How can I change $y=left(frac{1}{x}right)$ to have it land on $2$ specific points?
$begingroup$
I feel the need to prefix this by saying I've not touched maths for over $3$ years...
Because $y=left(frac{1}{x}right)$ goes from practically vertical to practically horizontal, surely this means some part of it can be positioned between $(1, 15)$ and $(300, 1)$?
How can I find the new equation of $y=left(frac{1}{x}right)$ once it has been moved? Or if it's easy to do, what is it?
graphing-functions
asked Mar 13 '17 at 2:13
Toby SmithToby Smith
1184
$endgroup$
|
show 2 more comments
$begingroup$
I feel the need to prefix this by saying I've not touched maths for over $3$ years...
Because $y=left(frac{1}{x}right)$ goes from practically vertical to practically horizontal, surely this means some part of it can be positioned between $(1, 15)$ and $(300, 1)$?
How can I find the new equation of $y=left(frac{1}{x}right)$ once it has been moved? Or if it's easy to do, what is it?
graphing-functions
asked Mar 13 '17 at 2:13
Toby SmithToby Smith
1184
$endgroup$
$begingroup$
what? Are you wanting to translate the graph of 1/x so it crosses both of those points?
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:15
$begingroup$
I guess so, yeah
$endgroup$
– Toby Smith
Mar 13 '17 at 2:16
$begingroup$
if you google "graph shifting" you should find what you need. you need to add or subtract from x and y in the equation.
$endgroup$
– The Count
Mar 13 '17 at 2:34
2
$begingroup$
You could use $frac{4200}{299}frac{1}{x}+frac{285}{299}$ which one gets from solving $frac ax + b = y$ for the points $(1,15)$ and $(300,1)$
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:35
$begingroup$
@BrevanEllefsen Thank you for your help, but is it me or does that not go through either of the points though?
$endgroup$
– Toby Smith
Mar 13 '17 at 15:26
|
show 2 more comments
$begingroup$
I feel the need to prefix this by saying I've not touched maths for over $3$ years...
Because $y=left(frac{1}{x}right)$ goes from practically vertical to practically horizontal, surely this means some part of it can be positioned between $(1, 15)$ and $(300, 1)$?
How can I find the new equation of $y=left(frac{1}{x}right)$ once it has been moved? Or if it's easy to do, what is it?
graphing-functions
asked Mar 13 '17 at 2:13
Toby SmithToby Smith
1184
$endgroup$
I feel the need to prefix this by saying I've not touched maths for over $3$ years...
Because $y=left(frac{1}{x}right)$ goes from practically vertical to practically horizontal, surely this means some part of it can be positioned between $(1, 15)$ and $(300, 1)$?
How can I find the new equation of $y=left(frac{1}{x}right)$ once it has been moved? Or if it's easy to do, what is it?
graphing-functions
graphing-functions
asked Mar 13 '17 at 2:13
Toby SmithToby Smith
1184
asked Mar 13 '17 at 2:13
Toby SmithToby Smith
1184
edited Mar 13 '17 at 2:41
user409521
asked Mar 13 '17 at 2:13
Toby SmithToby Smith
1184
asked Mar 13 '17 at 2:13
Toby SmithToby Smith
1184
asked Mar 13 '17 at 2:13
Toby SmithToby Smith
1184
1184
$begingroup$
what? Are you wanting to translate the graph of 1/x so it crosses both of those points?
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:15
$begingroup$
I guess so, yeah
$endgroup$
– Toby Smith
Mar 13 '17 at 2:16
$begingroup$
if you google "graph shifting" you should find what you need. you need to add or subtract from x and y in the equation.
$endgroup$
– The Count
Mar 13 '17 at 2:34
2
$begingroup$
You could use $frac{4200}{299}frac{1}{x}+frac{285}{299}$ which one gets from solving $frac ax + b = y$ for the points $(1,15)$ and $(300,1)$
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:35
$begingroup$
@BrevanEllefsen Thank you for your help, but is it me or does that not go through either of the points though?
$endgroup$
– Toby Smith
Mar 13 '17 at 15:26
|
show 2 more comments
$begingroup$
what? Are you wanting to translate the graph of 1/x so it crosses both of those points?
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:15
$begingroup$
I guess so, yeah
$endgroup$
– Toby Smith
Mar 13 '17 at 2:16
$begingroup$
if you google "graph shifting" you should find what you need. you need to add or subtract from x and y in the equation.
$endgroup$
– The Count
Mar 13 '17 at 2:34
2
$begingroup$
You could use $frac{4200}{299}frac{1}{x}+frac{285}{299}$ which one gets from solving $frac ax + b = y$ for the points $(1,15)$ and $(300,1)$
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:35
$begingroup$
@BrevanEllefsen Thank you for your help, but is it me or does that not go through either of the points though?
