Why when it comes to translating functions (x-b) negative? [closed]
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Quick question: when in the formula:(x-b), if it was (x-2) moves to the right but when it is (x-b)+c, you have (x-2)+3 then it moves up 3. I do not understand why it is a minus sign instead of a plus sign so instead be: (x+b)+c then (x+2)+3 would move to the right 2 and up 3
graphing-functions transformation
edited Dec 17 '18 at 20:37
Abdul Hadi Khan
518419
asked Dec 17 '18 at 20:11
user8980526user8980526
31
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closed as unclear what you're asking by mrtaurho, José Carlos Santos, Carl Christian, Cesareo, Brahadeesh Dec 18 '18 at 10:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Quick question: when in the formula:(x-b), if it was (x-2) moves to the right but when it is (x-b)+c, you have (x-2)+3 then it moves up 3. I do not understand why it is a minus sign instead of a plus sign so instead be: (x+b)+c then (x+2)+3 would move to the right 2 and up 3
graphing-functions transformation
edited Dec 17 '18 at 20:37
Abdul Hadi Khan
518419
asked Dec 17 '18 at 20:11
user8980526user8980526
31
$endgroup$
closed as unclear what you're asking by mrtaurho, José Carlos Santos, Carl Christian, Cesareo, Brahadeesh Dec 18 '18 at 10:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
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replacing $x$ with $x-3$ moves the graph right (direction of increasing $x$) by 3. Replacing $y$ with $y-3$ moves it up (direction of increasing $y$) by 3. So it works the same in both dimensions if you do the replacements the same way. It goes the opposite direction from what you might think because in the new equation, the $x$ that works must 3 bigger the it used to be to make the new formula with $x-3$ be like the old formula with $x$.
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– Ned
Dec 17 '18 at 20:19
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See math.stackexchange.com/a/3029278/265466.
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– amd
Dec 17 '18 at 21:29
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@Ned You should write that up as an answer.
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– amd
Dec 18 '18 at 1:16
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A duplicate of math.stackexchange.com/questions/2634587/….
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– amd
Dec 23 '18 at 0:50
add a comment |
$begingroup$
Quick question: when in the formula:(x-b), if it was (x-2) moves to the right but when it is (x-b)+c, you have (x-2)+3 then it moves up 3. I do not understand why it is a minus sign instead of a plus sign so instead be: (x+b)+c then (x+2)+3 would move to the right 2 and up 3
graphing-functions transformation
edited Dec 17 '18 at 20:37
Abdul Hadi Khan
518419
asked Dec 17 '18 at 20:11
user8980526user8980526
31
$endgroup$
Quick question: when in the formula:(x-b), if it was (x-2) moves to the right but when it is (x-b)+c, you have (x-2)+3 then it moves up 3. I do not understand why it is a minus sign instead of a plus sign so instead be: (x+b)+c then (x+2)+3 would move to the right 2 and up 3
graphing-functions transformation
graphing-functions transformation
edited Dec 17 '18 at 20:37
Abdul Hadi Khan
518419
asked Dec 17 '18 at 20:11
user8980526user8980526
31
edited Dec 17 '18 at 20:37
Abdul Hadi Khan
518419
asked Dec 17 '18 at 20:11
user8980526user8980526
31
edited Dec 17 '18 at 20:37
Abdul Hadi Khan
518419
edited Dec 17 '18 at 20:37
Abdul Hadi Khan
518419
edited Dec 17 '18 at 20:37
Abdul Hadi Khan
518419
518419
asked Dec 17 '18 at 20:11
user8980526user8980526
31
asked Dec 17 '18 at 20:11
user8980526user8980526
31
asked Dec 17 '18 at 20:11
user8980526user8980526
31
31
closed as unclear what you're asking by mrtaurho, José Carlos Santos, Carl Christian, Cesareo, Brahadeesh Dec 18 '18 at 10:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by mrtaurho, José Carlos Santos, Carl Christian, Cesareo, Brahadeesh Dec 18 '18 at 10:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
replacing $x$ with $x-3$ moves the graph right (direction of increasing $x$) by 3. Replacing $y$ with $y-3$ moves it up (direction of increasing $y$) by 3. So it works the same in both dimensions if you do the replacements the same way. It goes the opposite direction from what you might think because in the new equation, the $x$ that works must 3 bigger the it used to be to make the new formula with $x-3$ be like the old formula with $x$.
