How to write formula of a concave hyperbola
$begingroup$
I started with plotting this equation of a convex hyperbola:
y = 2/x
Which of course looks like this (I'm only interested in 1st quadrant):
Now, I want to plot the same function, but have it flipped about, such that it becomes concave. What is the formula for that? It should look concave, like this:
y = ???
Note: I made these graphs myself, don't scrutinize them too much!
functions graphing-functions
asked Dec 17 '18 at 12:47
Arash HowaidaArash Howaida
1084
$endgroup$
add a comment |
$begingroup$
I started with plotting this equation of a convex hyperbola:
y = 2/x
Which of course looks like this (I'm only interested in 1st quadrant):
Now, I want to plot the same function, but have it flipped about, such that it becomes concave. What is the formula for that? It should look concave, like this:
y = ???
Note: I made these graphs myself, don't scrutinize them too much!
functions graphing-functions
asked Dec 17 '18 at 12:47
Arash HowaidaArash Howaida
1084
$endgroup$
$begingroup$
y = k + 1/(x-c)
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:03
$begingroup$
@GNUSupporter8964民主女神地下教會 Ok, hmm, what's k? is k = x or is it a constant like c?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:04
$begingroup$
They are constants. You may try Desmos, which enables graph sharing.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:07
$begingroup$
@GNUSupporter8964民主女神地下教會 Just to double check, to flip my example equationy=2/xto concave, I would input:y=2+1/(x-2)?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:10
$begingroup$
You can check that on a graphical calculator like Desmos to be reassured of your observation. Actually, that would give the required shape, but I won't call that a "flipping" of 1/x. That's in fact a translation and clipping.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:15
add a comment |
$begingroup$
I started with plotting this equation of a convex hyperbola:
y = 2/x
Which of course looks like this (I'm only interested in 1st quadrant):
Now, I want to plot the same function, but have it flipped about, such that it becomes concave. What is the formula for that? It should look concave, like this:
y = ???
Note: I made these graphs myself, don't scrutinize them too much!
functions graphing-functions
asked Dec 17 '18 at 12:47
Arash HowaidaArash Howaida
1084
$endgroup$
I started with plotting this equation of a convex hyperbola:
y = 2/x
Which of course looks like this (I'm only interested in 1st quadrant):
Now, I want to plot the same function, but have it flipped about, such that it becomes concave. What is the formula for that? It should look concave, like this:
y = ???
Note: I made these graphs myself, don't scrutinize them too much!
functions graphing-functions
functions graphing-functions
asked Dec 17 '18 at 12:47
Arash HowaidaArash Howaida
1084
asked Dec 17 '18 at 12:47
Arash HowaidaArash Howaida
1084
edited Dec 17 '18 at 12:56
Arash Howaida
asked Dec 17 '18 at 12:47
Arash HowaidaArash Howaida
1084
asked Dec 17 '18 at 12:47
Arash HowaidaArash Howaida
1084
asked Dec 17 '18 at 12:47
Arash HowaidaArash Howaida
1084
1084
$begingroup$
y = k + 1/(x-c)
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:03
$begingroup$
@GNUSupporter8964民主女神地下教會 Ok, hmm, what's k? is k = x or is it a constant like c?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:04
$begingroup$
They are constants. You may try Desmos, which enables graph sharing.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:07
$begingroup$
@GNUSupporter8964民主女神地下教會 Just to double check, to flip my example equationy=2/xto concave, I would input:y=2+1/(x-2)?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:10
$begingroup$
You can check that on a graphical calculator like Desmos to be reassured of your observation. Actually, that would give the required shape, but I won't call that a "flipping" of 1/x. That's in fact a translation and clipping.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:15
add a comment |
$begingroup$
y = k + 1/(x-c)
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:03
$begingroup$
@GNUSupporter8964民主女神地下教會 Ok, hmm, what's k? is k = x or is it a constant like c?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:04
$begingroup$
They are constants. You may try Desmos, which enables graph sharing.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:07
$begingroup$
@GNUSupporter8964民主女神地下教會 Just to double check, to flip my example equationy=2/xto concave, I would input:y=2+1/(x-2)?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:10
$begingroup$
