. Create a table of values for the parent graph of $y=sqrt[3]x$ using easy, reasonable x values $(-8, -1, 0, 1, 8)$.

  3. To this parent graph table, create 2 new columns and modify the x and y values with the stretch and compression values.

  4. From the plotted ordered pair of $(1,4)$ move over and up according to the new modified values.
    This works and creates correct ordered pairs.


Teacher 2 method:
1. Create a mapping vector of $<1/2 x+1, 2y+4>$




  1. Simply apply these to the parent graph ordered pairs.


teacher 2 method seems simpler. But then I ran into a function that doesn't seem to work with teacher 1 as follows:



$y=2^left(x+1right)-3$.



When I try to create the transformation vector of $<-1,-3>$ I see that this ordered pair is NOT on the graph. So applying the parent graph values to a point NOT on the graph clearly produce erroneous results.

When I use the method of creating $<x-1, y-3$ this works properly.



Why doesn't Teacher method 1 work for this simple exponential?






linear-algebra algebra-precalculus







share|cite|improve this question









$endgroup$




Teacher 1 method to graph $y=2sqrt[3]{2(x-1)}+4$ says:




  1. Create the transformation vector of $<1,4>$ and plot this point as an ordered pair.

  2. Create a table of values for the parent graph of $y=sqrt[3]x$ using easy, reasonable x values $(-8, -1, 0, 1, 8)$.

  3. To this parent graph table, create 2 new columns and modify the x and y values with the stretch and compression values.

  4. From the plotted ordered pair of $(1,4)$ move over and up according to the new modified values.
    This works and creates correct ordered pairs.


Teacher 2 method:
1. Create a mapping vector of $<1/2 x+1, 2y+4>$




  1. Simply apply these to the parent graph ordered pairs.


teacher 2 method seems simpler. But then I ran into a function that doesn't seem to work with teacher 1 as follows:



$y=2^left(x+1right)-3$.



When I try to create the transformation vector of $<-1,-3>$ I see that this ordered pair is NOT on the graph. So applying the parent graph values to a point NOT on the graph clearly produce erroneous results.

When I use the method of creating $<x-1, y-3$ this works properly.



Why doesn't Teacher method 1 work for this simple exponential?






linear-algebra algebra-precalculus




linear-algebra algebra-precalculus






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Dec 6 '18 at 0:58









user163862user163862

86521016




86521016












  • $begingroup$
    Isn't this the same as math.stackexchange.com/questions/3027869/… ?
    $endgroup$
    – Gerry Myerson
    Dec 6 '18 at 1:50






  • 1




    $begingroup$
    I deleted the duplicate.
    $endgroup$
    – user163862
    Dec 6 '18 at 6:51


















  • $begingroup$
    Isn't this the same as math.stackexchange.com/questions/3027869/… ?
    $endgroup$
    – Gerry Myerson
    Dec 6 '18 at 1:50






  • 1




    $begingroup$
    I deleted the duplicate.
    $endgroup$
    – user163862
    Dec 6 '18 at 6:51
















$begingroup$
Isn't this the same as math.stackexchange.com/questions/3027869/… ?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:50




$begingroup$
Isn't this the same as math.stackexchange.com/questions/3027869/… ?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:50




1




1




$begingroup$
I deleted the duplicate.
$endgroup$
– user163862
Dec 6 '18 at 6:51




$begingroup$
I deleted the duplicate.
$endgroup$
– user163862
Dec 6 '18 at 6:51










1 Answer
1






active

oldest

votes


















1












$begingroup$

I think you know why method 1 didn't work for the exponential. It's because, exactly as you say, $(-1,-3)$ is not on the graph. Use the transformation vector $(-1,-3)$, but apply the parent graph values to the point $(-1,-2)$, which is on the graph, and see if that doesn't work.



Basically, when you strip away everything from the first equation, you're left with $y=root3of x$, which goes through the origin; but when you do the same for the second equation, you get $y=2^x$, which doesn't. That's the difference.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I think I will use the second method which always works.
    $endgroup$
    – user163862
    Dec 6 '18 at 17:59











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1 Answer
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1 Answer
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$begingroup$

I think you know why method 1 didn't work for the exponential. It's because, exactly as you say, $(-1,-3)$ is not on the graph. Use the transformation vector $(-1,-3)$, but apply the parent graph values to the point $(-1,-2)$, which is on the graph, and see if that doesn't work.



Basically, when you strip away everything from the first equation, you're left with $y=root3of x$, which goes through the origin; but when you do the same for the second equation, you get $y=2^x$, which doesn't. That's the difference.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I think I will use the second method which always works.
    $endgroup$
    – user163862
    Dec 6 '18 at 17:59
















1












$begingroup$

I think you know why method 1 didn't work for the exponential. It's because, exactly as you say, $(-1,-3)$ is not on the graph. Use the transformation vector $(-1,-3)$, but apply the parent graph values to the point $(-1,-2)$, which is on the graph, and see if that doesn't work.



Basically, when you strip away everything from the first equation, you're left with $y=root3of x$, which goes through the origin; but when you do the same for the second equation, you get $y=2^x$, which doesn't. That's the difference.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I think I will use the second method which always works.
    $endgroup$
    – user163862
    Dec 6 '18 at 17:59














1












1








1





$begingroup$

I think you know why method 1 didn't work for the exponential. It's because, exactly as you say, $(-1,-3)$ is not on the graph. Use the transformation vector $(-1,-3)$, but apply the parent graph values to the point $(-1,-2)$, which is on the graph, and see if that doesn't work.



Basically, when you strip away everything from the first equation, you're left with $y=root3of x$, which goes through the origin; but when you do the same for the second equation, you get $y=2^x$, which doesn't. That's the difference.






share|cite|improve this answer









$endgroup$



I think you know why method 1 didn't work for the exponential. It's because, exactly as you say, $(-1,-3)$ is not on the graph. Use the transformation vector $(-1,-3)$, but apply the parent graph values to the point $(-1,-2)$, which is on the graph, and see if that doesn't work.



Basically, when you strip away everything from the first equation, you're left with $y=root3of x$, which goes through the origin; but when you do the same for the second equation, you get $y=2^x$, which doesn't. That's the difference.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 10:07









Gerry MyersonGerry Myerson

146k8147299




146k8147299












  • $begingroup$
    Thank you. I think I will use the second method which always works.
    $endgroup$
    – user163862
    Dec 6 '18 at 17:59


















  • $begingroup$
    Thank you. I think I will use the second method which always works.
    $endgroup$
    – user163862
    Dec 6 '18 at 17:59
















$begingroup$
Thank you. I think I will use the second method which always works.
$endgroup$
– user163862
Dec 6 '18 at 17:59




$begingroup$
Thank you. I think I will use the second method which always works.
$endgroup$
– user163862
Dec 6 '18 at 17:59


















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