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– Ned
Dec 17 '18 at 20:19
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See math.stackexchange.com/a/3029278/265466.
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– amd
Dec 17 '18 at 21:29
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@Ned You should write that up as an answer.
$endgroup$
– amd
Dec 18 '18 at 1:16
$begingroup$
A duplicate of math.stackexchange.com/questions/2634587/….
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– amd
Dec 23 '18 at 0:50
add a comment |
2
$begingroup$
replacing $x$ with $x-3$ moves the graph right (direction of increasing $x$) by 3. Replacing $y$ with $y-3$ moves it up (direction of increasing $y$) by 3. So it works the same in both dimensions if you do the replacements the same way. It goes the opposite direction from what you might think because in the new equation, the $x$ that works must 3 bigger the it used to be to make the new formula with $x-3$ be like the old formula with $x$.
$endgroup$
– Ned
Dec 17 '18 at 20:19
$begingroup$
See math.stackexchange.com/a/3029278/265466.
$endgroup$
– amd
Dec 17 '18 at 21:29
$begingroup$
@Ned You should write that up as an answer.
$endgroup$
– amd
Dec 18 '18 at 1:16
$begingroup$
A duplicate of math.stackexchange.com/questions/2634587/….
$endgroup$
– amd
Dec 23 '18 at 0:50
2
2
$begingroup$
replacing $x$ with $x-3$ moves the graph right (direction of increasing $x$) by 3. Replacing $y$ with $y-3$ moves it up (direction of increasing $y$) by 3. So it works the same in both dimensions if you do the replacements the same way. It goes the opposite direction from what you might think because in the new equation, the $x$ that works must 3 bigger the it used to be to make the new formula with $x-3$ be like the old formula with $x$.
$endgroup$
– Ned
Dec 17 '18 at 20:19
$begingroup$
replacing $x$ with $x-3$ moves the graph right (direction of increasing $x$) by 3. Replacing $y$ with $y-3$ moves it up (direction of increasing $y$) by 3. So it works the same in both dimensions if you do the replacements the same way. It goes the opposite direction from what you might think because in the new equation, the $x$ that works must 3 bigger the it used to be to make the new formula with $x-3$ be like the old formula with $x$.
$endgroup$
– Ned
Dec 17 '18 at 20:19
$begingroup$
See math.stackexchange.com/a/3029278/265466.
$endgroup$
– amd
Dec 17 '18 at 21:29
$begingroup$
See math.stackexchange.com/a/3029278/265466.
$endgroup$
– amd
Dec 17 '18 at 21:29
$begingroup$
@Ned You should write that up as an answer.
$endgroup$
– amd
Dec 18 '18 at 1:16
$begingroup$
@Ned You should write that up as an answer.
$endgroup$
– amd
Dec 18 '18 at 1:16
$begingroup$
A duplicate of math.stackexchange.com/questions/2634587/….
$endgroup$
– amd
Dec 23 '18 at 0:50
$begingroup$
A duplicate of math.stackexchange.com/questions/2634587/….
$endgroup$
– amd
Dec 23 '18 at 0:50
add a comment |
1 Answer
1
active
oldest
votes
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Comment made into an answer:
Replacing $x$ with $x-3$ moves the graph $3$ to the right (direction of increasing $x$).
Replacing $y$ with $y-3$ moves the graph $3$ up (direction of increasing $y$). So it works the same in both dimensions if you do the replacements the same way.