You can check that on a graphical calculator like Desmos to be reassured of your observation. Actually, that would give the required shape, but I won't call that a "flipping" of 1/x. That's in fact a translation and clipping.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:15
$begingroup$
y = k + 1/(x-c)
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:03
$begingroup$
y = k + 1/(x-c)
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:03
$begingroup$
@GNUSupporter8964民主女神地下教會 Ok, hmm, what's k? is k = x or is it a constant like c?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:04
$begingroup$
@GNUSupporter8964民主女神地下教會 Ok, hmm, what's k? is k = x or is it a constant like c?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:04
$begingroup$
They are constants. You may try Desmos, which enables graph sharing.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:07
$begingroup$
They are constants. You may try Desmos, which enables graph sharing.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:07
$begingroup$
@GNUSupporter8964民主女神地下教會 Just to double check, to flip my example equation
y=2/x to concave, I would input: y=2+1/(x-2)?$endgroup$
– Arash Howaida
Dec 17 '18 at 13:10
$begingroup$
@GNUSupporter8964民主女神地下教會 Just to double check, to flip my example equation
y=2/x to concave, I would input: y=2+1/(x-2)?$endgroup$
– Arash Howaida
Dec 17 '18 at 13:10
$begingroup$
You can check that on a graphical calculator like Desmos to be reassured of your observation. Actually, that would give the required shape, but I won't call that a "flipping" of 1/x. That's in fact a translation and clipping.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:15
$begingroup$
You can check that on a graphical calculator like Desmos to be reassured of your observation. Actually, that would give the required shape, but I won't call that a "flipping" of 1/x. That's in fact a translation and clipping.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$y=frac{2}{x}$ is its own reflection across the line $y=-x$ (it is already 'concave' for $xleq0$). You only need to shift the viewing window to the lower-left quadrant or translate the function right and up by some suitable distance.
The curve $y-a=f(x-b)$ is the translation of the curve $y=f(x)$ upwards by $a$ and to the right by $b$. This is why GNUSupporter8964民主女神地下教會 commented $y=k+frac{1}{x-c}$. If you want to shift the graph upwards by $k$ and to the right by $c$, you would solve for $y$ in the equation $y-k=f(x-c)$ where $f(x)=your function=frac{2}{x}$.
$$f(x)=frac{2}{x};quad y=f(x);quad y-k=f(x-c)implies y=frac{2}{x-c}+k$$
answered Dec 17 '18 at 15:10
R. BurtonR. Burton
45619
$endgroup$
$begingroup$
thanks, that helps a bit. Would you mind sharing what values I should put for c and k to translate the hyperbola up to the correct position? I tried tinkering with it using wolframalpha. You can see I didn't quite shift it to the right place: wolframalpha.com/input/?i=y%3D2%2F(x-2)%2B2. When I try plugging in larger constants, all that changes is the function gets more stretched out. I'm not sure why, hopefully I'm not doing something wrong.
$endgroup$
– Arash Howaida
Dec 17 '18 at 15:29
1
$begingroup$
The graph isn't being stretched out, Wolfram Alpha scales the viewing window automatically when you plot a function. The values for $c$ and $k$ depend on the size of your viewing window and where it's centered. If your viewing window is $x=0ldots u,y=0ldots v$, then you'll want to have $c=u$ and $k=v$.
$endgroup$
– R. Burton
Dec 17 '18 at 15:40
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
$y=frac{2}{x}$ is its own reflection across the line $y=-x$ (it is already 'concave' for $xleq0$). You only need to shift the viewing window to the lower-left quadrant or translate the function right and up by some suitable distance.
The curve $y-a=f(x-b)$ is the translation of the curve $y=f(x)$ upwards by $a$ and to the right by $b$. This is why GNUSupporter8964民主女神地下教會 commented $y=k+frac{1}{x-c}$. If you want to shift the graph upwards by $k$ and to the right by $c$, you would solve for $y$ in the equation $y-k=f(x-c)$ where $f(x)=your function=frac{2}{x}$.
$$f(x)=frac{2}{x};quad y=f(x);quad y-k=f(x-c)implies y=frac{2}{x-c}+k$$
answered Dec 17 '18 at 15:10
R. BurtonR. Burton
45619
$endgroup$
$begingroup$
thanks, that helps a bit. Would you mind sharing what values I should put for c and k to translate the hyperbola up to the correct position? I tried tinkering with it using wolframalpha. You can see I didn't quite shift it to the right place: wolframalpha.com/input/?i=y%3D2%2F(x-2)%2B2. When I try plugging in larger constants, all that changes is the function gets more stretched out. I'm not sure why, hopefully I'm not doing something wrong.
