How do you tell the difference between a left shift and a right shift of a periodic function?
$begingroup$
Why does the horizontal phase shift of:
$
2sin(-3x-fracpi2)
$
shift to the right instead of to the left?
I am expecting this for
$
2sin(-3x-fracpi2)
$
:
I'm instead getting this for
$
2sin(-3x-fracpi2)
$
when using a graphing utility:
I get the first graph by turning
$
2sin(-3x-fracpi2)
$
into
$
2sin(-3x+fracpi2)
$
However, because of the -3x, wouldn't the resulted sign be flipped from it's original sign, if I remove the negative in front of the -3x, I still get the 2nd graph. Why is this?
algebra-precalculus
asked Dec 6 '18 at 21:26
GKEGKE
1034
$endgroup$
add a comment |
$begingroup$
Why does the horizontal phase shift of:
$
2sin(-3x-fracpi2)
$
shift to the right instead of to the left?
I am expecting this for
$
2sin(-3x-fracpi2)
$
:
I'm instead getting this for
$
2sin(-3x-fracpi2)
$
when using a graphing utility:
I get the first graph by turning
$
2sin(-3x-fracpi2)
$
into
$
2sin(-3x+fracpi2)
$
However, because of the -3x, wouldn't the resulted sign be flipped from it's original sign, if I remove the negative in front of the -3x, I still get the 2nd graph. Why is this?
algebra-precalculus
asked Dec 6 '18 at 21:26
GKEGKE
1034
$endgroup$
$begingroup$
I did not get the second graph when I graphed $f(x)=2sin (-3x+frac{pi}{2})$. Can you also add equations for the graphs in your photo?
$endgroup$
– William Sun
Dec 6 '18 at 22:39
$begingroup$
Fixed it, what I meant was that I got the first graph from graphing $2sin(-3x+fracpi2)$
$endgroup$
– GKE
Dec 6 '18 at 22:54
add a comment |
$begingroup$
Why does the horizontal phase shift of:
$
2sin(-3x-fracpi2)
$
shift to the right instead of to the left?
I am expecting this for
$
2sin(-3x-fracpi2)
$
:
I'm instead getting this for
$
2sin(-3x-fracpi2)
$
when using a graphing utility:
I get the first graph by turning
$
2sin(-3x-fracpi2)
$
into
$
2sin(-3x+fracpi2)
$
However, because of the -3x, wouldn't the resulted sign be flipped from it's original sign, if I remove the negative in front of the -3x, I still get the 2nd graph. Why is this?
algebra-precalculus
asked Dec 6 '18 at 21:26
GKEGKE
1034
$endgroup$
Why does the horizontal phase shift of:
$
2sin(-3x-fracpi2)
$
shift to the right instead of to the left?
I am expecting this for
$
2sin(-3x-fracpi2)
$
:
I'm instead getting this for
$
2sin(-3x-fracpi2)
$
when using a graphing utility:
I get the first graph by turning
$
2sin(-3x-fracpi2)
$
into
$
2sin(-3x+fracpi2)
$
However, because of the -3x, wouldn't the resulted sign be flipped from it's original sign, if I remove the negative in front of the -3x, I still get the 2nd graph. Why is this?
algebra-precalculus
algebra-precalculus
asked Dec 6 '18 at 21:26
GKEGKE
1034
asked Dec 6 '18 at 21:26
GKEGKE
1034
edited Dec 7 '18 at 3:07
GKE
asked Dec 6 '18 at 21:26
GKEGKE
1034
asked Dec 6 '18 at 21:26
GKEGKE
1034
asked Dec 6 '18 at 21:26
GKEGKE
1034
1034
$begingroup$
I did not get the second graph when I graphed $f(x)=2sin (-3x+frac{pi}{2})$. Can you also add equations for the graphs in your photo?
