Graph of the function in the form: sqrt(-x+a) through manipulation of the function sqrt(x)?
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In class we learned that certain simple functions $(x, x^2, sqrt x)$ etc. can be manipulated to easily find the graphs of more complicated versions of these original functions...
$f(x+1) Longrightarrow$ shift to the left/ $f(x-1)Longrightarrow$ shift to right, etc.
However, I'm having trouble applying that to the function:
$f(x) = sqrt{(-x+a)}.$
So far I understand this much:
$sqrt x Longrightarrow sqrt{(-x)}$ results in reflection across $y-$axis.
$sqrt x Longrightarrow sqrt{(xpm a)}$ results in shift of original graph a units to the left/right, respectively.
But when you apply the principles to $sqrt{(-x+a)}$ or $sqrt{(-x-a)},$ it doesn't respond appropriately. The graph I get for $sqrt{(-x+a)}$ is the graph of $sqrt{(-x)}$ shifted a units to the right , not the left.
And the opposite goes for $sqrt{(-x-a)}.$
Why is this? Can normal manipulation of the graph not be applied to functions of this form?
functions graphing-functions
edited Nov 15 at 9:37
user376343
2,4581718
asked Oct 31 '15 at 23:46
RiddleMeThis
1
add a comment |
up vote
0
down vote
favorite
In class we learned that certain simple functions $(x, x^2, sqrt x)$ etc. can be manipulated to easily find the graphs of more complicated versions of these original functions...
$f(x+1) Longrightarrow$ shift to the left/ $f(x-1)Longrightarrow$ shift to right, etc.
However, I'm having trouble applying that to the function:
$f(x) = sqrt{(-x+a)}.$
So far I understand this much:
$sqrt x Longrightarrow sqrt{(-x)}$ results in reflection across $y-$axis.
$sqrt x Longrightarrow sqrt{(xpm a)}$ results in shift of original graph a units to the left/right, respectively.
But when you apply the principles to $sqrt{(-x+a)}$ or $sqrt{(-x-a)},$ it doesn't respond appropriately. The graph I get for $sqrt{(-x+a)}$ is the graph of $sqrt{(-x)}$ shifted a units to the right , not the left.
And the opposite goes for $sqrt{(-x-a)}.$
Why is this? Can normal manipulation of the graph not be applied to functions of this form?
functions graphing-functions
edited Nov 15 at 9:37
user376343
2,4581718
asked Oct 31 '15 at 23:46
RiddleMeThis
1
3
$f(-x+a)=f(-(x-a))=g(x-a)$ where $g(x)=f(-x)$. So, one has to shift $g$ to the right and $g$ is the reflection of $f$ over $y$. The correct order: (1) reflecting $f$ (2) then shifting the reflection to the right.
– zoli
Oct 31 '15 at 23:53
Ok, so reflect f across y, then shift the reflection to the right. So this means that f(x) = sqrt(x-a)?
– RiddleMeThis
Nov 1 '15 at 4:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In class we learned that certain simple functions $(x, x^2, sqrt x)$ etc. can be manipulated to easily find the graphs of more complicated versions of these original functions...
$f(x+1) Longrightarrow$ shift to the left/ $f(x-1)Longrightarrow$ shift to right, etc.
However, I'm having trouble applying that to the function:
$f(x) = sqrt{(-x+a)}.$
So far I understand this much:
$sqrt x Longrightarrow sqrt{(-x)}$ results in reflection across $y-$axis.
$sqrt x Longrightarrow sqrt{(xpm a)}$ results in shift of original graph a units to the left/right, respectively.
But when you apply the principles to $sqrt{(-x+a)}$ or $sqrt{(-x-a)},$ it doesn't respond appropriately. The graph I get for $sqrt{(-x+a)}$ is the graph of $sqrt{(-x)}$ shifted a units to the right , not the left.
And the opposite goes for $sqrt{(-x-a)}.$
Why is this? Can normal manipulation of the graph not be applied to functions of this form?
functions graphing-functions
edited Nov 15 at 9:37
user376343
2,4581718
asked Oct 31 '15 at 23:46
RiddleMeThis
1
In class we learned that certain simple functions $(x, x^2, sqrt x)$ etc. can be manipulated to easily find the graphs of more complicated versions of these original functions...
$f(x+1) Longrightarrow$ shift to the left/ $f(x-1)Longrightarrow$ shift to right, etc.
However, I'm having trouble applying that to the function:
$f(x) = sqrt{(-x+a)}.$
So far I understand this much:
$sqrt x Longrightarrow sqrt{(-x)}$ results in reflection across $y-$axis.
$sqrt x Longrightarrow sqrt{(xpm a)}$ results in shift of original graph a units to the left/right, respectively.
