What is the main difference between shift and a stretch in Transformation of functions?
$begingroup$
Recall The shift in a nutshell
$$y = f(x) + c $$
and stretch
$$y = f(x) * c $$
We actually not do any nothing except we duplicate the + many of times in stretching!
Is there is anything special about stretching the function ?
algebra-precalculus functions graphing-functions
asked Jan 17 at 19:07
Ammar BamhdiAmmar Bamhdi
325
$endgroup$
add a comment |
$begingroup$
Recall The shift in a nutshell
$$y = f(x) + c $$
and stretch
$$y = f(x) * c $$
We actually not do any nothing except we duplicate the + many of times in stretching!
Is there is anything special about stretching the function ?
algebra-precalculus functions graphing-functions
asked Jan 17 at 19:07
Ammar BamhdiAmmar Bamhdi
325
$endgroup$
add a comment |
$begingroup$
Recall The shift in a nutshell
$$y = f(x) + c $$
and stretch
$$y = f(x) * c $$
We actually not do any nothing except we duplicate the + many of times in stretching!
Is there is anything special about stretching the function ?
algebra-precalculus functions graphing-functions
asked Jan 17 at 19:07
Ammar BamhdiAmmar Bamhdi
325
$endgroup$
Recall The shift in a nutshell
$$y = f(x) + c $$
and stretch
$$y = f(x) * c $$
We actually not do any nothing except we duplicate the + many of times in stretching!
Is there is anything special about stretching the function ?
algebra-precalculus functions graphing-functions
algebra-precalculus functions graphing-functions
asked Jan 17 at 19:07
Ammar BamhdiAmmar Bamhdi
325
asked Jan 17 at 19:07
Ammar BamhdiAmmar Bamhdi
325
asked Jan 17 at 19:07
Ammar BamhdiAmmar Bamhdi
325
asked Jan 17 at 19:07
Ammar BamhdiAmmar Bamhdi
325
asked Jan 17 at 19:07
Ammar BamhdiAmmar Bamhdi
325
325
add a comment |
add a comment |
2 Answers
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$begingroup$
well when you stretch a graph you're moving every point by a different amount, whereas shifting a graph you're moving every point by the same amount.
if you were to take a standard parabola and stretch it by a factor of $2$ the vertex would stay at $(0,0)$ but $x=2$ which previously would have been at $y=4$ is now at $y=8$, $x=3$ is at $y=18$ instead of $y=9$ and so on. if you were to take that same parabola and shift it up by $1$ then every point would move by the same amount. $x=0$ would be at $y=1$, $x=2$ would be at $y=5$ and so on.
answered Jan 17 at 19:17
WillWill
465
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Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
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– dantopa
Jan 17 at 19:24
add a comment |
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While you are correct in saying that multiplication is repeated addition, you should also think about what is being added in each case. Let's consider the case where $c = 3$. Then in the shift, $$f(x) + 3,$$ we are taking every point $(x,y)$ and moving it to the point $(x,y+3)$. In particular this means that the shape of the graph is the same, it's just move up by three units. If we consider the stretch $$f(x)cdot 3$$ we get something different. Viewing this as repeated addition we can write $$f(x)cdot 3 = f(x) + f(x) + f(x).$$ This means that we take every point $(x,y)$ and move it to the point $(x,y + y + y)$ or $(x,3y)$. Now, depending on what $y$ is for a particular $x$ this new point could any distance away from the original point. If $y = 1$ then the point would move up to $y = 3$, whereas if the point was originally at $y = 3$ then it would move up to $y = 9$. If $y$ were negative the point would move down. One key distinction here is that with a stretch the shape of the graph will change, whereas with a shift only the location of the graph will change.
answered Jan 17 at 19:43
DMcMorDMcMor
2,74521328
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2 Answers
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$begingroup$
well when you stretch a graph you're moving every point by a different amount, whereas shifting a graph you're moving every point by the same amount.
if you were to take a standard parabola and stretch it by a factor of $2$ the vertex would stay at $(0,0)$ but $x=2$ which previously would have been at $y=4$ is now at $y=8$, $x=3$ is at $y=18$ instead of $y=9$ and so on. if you were to take that same parabola and shift it up by $1$ then every point would move by the same amount. $x=0$ would be at $y=1$, $x=2$ would be at $y=5$ and so on.
answered Jan 17 at 19:17
WillWill
465
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 17 at 19:24
add a comment |
$begingroup$
well when you stretch a graph you're moving every point by a different amount, whereas shifting a graph you're moving every point by the same amount.
if you were to take a standard parabola and stretch it by a factor of $2$ the vertex would stay at $(0,0)$ but $x=2$ which previously would have been at $y=4$ is now at $y=8$, $x=3$ is at $y=18$ instead of $y=9$ and so on. if you were to take that same parabola and shift it up by $1$ then every point would move by the same amount. $x=0$ would be at $y=1$, $x=2$ would be at $y=5$ and so on.
answered Jan 17 at 19:17
WillWill
465
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 17 at 19:24
add a comment |
$begingroup$
well when you stretch a graph you're moving every point by a different amount, whereas shifting a graph you're moving every point by the same amount.
if you were to take a standard parabola and stretch it by a factor of $2$ the vertex would stay at $(0,0)$ but $x=2$ which previously would have been at $y=4$ is now at $y=8$, $x=3$ is at $y=18$ instead of $y=9$ and so on. if you were to take that same parabola and shift it up by $1$ then every point would move by the same amount. $x=0$ would be at $y=1$, $x=2$ would be at $y=5$ and so on.