$endgroup$
– Toby Smith
Mar 13 '17 at 15:26
$begingroup$
what? Are you wanting to translate the graph of 1/x so it crosses both of those points?
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:15
$begingroup$
what? Are you wanting to translate the graph of 1/x so it crosses both of those points?
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:15
$begingroup$
I guess so, yeah
$endgroup$
– Toby Smith
Mar 13 '17 at 2:16
$begingroup$
I guess so, yeah
$endgroup$
– Toby Smith
Mar 13 '17 at 2:16
$begingroup$
if you google "graph shifting" you should find what you need. you need to add or subtract from x and y in the equation.
$endgroup$
– The Count
Mar 13 '17 at 2:34
$begingroup$
if you google "graph shifting" you should find what you need. you need to add or subtract from x and y in the equation.
$endgroup$
– The Count
Mar 13 '17 at 2:34
2
2
$begingroup$
You could use $frac{4200}{299}frac{1}{x}+frac{285}{299}$ which one gets from solving $frac ax + b = y$ for the points $(1,15)$ and $(300,1)$
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:35
$begingroup$
You could use $frac{4200}{299}frac{1}{x}+frac{285}{299}$ which one gets from solving $frac ax + b = y$ for the points $(1,15)$ and $(300,1)$
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:35
$begingroup$
@BrevanEllefsen Thank you for your help, but is it me or does that not go through either of the points though?
$endgroup$
– Toby Smith
Mar 13 '17 at 15:26
$begingroup$
@BrevanEllefsen Thank you for your help, but is it me or does that not go through either of the points though?
$endgroup$
– Toby Smith
Mar 13 '17 at 15:26
|
show 2 more comments
1 Answer
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$begingroup$
From Brevan Ellefsen's comment:
The graph of the equation $$y=frac{4200}{299}frac{1}{x}+frac{285}{299}$$ passes through $(1,15)$ and $(300,1)$. That equation can be found by solving a system of two equations of the form $y=frac{a}{x}+b$ for $a$ and $b$ after plugging in the coordinates of the two given points for $x$ and $y$.
$endgroup$
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$begingroup$
From Brevan Ellefsen's comment:
The graph of the equation $$y=frac{4200}{299}frac{1}{x}+frac{285}{299}$$ passes through $(1,15)$ and $(300,1)$. That equation can be found by solving a system of two equations of the form $y=frac{a}{x}+b$ for $a$ and $b$ after plugging in the coordinates of the two given points for $x$ and $y$.
$endgroup$
add a comment |
$begingroup$
From Brevan Ellefsen's comment:
The graph of the equation $$y=frac{4200}{299}frac{1}{x}+frac{285}{299}$$ passes through $(1,15)$ and $(300,1)$. That equation can be found by solving a system of two equations of the form $y=frac{a}{x}+b$ for $a$ and $b$ after plugging in the coordinates of the two given points for $x$ and $y$.
$endgroup$
add a comment |
$begingroup$
From Brevan Ellefsen's comment:
The graph of the equation $$y=frac{4200}{299}frac{1}{x}+frac{285}{299}$$ passes through $(1,15)$ and $(300,1)$. That equation can be found by solving a system of two equations of the form $y=frac{a}{x}+b$ for $a$ and $b$ after plugging in the coordinates of the two given points for $x$ and $y$.
$endgroup$
From Brevan Ellefsen's comment:
The graph of the equation $$y=frac{4200}{299}frac{1}{x}+frac{285}{299}$$ passes through $(1,15)$ and $(300,1)$. That equation can be found by solving a system of two equations of the form $y=frac{a}{x}+b$ for $a$ and $b$ after plugging in the coordinates of the two given points for $x$ and $y$.
answered Dec 5 '18 at 6:45
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Robert Howard
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$begingroup$
what? Are you wanting to translate the graph of 1/x so it crosses both of those points?
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:15
$begingroup$
I guess so, yeah
$endgroup$
– Toby Smith
Mar 13 '17 at 2:16
$begingroup$
if you google "graph shifting" you should find what you need. you need to add or subtract from x and y in the equation.
$endgroup$
– The Count
Mar 13 '17 at 2:34
2
$begingroup$
You could use $frac{4200}{299}frac{1}{x}+frac{285}{299}$ which one gets from solving $frac ax + b = y$ for the points $(1,15)$ and $(300,1)$
$endgroup$
– Brevan Ellefsen
Mar 13 '17 at 2:35
$begingroup$
@BrevanEllefsen Thank you for your help, but is it me or does that not go through either of the points though?
$endgroup$
– Toby Smith
Mar 13 '17 at 15:26