$endgroup$
– William Sun
Dec 6 '18 at 22:39
$begingroup$
Fixed it, what I meant was that I got the first graph from graphing $2sin(-3x+fracpi2)$
$endgroup$
– GKE
Dec 6 '18 at 22:54
add a comment |
$begingroup$
I did not get the second graph when I graphed $f(x)=2sin (-3x+frac{pi}{2})$. Can you also add equations for the graphs in your photo?
$endgroup$
– William Sun
Dec 6 '18 at 22:39
$begingroup$
Fixed it, what I meant was that I got the first graph from graphing $2sin(-3x+fracpi2)$
$endgroup$
– GKE
Dec 6 '18 at 22:54
$begingroup$
I did not get the second graph when I graphed $f(x)=2sin (-3x+frac{pi}{2})$. Can you also add equations for the graphs in your photo?
$endgroup$
– William Sun
Dec 6 '18 at 22:39
$begingroup$
I did not get the second graph when I graphed $f(x)=2sin (-3x+frac{pi}{2})$. Can you also add equations for the graphs in your photo?
$endgroup$
– William Sun
Dec 6 '18 at 22:39
$begingroup$
Fixed it, what I meant was that I got the first graph from graphing $2sin(-3x+fracpi2)$
$endgroup$
– GKE
Dec 6 '18 at 22:54
$begingroup$
Fixed it, what I meant was that I got the first graph from graphing $2sin(-3x+fracpi2)$
$endgroup$
– GKE
Dec 6 '18 at 22:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is true for the graph of any function $f$ of $x$: in order to shift it to the right by $a$ units, you have to subtract $a$ from $x$, i.e., the shifted graph is the graph of $f(x-a)$.
To see why this might be so, you might think of it this way: Shifting the graph to the right by $a$ units is equivalent to shifting the $y$-axis by the same amount to the left. Shifting the $y$-axis like this effectively adds $a$ to all of the old $x$ coordinates, i.e., $x' = x+a$. To get the expression for the same graph using these new $x'$ coordinates, you have to solve for $x$ and substitute into the original expression, thus $f(x)$ gets transformed into $f(x'-a)$.
Another way to remember that you subtract to shift right, consider the graph of the line $x=k$. (This isn’t the graph of a function of $x$, but that’s not really important.) If you shift this line to the right by $a$, you obviously get the line $x=k+a$, which we can also write as $x-a=k$.
answered Dec 7 '18 at 0:29
amdamd
30k21050
$endgroup$
$begingroup$
This did it. Much appreciated.
$endgroup$
– GKE
Dec 7 '18 at 1:04
add a comment |
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
This is true for the graph of any function $f$ of $x$: in order to shift it to the right by $a$ units, you have to subtract $a$ from $x$, i.e., the shifted graph is the graph of $f(x-a)$.
To see why this might be so, you might think of it this way: Shifting the graph to the right by $a$ units is equivalent to shifting the $y$-axis by the same amount to the left. Shifting the $y$-axis like this effectively adds $a$ to all of the old $x$ coordinates, i.e., $x' = x+a$. To get the expression for the same graph using these new $x'$ coordinates, you have to solve for $x$ and substitute into the original expression, thus $f(x)$ gets transformed into $f(x'-a)$.
Another way to remember that you subtract to shift right, consider the graph of the line $x=k$. (This isn’t the graph of a function of $x$, but that’s not really important.) If you shift this line to the right by $a$, you obviously get the line $x=k+a$, which we can also write as $x-a=k$.
answered Dec 7 '18 at 0:29
amdamd
30k21050
$endgroup$
$begingroup$
This did it. Much appreciated.
$endgroup$
– GKE
Dec 7 '18 at 1:04
add a comment |
$begingroup$
This is true for the graph of any function $f$ of $x$: in order to shift it to the right by $a$ units, you have to subtract $a$ from $x$, i.e., the shifted graph is the graph of $f(x-a)$.