But when you apply the principles to $sqrt{(-x+a)}$ or $sqrt{(-x-a)},$ it doesn't respond appropriately. The graph I get for $sqrt{(-x+a)}$ is the graph of $sqrt{(-x)}$ shifted a units to the right , not the left.
And the opposite goes for $sqrt{(-x-a)}.$
Why is this? Can normal manipulation of the graph not be applied to functions of this form?
functions graphing-functions
functions graphing-functions
edited Nov 15 at 9:37
user376343
2,4581718
asked Oct 31 '15 at 23:46
RiddleMeThis
1
edited Nov 15 at 9:37
user376343
2,4581718
asked Oct 31 '15 at 23:46
RiddleMeThis
1
edited Nov 15 at 9:37
user376343
2,4581718
edited Nov 15 at 9:37
user376343
2,4581718
edited Nov 15 at 9:37
user376343
2,4581718
2,4581718
asked Oct 31 '15 at 23:46
RiddleMeThis
1
asked Oct 31 '15 at 23:46
RiddleMeThis
1
asked Oct 31 '15 at 23:46
RiddleMeThis
1
1
3
$f(-x+a)=f(-(x-a))=g(x-a)$ where $g(x)=f(-x)$. So, one has to shift $g$ to the right and $g$ is the reflection of $f$ over $y$. The correct order: (1) reflecting $f$ (2) then shifting the reflection to the right.
– zoli
Oct 31 '15 at 23:53
Ok, so reflect f across y, then shift the reflection to the right. So this means that f(x) = sqrt(x-a)?
– RiddleMeThis
Nov 1 '15 at 4:45
add a comment |
3
$f(-x+a)=f(-(x-a))=g(x-a)$ where $g(x)=f(-x)$. So, one has to shift $g$ to the right and $g$ is the reflection of $f$ over $y$. The correct order: (1) reflecting $f$ (2) then shifting the reflection to the right.
– zoli
Oct 31 '15 at 23:53
Ok, so reflect f across y, then shift the reflection to the right. So this means that f(x) = sqrt(x-a)?
– RiddleMeThis
Nov 1 '15 at 4:45
3
3
$f(-x+a)=f(-(x-a))=g(x-a)$ where $g(x)=f(-x)$. So, one has to shift $g$ to the right and $g$ is the reflection of $f$ over $y$. The correct order: (1) reflecting $f$ (2) then shifting the reflection to the right.
– zoli
Oct 31 '15 at 23:53
$f(-x+a)=f(-(x-a))=g(x-a)$ where $g(x)=f(-x)$. So, one has to shift $g$ to the right and $g$ is the reflection of $f$ over $y$. The correct order: (1) reflecting $f$ (2) then shifting the reflection to the right.
– zoli
Oct 31 '15 at 23:53
Ok, so reflect f across y, then shift the reflection to the right. So this means that f(x) = sqrt(x-a)?
– RiddleMeThis
Nov 1 '15 at 4:45
Ok, so reflect f across y, then shift the reflection to the right. So this means that f(x) = sqrt(x-a)?
– RiddleMeThis
Nov 1 '15 at 4:45
add a comment |
1 Answer
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I would like to illustrate my comments in the following figure. I hope that this will explain everything.
- the dark blue line is the graph of $color{blue}{sqrt{(x)}}$ -- note that $sqrt{(x)}$ is not defined on $(-infty,0)$.
- the purple line is the graph of $color{purple}{g(x)=sqrt{(-x)}}$ -- note that this function is not defined on $(0,infty)$.
- The green line is the graph of $color{green}{g(x-1)=sqrt{-(x-1)}=sqrt{-x+1}}$ -- note that this function is not defined if $-(x-1)<0$ or if $x>1$, that is over the interval $(1,infty)$.
answered Nov 1 '15 at 8:11
zoli
16.4k41643
ok. The graphs of sqrt(x) and sqrt(-x) make sense to me because they are each reflections of each other across the y-axis. I also understand how the graph of sqrt(x+1) would look like sqrt(x) shifted 1 unit to the left on x-axis. Then you negate every x in that function, by doing f(-x) and this becomes a reflection across y-axis to yield what you drew with the green line. The thing I was doing wrong was that you can't replace x with -x+1. You can only replace x in sqrt(x+1) with -x. So sqrt(-x+1) is a reflection of sqrt(x+1). Is this correct?
– RiddleMeThis
Nov 1 '15 at 16:39
@RiddleMeThis: The rule (my rule) is that if you have a function then you have only three legal transformations $f(x+a)$, $f(x-a)$, and $f(-x)$. So, $f(-xpm a)$ is not legal. You can make it legal via the following operation $f(-xpm a)=f(-(xmp a))$ but then you've created another function: $g(x)=f(-x)$.