$endgroup$
– Arash Howaida
Dec 17 '18 at 15:29
1
$begingroup$
The graph isn't being stretched out, Wolfram Alpha scales the viewing window automatically when you plot a function. The values for $c$ and $k$ depend on the size of your viewing window and where it's centered. If your viewing window is $x=0ldots u,y=0ldots v$, then you'll want to have $c=u$ and $k=v$.
$endgroup$
– R. Burton
Dec 17 '18 at 15:40
add a comment |
$begingroup$
$y=frac{2}{x}$ is its own reflection across the line $y=-x$ (it is already 'concave' for $xleq0$). You only need to shift the viewing window to the lower-left quadrant or translate the function right and up by some suitable distance.
The curve $y-a=f(x-b)$ is the translation of the curve $y=f(x)$ upwards by $a$ and to the right by $b$. This is why GNUSupporter8964民主女神地下教會 commented $y=k+frac{1}{x-c}$. If you want to shift the graph upwards by $k$ and to the right by $c$, you would solve for $y$ in the equation $y-k=f(x-c)$ where $f(x)=your function=frac{2}{x}$.
$$f(x)=frac{2}{x};quad y=f(x);quad y-k=f(x-c)implies y=frac{2}{x-c}+k$$
answered Dec 17 '18 at 15:10
R. BurtonR. Burton
45619
$endgroup$
$begingroup$
thanks, that helps a bit. Would you mind sharing what values I should put for c and k to translate the hyperbola up to the correct position? I tried tinkering with it using wolframalpha. You can see I didn't quite shift it to the right place: wolframalpha.com/input/?i=y%3D2%2F(x-2)%2B2. When I try plugging in larger constants, all that changes is the function gets more stretched out. I'm not sure why, hopefully I'm not doing something wrong.
$endgroup$
– Arash Howaida
Dec 17 '18 at 15:29
1
$begingroup$
The graph isn't being stretched out, Wolfram Alpha scales the viewing window automatically when you plot a function. The values for $c$ and $k$ depend on the size of your viewing window and where it's centered. If your viewing window is $x=0ldots u,y=0ldots v$, then you'll want to have $c=u$ and $k=v$.
$endgroup$
– R. Burton
Dec 17 '18 at 15:40
add a comment |
$begingroup$
$y=frac{2}{x}$ is its own reflection across the line $y=-x$ (it is already 'concave' for $xleq0$). You only need to shift the viewing window to the lower-left quadrant or translate the function right and up by some suitable distance.
The curve $y-a=f(x-b)$ is the translation of the curve $y=f(x)$ upwards by $a$ and to the right by $b$. This is why GNUSupporter8964民主女神地下教會 commented $y=k+frac{1}{x-c}$. If you want to shift the graph upwards by $k$ and to the right by $c$, you would solve for $y$ in the equation $y-k=f(x-c)$ where $f(x)=your function=frac{2}{x}$.
$$f(x)=frac{2}{x};quad y=f(x);quad y-k=f(x-c)implies y=frac{2}{x-c}+k$$
answered Dec 17 '18 at 15:10
R. BurtonR. Burton
45619
$endgroup$
$y=frac{2}{x}$ is its own reflection across the line $y=-x$ (it is already 'concave' for $xleq0$). You only need to shift the viewing window to the lower-left quadrant or translate the function right and up by some suitable distance.
The curve $y-a=f(x-b)$ is the translation of the curve $y=f(x)$ upwards by $a$ and to the right by $b$. This is why GNUSupporter8964民主女神地下教會 commented $y=k+frac{1}{x-c}$. If you want to shift the graph upwards by $k$ and to the right by $c$, you would solve for $y$ in the equation $y-k=f(x-c)$ where $f(x)=your function=frac{2}{x}$.
$$f(x)=frac{2}{x};quad y=f(x);quad y-k=f(x-c)implies y=frac{2}{x-c}+k$$
answered Dec 17 '18 at 15:10
R. BurtonR. Burton
45619
answered Dec 17 '18 at 15:10
R. BurtonR. Burton
45619
answered Dec 17 '18 at 15:10
R. BurtonR. Burton
45619
answered Dec 17 '18 at 15:10
R. BurtonR. Burton
45619
45619
$begingroup$
thanks, that helps a bit. Would you mind sharing what values I should put for c and k to translate the hyperbola up to the correct position? I tried tinkering with it using wolframalpha. You can see I didn't quite shift it to the right place: wolframalpha.com/input/?i=y%3D2%2F(x-2)%2B2. When I try plugging in larger constants, all that changes is the function gets more stretched out. I'm not sure why, hopefully I'm not doing something wrong.