answered Jan 17 at 19:17
WillWill
465
$endgroup$
well when you stretch a graph you're moving every point by a different amount, whereas shifting a graph you're moving every point by the same amount.
if you were to take a standard parabola and stretch it by a factor of $2$ the vertex would stay at $(0,0)$ but $x=2$ which previously would have been at $y=4$ is now at $y=8$, $x=3$ is at $y=18$ instead of $y=9$ and so on. if you were to take that same parabola and shift it up by $1$ then every point would move by the same amount. $x=0$ would be at $y=1$, $x=2$ would be at $y=5$ and so on.
answered Jan 17 at 19:17
WillWill
465
edited Jan 17 at 19:33
answered Jan 17 at 19:17
WillWill
465
answered Jan 17 at 19:17
WillWill
465
answered Jan 17 at 19:17
WillWill
465
465
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
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– dantopa
Jan 17 at 19:24
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 17 at 19:24
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 17 at 19:24
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 17 at 19:24
add a comment |
$begingroup$
While you are correct in saying that multiplication is repeated addition, you should also think about what is being added in each case. Let's consider the case where $c = 3$. Then in the shift, $$f(x) + 3,$$ we are taking every point $(x,y)$ and moving it to the point $(x,y+3)$. In particular this means that the shape of the graph is the same, it's just move up by three units. If we consider the stretch $$f(x)cdot 3$$ we get something different. Viewing this as repeated addition we can write $$f(x)cdot 3 = f(x) + f(x) + f(x).$$ This means that we take every point $(x,y)$ and move it to the point $(x,y + y + y)$ or $(x,3y)$. Now, depending on what $y$ is for a particular $x$ this new point could any distance away from the original point. If $y = 1$ then the point would move up to $y = 3$, whereas if the point was originally at $y = 3$ then it would move up to $y = 9$. If $y$ were negative the point would move down. One key distinction here is that with a stretch the shape of the graph will change, whereas with a shift only the location of the graph will change.
answered Jan 17 at 19:43
DMcMorDMcMor
2,74521328
$endgroup$
add a comment |
$begingroup$
While you are correct in saying that multiplication is repeated addition, you should also think about what is being added in each case. Let's consider the case where $c = 3$. Then in the shift, $$f(x) + 3,$$ we are taking every point $(x,y)$ and moving it to the point $(x,y+3)$. In particular this means that the shape of the graph is the same, it's just move up by three units. If we consider the stretch $$f(x)cdot 3$$ we get something different. Viewing this as repeated addition we can write $$f(x)cdot 3 = f(x) + f(x) + f(x).$$ This means that we take every point $(x,y)$ and move it to the point $(x,y + y + y)$ or $(x,3y)$. Now, depending on what $y$ is for a particular $x$ this new point could any distance away from the original point. If $y = 1$ then the point would move up to $y = 3$, whereas if the point was originally at $y = 3$ then it would move up to $y = 9$. If $y$ were negative the point would move down. One key distinction here is that with a stretch the shape of the graph will change, whereas with a shift only the location of the graph will change.
answered Jan 17 at 19:43
DMcMorDMcMor
2,74521328
$endgroup$
add a comment |
$begingroup$
While you are correct in saying that multiplication is repeated addition, you should also think about what is being added in each case. Let's consider the case where $c = 3$. Then in the shift, $$f(x) + 3,$$ we are taking every point $(x,y)$ and moving it to the point $(x,y+3)$. In particular this means that the shape of the graph is the same, it's just move up by three units. If we consider the stretch $$f(x)cdot 3$$ we get something different. Viewing this as repeated addition we can write $$f(x)cdot 3 = f(x) + f(x) + f(x).$$ This means that we take every point $(x,y)$ and move it to the point $(x,y + y + y)$ or $(x,3y)$. Now, depending on what $y$ is for a particular $x$ this new point could any distance away from the original point. If $y = 1$ then the point would move up to $y = 3$, whereas if the point was originally at $y = 3$ then it would move up to $y = 9$. If $y$ were negative the point would move down. One key distinction here is that with a stretch the shape of the graph will change, whereas with a shift only the location of the graph will change.
answered Jan 17 at 19:43
DMcMorDMcMor
2,74521328
$endgroup$
While you are correct in saying that multiplication is repeated addition, you should also think about what is being added in each case. Let's consider the case where $c = 3$. Then in the shift, $$f(x) + 3,$$ we are taking every point $(x,y)$ and moving it to the point $(x,y+3)$. In particular this means that the shape of the graph is the same, it's just move up by three units. If we consider the stretch $$f(x)cdot 3$$ we get something different. Viewing this as repeated addition we can write $$f(x)cdot 3 = f(x) + f(x) + f(x).$$ This means that we take every point $(x,y)$ and move it to the point $(x,y + y + y)$ or $(x,3y)$. Now, depending on what $y$ is for a particular $x$ this new point could any distance away from the original point. If $y = 1$ then the point would move up to $y = 3$, whereas if the point was originally at $y = 3$ then it would move up to $y = 9$. If $y$ were negative the point would move down. One key distinction here is that with a stretch the shape of the graph will change, whereas with a shift only the location of the graph will change.
answered Jan 17 at 19:43
DMcMorDMcMor
2,74521328
answered Jan 17 at 19:43
DMcMorDMcMor
2,74521328
answered Jan 17 at 19:43
DMcMorDMcMor
2,74521328
answered Jan 17 at 19:43
DMcMorDMcMor
2,74521328
2,74521328
add a comment |
add a comment |
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