– zoli
Nov 1 '15 at 21:14
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I would like to illustrate my comments in the following figure. I hope that this will explain everything.
- the dark blue line is the graph of $color{blue}{sqrt{(x)}}$ -- note that $sqrt{(x)}$ is not defined on $(-infty,0)$.
- the purple line is the graph of $color{purple}{g(x)=sqrt{(-x)}}$ -- note that this function is not defined on $(0,infty)$.
- The green line is the graph of $color{green}{g(x-1)=sqrt{-(x-1)}=sqrt{-x+1}}$ -- note that this function is not defined if $-(x-1)<0$ or if $x>1$, that is over the interval $(1,infty)$.
answered Nov 1 '15 at 8:11
zoli
16.4k41643
ok. The graphs of sqrt(x) and sqrt(-x) make sense to me because they are each reflections of each other across the y-axis. I also understand how the graph of sqrt(x+1) would look like sqrt(x) shifted 1 unit to the left on x-axis. Then you negate every x in that function, by doing f(-x) and this becomes a reflection across y-axis to yield what you drew with the green line. The thing I was doing wrong was that you can't replace x with -x+1. You can only replace x in sqrt(x+1) with -x. So sqrt(-x+1) is a reflection of sqrt(x+1). Is this correct?
– RiddleMeThis
Nov 1 '15 at 16:39
@RiddleMeThis: The rule (my rule) is that if you have a function then you have only three legal transformations $f(x+a)$, $f(x-a)$, and $f(-x)$. So, $f(-xpm a)$ is not legal. You can make it legal via the following operation $f(-xpm a)=f(-(xmp a))$ but then you've created another function: $g(x)=f(-x)$.
– zoli
Nov 1 '15 at 21:14
add a comment |
up vote
1
down vote
I would like to illustrate my comments in the following figure. I hope that this will explain everything.
- the dark blue line is the graph of $color{blue}{sqrt{(x)}}$ -- note that $sqrt{(x)}$ is not defined on $(-infty,0)$.
- the purple line is the graph of $color{purple}{g(x)=sqrt{(-x)}}$ -- note that this function is not defined on $(0,infty)$.
- The green line is the graph of $color{green}{g(x-1)=sqrt{-(x-1)}=sqrt{-x+1}}$ -- note that this function is not defined if $-(x-1)<0$ or if $x>1$, that is over the interval $(1,infty)$.
answered Nov 1 '15 at 8:11
zoli
16.4k41643
ok. The graphs of sqrt(x) and sqrt(-x) make sense to me because they are each reflections of each other across the y-axis. I also understand how the graph of sqrt(x+1) would look like sqrt(x) shifted 1 unit to the left on x-axis. Then you negate every x in that function, by doing f(-x) and this becomes a reflection across y-axis to yield what you drew with the green line. The thing I was doing wrong was that you can't replace x with -x+1. You can only replace x in sqrt(x+1) with -x. So sqrt(-x+1) is a reflection of sqrt(x+1). Is this correct?
– RiddleMeThis
Nov 1 '15 at 16:39
@RiddleMeThis: The rule (my rule) is that if you have a function then you have only three legal transformations $f(x+a)$, $f(x-a)$, and $f(-x)$. So, $f(-xpm a)$ is not legal. You can make it legal via the following operation $f(-xpm a)=f(-(xmp a))$ but then you've created another function: $g(x)=f(-x)$.
– zoli
Nov 1 '15 at 21:14
add a comment |
up vote
1
down vote
up vote
1
down vote
I would like to illustrate my comments in the following figure. I hope that this will explain everything.
- the dark blue line is the graph of $color{blue}{sqrt{(x)}}$ -- note that $sqrt{(x)}$ is not defined on $(-infty,0)$.
- the purple line is the graph of $color{purple}{g(x)=sqrt{(-x)}}$ -- note that this function is not defined on $(0,infty)$.
- The green line is the graph of $color{green}{g(x-1)=sqrt{-(x-1)}=sqrt{-x+1}}$ -- note that this function is not defined if $-(x-1)<0$ or if $x>1$, that is over the interval $(1,infty)$.
answered Nov 1 '15 at 8:11
zoli
16.4k41643
I would like to illustrate my comments in the following figure. I hope that this will explain everything.
- the dark blue line is the graph of $color{blue}{sqrt{(x)}}$ -- note that $sqrt{(x)}$ is not defined on $(-infty,0)$.
- the purple line is the graph of $color{purple}{g(x)=sqrt{(-x)}}$ -- note that this function is not defined on $(0,infty)$.