$endgroup$
– Arash Howaida
Dec 17 '18 at 15:29
1
$begingroup$
The graph isn't being stretched out, Wolfram Alpha scales the viewing window automatically when you plot a function. The values for $c$ and $k$ depend on the size of your viewing window and where it's centered. If your viewing window is $x=0ldots u,y=0ldots v$, then you'll want to have $c=u$ and $k=v$.
$endgroup$
– R. Burton
Dec 17 '18 at 15:40
add a comment |
$begingroup$
thanks, that helps a bit. Would you mind sharing what values I should put for c and k to translate the hyperbola up to the correct position? I tried tinkering with it using wolframalpha. You can see I didn't quite shift it to the right place: wolframalpha.com/input/?i=y%3D2%2F(x-2)%2B2. When I try plugging in larger constants, all that changes is the function gets more stretched out. I'm not sure why, hopefully I'm not doing something wrong.
$endgroup$
– Arash Howaida
Dec 17 '18 at 15:29
1
$begingroup$
The graph isn't being stretched out, Wolfram Alpha scales the viewing window automatically when you plot a function. The values for $c$ and $k$ depend on the size of your viewing window and where it's centered. If your viewing window is $x=0ldots u,y=0ldots v$, then you'll want to have $c=u$ and $k=v$.
$endgroup$
– R. Burton
Dec 17 '18 at 15:40
$begingroup$
thanks, that helps a bit. Would you mind sharing what values I should put for c and k to translate the hyperbola up to the correct position? I tried tinkering with it using wolframalpha. You can see I didn't quite shift it to the right place: wolframalpha.com/input/?i=y%3D2%2F(x-2)%2B2. When I try plugging in larger constants, all that changes is the function gets more stretched out. I'm not sure why, hopefully I'm not doing something wrong.
$endgroup$
– Arash Howaida
Dec 17 '18 at 15:29
$begingroup$
thanks, that helps a bit. Would you mind sharing what values I should put for c and k to translate the hyperbola up to the correct position? I tried tinkering with it using wolframalpha. You can see I didn't quite shift it to the right place: wolframalpha.com/input/?i=y%3D2%2F(x-2)%2B2. When I try plugging in larger constants, all that changes is the function gets more stretched out. I'm not sure why, hopefully I'm not doing something wrong.
$endgroup$
– Arash Howaida
Dec 17 '18 at 15:29
1
1
$begingroup$
The graph isn't being stretched out, Wolfram Alpha scales the viewing window automatically when you plot a function. The values for $c$ and $k$ depend on the size of your viewing window and where it's centered. If your viewing window is $x=0ldots u,y=0ldots v$, then you'll want to have $c=u$ and $k=v$.
$endgroup$
– R. Burton
Dec 17 '18 at 15:40
$begingroup$
The graph isn't being stretched out, Wolfram Alpha scales the viewing window automatically when you plot a function. The values for $c$ and $k$ depend on the size of your viewing window and where it's centered. If your viewing window is $x=0ldots u,y=0ldots v$, then you'll want to have $c=u$ and $k=v$.
$endgroup$
– R. Burton
Dec 17 '18 at 15:40
add a comment |
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$begingroup$
y = k + 1/(x-c)
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:03
$begingroup$
@GNUSupporter8964民主女神地下教會 Ok, hmm, what's k? is k = x or is it a constant like c?
$endgroup$
– Arash Howaida
Dec 17 '18 at 13:04
$begingroup$
They are constants. You may try Desmos, which enables graph sharing.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:07
$begingroup$
@GNUSupporter8964民主女神地下教會 Just to double check, to flip my example equation
y=2/xto concave, I would input:y=2+1/(x-2)?$endgroup$
– Arash Howaida
Dec 17 '18 at 13:10
$begingroup$
You can check that on a graphical calculator like Desmos to be reassured of your observation. Actually, that would give the required shape, but I won't call that a "flipping" of 1/x. That's in fact a translation and clipping.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 13:15