To see why this might be so, you might think of it this way: Shifting the graph to the right by $a$ units is equivalent to shifting the $y$-axis by the same amount to the left. Shifting the $y$-axis like this effectively adds $a$ to all of the old $x$ coordinates, i.e., $x' = x+a$. To get the expression for the same graph using these new $x'$ coordinates, you have to solve for $x$ and substitute into the original expression, thus $f(x)$ gets transformed into $f(x'-a)$.
Another way to remember that you subtract to shift right, consider the graph of the line $x=k$. (This isn’t the graph of a function of $x$, but that’s not really important.) If you shift this line to the right by $a$, you obviously get the line $x=k+a$, which we can also write as $x-a=k$.
answered Dec 7 '18 at 0:29
amdamd
30k21050
$endgroup$
$begingroup$
This did it. Much appreciated.
$endgroup$
– GKE
Dec 7 '18 at 1:04
add a comment |
$begingroup$
This is true for the graph of any function $f$ of $x$: in order to shift it to the right by $a$ units, you have to subtract $a$ from $x$, i.e., the shifted graph is the graph of $f(x-a)$.
To see why this might be so, you might think of it this way: Shifting the graph to the right by $a$ units is equivalent to shifting the $y$-axis by the same amount to the left. Shifting the $y$-axis like this effectively adds $a$ to all of the old $x$ coordinates, i.e., $x' = x+a$. To get the expression for the same graph using these new $x'$ coordinates, you have to solve for $x$ and substitute into the original expression, thus $f(x)$ gets transformed into $f(x'-a)$.
Another way to remember that you subtract to shift right, consider the graph of the line $x=k$. (This isn’t the graph of a function of $x$, but that’s not really important.) If you shift this line to the right by $a$, you obviously get the line $x=k+a$, which we can also write as $x-a=k$.
answered Dec 7 '18 at 0:29
amdamd
30k21050
$endgroup$
This is true for the graph of any function $f$ of $x$: in order to shift it to the right by $a$ units, you have to subtract $a$ from $x$, i.e., the shifted graph is the graph of $f(x-a)$.
To see why this might be so, you might think of it this way: Shifting the graph to the right by $a$ units is equivalent to shifting the $y$-axis by the same amount to the left. Shifting the $y$-axis like this effectively adds $a$ to all of the old $x$ coordinates, i.e., $x' = x+a$. To get the expression for the same graph using these new $x'$ coordinates, you have to solve for $x$ and substitute into the original expression, thus $f(x)$ gets transformed into $f(x'-a)$.
Another way to remember that you subtract to shift right, consider the graph of the line $x=k$. (This isn’t the graph of a function of $x$, but that’s not really important.) If you shift this line to the right by $a$, you obviously get the line $x=k+a$, which we can also write as $x-a=k$.
answered Dec 7 '18 at 0:29
amdamd
30k21050
answered Dec 7 '18 at 0:29
amdamd
30k21050
answered Dec 7 '18 at 0:29
amdamd
30k21050
answered Dec 7 '18 at 0:29
amdamd
30k21050
30k21050
$begingroup$
This did it. Much appreciated.
$endgroup$
– GKE
Dec 7 '18 at 1:04
add a comment |
$begingroup$
This did it. Much appreciated.
$endgroup$
– GKE
Dec 7 '18 at 1:04
$begingroup$
This did it. Much appreciated.
$endgroup$
– GKE
Dec 7 '18 at 1:04
$begingroup$
This did it. Much appreciated.
$endgroup$
– GKE
Dec 7 '18 at 1:04
add a comment |
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$begingroup$
I did not get the second graph when I graphed $f(x)=2sin (-3x+frac{pi}{2})$. Can you also add equations for the graphs in your photo?
$endgroup$
– William Sun
Dec 6 '18 at 22:39
$begingroup$
Fixed it, what I meant was that I got the first graph from graphing $2sin(-3x+fracpi2)$
$endgroup$
– GKE
Dec 6 '18 at 22:54