- The green line is the graph of $color{green}{g(x-1)=sqrt{-(x-1)}=sqrt{-x+1}}$ -- note that this function is not defined if $-(x-1)<0$ or if $x>1$, that is over the interval $(1,infty)$.
answered Nov 1 '15 at 8:11
zoli
16.4k41643
edited Nov 1 '15 at 10:56
answered Nov 1 '15 at 8:11
zoli
16.4k41643
answered Nov 1 '15 at 8:11
zoli
16.4k41643
answered Nov 1 '15 at 8:11
zoli
16.4k41643
16.4k41643
ok. The graphs of sqrt(x) and sqrt(-x) make sense to me because they are each reflections of each other across the y-axis. I also understand how the graph of sqrt(x+1) would look like sqrt(x) shifted 1 unit to the left on x-axis. Then you negate every x in that function, by doing f(-x) and this becomes a reflection across y-axis to yield what you drew with the green line. The thing I was doing wrong was that you can't replace x with -x+1. You can only replace x in sqrt(x+1) with -x. So sqrt(-x+1) is a reflection of sqrt(x+1). Is this correct?
– RiddleMeThis
Nov 1 '15 at 16:39
@RiddleMeThis: The rule (my rule) is that if you have a function then you have only three legal transformations $f(x+a)$, $f(x-a)$, and $f(-x)$. So, $f(-xpm a)$ is not legal. You can make it legal via the following operation $f(-xpm a)=f(-(xmp a))$ but then you've created another function: $g(x)=f(-x)$.
– zoli
Nov 1 '15 at 21:14
add a comment |
ok. The graphs of sqrt(x) and sqrt(-x) make sense to me because they are each reflections of each other across the y-axis. I also understand how the graph of sqrt(x+1) would look like sqrt(x) shifted 1 unit to the left on x-axis. Then you negate every x in that function, by doing f(-x) and this becomes a reflection across y-axis to yield what you drew with the green line. The thing I was doing wrong was that you can't replace x with -x+1. You can only replace x in sqrt(x+1) with -x. So sqrt(-x+1) is a reflection of sqrt(x+1). Is this correct?
– RiddleMeThis
Nov 1 '15 at 16:39
@RiddleMeThis: The rule (my rule) is that if you have a function then you have only three legal transformations $f(x+a)$, $f(x-a)$, and $f(-x)$. So, $f(-xpm a)$ is not legal. You can make it legal via the following operation $f(-xpm a)=f(-(xmp a))$ but then you've created another function: $g(x)=f(-x)$.
– zoli
Nov 1 '15 at 21:14
ok. The graphs of sqrt(x) and sqrt(-x) make sense to me because they are each reflections of each other across the y-axis. I also understand how the graph of sqrt(x+1) would look like sqrt(x) shifted 1 unit to the left on x-axis. Then you negate every x in that function, by doing f(-x) and this becomes a reflection across y-axis to yield what you drew with the green line. The thing I was doing wrong was that you can't replace x with -x+1. You can only replace x in sqrt(x+1) with -x. So sqrt(-x+1) is a reflection of sqrt(x+1). Is this correct?
– RiddleMeThis
Nov 1 '15 at 16:39
ok. The graphs of sqrt(x) and sqrt(-x) make sense to me because they are each reflections of each other across the y-axis. I also understand how the graph of sqrt(x+1) would look like sqrt(x) shifted 1 unit to the left on x-axis. Then you negate every x in that function, by doing f(-x) and this becomes a reflection across y-axis to yield what you drew with the green line. The thing I was doing wrong was that you can't replace x with -x+1. You can only replace x in sqrt(x+1) with -x. So sqrt(-x+1) is a reflection of sqrt(x+1). Is this correct?
– RiddleMeThis
Nov 1 '15 at 16:39
@RiddleMeThis: The rule (my rule) is that if you have a function then you have only three legal transformations $f(x+a)$, $f(x-a)$, and $f(-x)$. So, $f(-xpm a)$ is not legal. You can make it legal via the following operation $f(-xpm a)=f(-(xmp a))$ but then you've created another function: $g(x)=f(-x)$.
– zoli
Nov 1 '15 at 21:14
@RiddleMeThis: The rule (my rule) is that if you have a function then you have only three legal transformations $f(x+a)$, $f(x-a)$, and $f(-x)$. So, $f(-xpm a)$ is not legal. You can make it legal via the following operation $f(-xpm a)=f(-(xmp a))$ but then you've created another function: $g(x)=f(-x)$.
– zoli
Nov 1 '15 at 21:14
add a comment |
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$f(-x+a)=f(-(x-a))=g(x-a)$ where $g(x)=f(-x)$. So, one has to shift $g$ to the right and $g$ is the reflection of $f$ over $y$. The correct order: (1) reflecting $f$ (2) then shifting the reflection to the right.
– zoli
Oct 31 '15 at 23:53
Ok, so reflect f across y, then shift the reflection to the right. So this means that f(x) = sqrt(x-a)?
– RiddleMeThis
Nov 1 '15 at 4:45