The movement is in the opposite direction to what you might think because when you replace $x$ by $x-3$, the value of $x$ that works in the new equation with $x-3$ must be $3$ $bigger$ than it used to be to make the new formula with $x-3$ come out the same as the old formula with $x$.
answered Dec 18 '18 at 3:54
NedNed
2,048910
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Comment made into an answer:
Replacing $x$ with $x-3$ moves the graph $3$ to the right (direction of increasing $x$).
Replacing $y$ with $y-3$ moves the graph $3$ up (direction of increasing $y$). So it works the same in both dimensions if you do the replacements the same way.
The movement is in the opposite direction to what you might think because when you replace $x$ by $x-3$, the value of $x$ that works in the new equation with $x-3$ must be $3$ $bigger$ than it used to be to make the new formula with $x-3$ come out the same as the old formula with $x$.
answered Dec 18 '18 at 3:54
NedNed
2,048910
$endgroup$
add a comment |
$begingroup$
Comment made into an answer:
Replacing $x$ with $x-3$ moves the graph $3$ to the right (direction of increasing $x$).
Replacing $y$ with $y-3$ moves the graph $3$ up (direction of increasing $y$). So it works the same in both dimensions if you do the replacements the same way.
The movement is in the opposite direction to what you might think because when you replace $x$ by $x-3$, the value of $x$ that works in the new equation with $x-3$ must be $3$ $bigger$ than it used to be to make the new formula with $x-3$ come out the same as the old formula with $x$.
answered Dec 18 '18 at 3:54
NedNed
2,048910
$endgroup$
add a comment |
$begingroup$
Comment made into an answer:
Replacing $x$ with $x-3$ moves the graph $3$ to the right (direction of increasing $x$).
Replacing $y$ with $y-3$ moves the graph $3$ up (direction of increasing $y$). So it works the same in both dimensions if you do the replacements the same way.
The movement is in the opposite direction to what you might think because when you replace $x$ by $x-3$, the value of $x$ that works in the new equation with $x-3$ must be $3$ $bigger$ than it used to be to make the new formula with $x-3$ come out the same as the old formula with $x$.
answered Dec 18 '18 at 3:54
NedNed
2,048910
$endgroup$
Comment made into an answer:
Replacing $x$ with $x-3$ moves the graph $3$ to the right (direction of increasing $x$).
Replacing $y$ with $y-3$ moves the graph $3$ up (direction of increasing $y$). So it works the same in both dimensions if you do the replacements the same way.
The movement is in the opposite direction to what you might think because when you replace $x$ by $x-3$, the value of $x$ that works in the new equation with $x-3$ must be $3$ $bigger$ than it used to be to make the new formula with $x-3$ come out the same as the old formula with $x$.
answered Dec 18 '18 at 3:54
NedNed
2,048910
answered Dec 18 '18 at 3:54
NedNed
2,048910
answered Dec 18 '18 at 3:54
NedNed
2,048910
answered Dec 18 '18 at 3:54
NedNed
2,048910
2,048910
add a comment |
add a comment |
2
$begingroup$
replacing $x$ with $x-3$ moves the graph right (direction of increasing $x$) by 3. Replacing $y$ with $y-3$ moves it up (direction of increasing $y$) by 3. So it works the same in both dimensions if you do the replacements the same way. It goes the opposite direction from what you might think because in the new equation, the $x$ that works must 3 bigger the it used to be to make the new formula with $x-3$ be like the old formula with $x$.
$endgroup$
– Ned
Dec 17 '18 at 20:19
$begingroup$
See math.stackexchange.com/a/3029278/265466.
$endgroup$
– amd
Dec 17 '18 at 21:29
$begingroup$
@Ned You should write that up as an answer.
$endgroup$
– amd
Dec 18 '18 at 1:16
$begingroup$
A duplicate of math.stackexchange.com/questions/2634587/….
$endgroup$
– amd
Dec 23 '18 at 0:50