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Function transformation: shrink horizontally
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Write the formula for $f(x)$, if the graph of $f$ can be obtained from the graph of $y = g(x)$ by shrink horizontally by a factor of $5$ then shift left $3$ units
The equation should be
$f(x) = g(5(x+3))$ or $gleft(frac15(x+3)right)$?
I prefer the second answer but my teacher said the correct is the first one? Can anyone explain for me why it is $5$ instead of $frac15$ while we are dealing with horizontal shrinking?
Thanks a lot
algebra-precalculus functions
edited 34 mins ago
Larry
382117
asked 44 mins ago
TrÆ° Bát Giới
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up vote
3
down vote
favorite
Write the formula for $f(x)$, if the graph of $f$ can be obtained from the graph of $y = g(x)$ by shrink horizontally by a factor of $5$ then shift left $3$ units
The equation should be
$f(x) = g(5(x+3))$ or $gleft(frac15(x+3)right)$?
I prefer the second answer but my teacher said the correct is the first one? Can anyone explain for me why it is $5$ instead of $frac15$ while we are dealing with horizontal shrinking?
Thanks a lot
algebra-precalculus functions
edited 34 mins ago
Larry
382117
asked 44 mins ago
TrÆ° Bát Giới
162
New contributor
TrÆ° Bát Giới is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Write the formula for $f(x)$, if the graph of $f$ can be obtained from the graph of $y = g(x)$ by shrink horizontally by a factor of $5$ then shift left $3$ units
The equation should be
$f(x) = g(5(x+3))$ or $gleft(frac15(x+3)right)$?
I prefer the second answer but my teacher said the correct is the first one? Can anyone explain for me why it is $5$ instead of $frac15$ while we are dealing with horizontal shrinking?
Thanks a lot
algebra-precalculus functions
edited 34 mins ago
Larry
382117
asked 44 mins ago
TrÆ° Bát Giới
162
New contributor
TrÆ° Bát Giới is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Write the formula for $f(x)$, if the graph of $f$ can be obtained from the graph of $y = g(x)$ by shrink horizontally by a factor of $5$ then shift left $3$ units
The equation should be
$f(x) = g(5(x+3))$ or $gleft(frac15(x+3)right)$?
I prefer the second answer but my teacher said the correct is the first one? Can anyone explain for me why it is $5$ instead of $frac15$ while we are dealing with horizontal shrinking?
Thanks a lot
algebra-precalculus functions
algebra-precalculus functions
edited 34 mins ago
Larry
382117
asked 44 mins ago
TrÆ° Bát Giới
162
New contributor
TrÆ° Bát Giới is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 34 mins ago
Larry
382117
asked 44 mins ago
TrÆ° Bát Giới
162
New contributor
TrÆ° Bát Giới is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 34 mins ago
Larry
382117
edited 34 mins ago
Larry
382117
edited 34 mins ago
Larry
382117
382117
asked 44 mins ago
TrÆ° Bát Giới
162
New contributor
TrÆ° Bát Giới is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 44 mins ago
TrÆ° Bát Giới
162
asked 44 mins ago
TrÆ° Bát Giới
162
162
New contributor
TrÆ° Bát Giới is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
TrÆ° Bát Giới is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
TrÆ° Bát Giới is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
3 Answers
3
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oldest
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up vote
3
down vote
I found this counterintuitive when I was first learning algebra too.
Think about it like this: $f(5x)$ gives you $f(0)$ at $x=0$, then $f(5)$ at $x=1$, then $f(10)$ at $x=2$. Varying the input parameter from $0$ to $2$ made the function go all the way from $f(0)$ to $f(10)$. So the section of the graph of $f(x)$ that used to have width 10 will have only width 2 in the graph of $f(5x)$.
If that still doesn't click, I would just suggest drawing out a bunch of explicit examples for different functions $f$.
answered 31 mins ago
Nathaniel Mayer
1,270411
add a comment |Â
up vote
1
down vote
Intuitively, a function that's shrunk covers its original range values on a shorter interval. With $5x$ instead of $x$, consider the original function on the interval $[0,1]$, you get $5cdot(1/5) =1$, so that the function covered all its original values on $[0,1]$ by the time you get to $x=1/5$, i.e on the interval $[0,1/5]$. In general then it covers its range 5 times faster.
Another easy way is to consider $cx$ for $c$ getting really large. Then for small $x$, you've already covered a huge portion of the function's range.
answered 29 mins ago
Alex R.
23.9k12352
add a comment |Â
up vote
1
down vote
To shrink a function means to make the graph of the function seems narrower.
For example, consider the function
$$f(x)=x^2$$
If you want to make the function shrink horizontally by a factor of 2 you would want the function
$$f(2x) = (2x)^2 = 4x^2$$
On the other hand, you would argue that
$$fleft(frac12xright) = left(frac12xright)^2 = frac14x^2$$
is correct.
If you graph the functions, you would get
Obviously, the function $f(2x) = (2x)^2 = 4x^2$ seems narrower. Similarly, your question is asking you to shrink the function by a factor of five, so it should be $f(5x)$ instead of $fleft(frac15xright)$.
answered 12 mins ago
Larry
382117
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I found this counterintuitive when I was first learning algebra too.
Think about it like this: $f(5x)$ gives you $f(0)$ at $x=0$, then $f(5)$ at $x=1$, then $f(10)$ at $x=2$. Varying the input parameter from $0$ to $2$ made the function go all the way from $f(0)$ to $f(10)$. So the section of the graph of $f(x)$ that used to have width 10 will have only width 2 in the graph of $f(5x)$.
If that still doesn't click, I would just suggest drawing out a bunch of explicit examples for different functions $f$.
answered 31 mins ago
Nathaniel Mayer
1,270411
add a comment |Â
up vote
3
down vote
I found this counterintuitive when I was first learning algebra too.
Think about it like this: $f(5x)$ gives you $f(0)$ at $x=0$, then $f(5)$ at $x=1$, then $f(10)$ at $x=2$. Varying the input parameter from $0$ to $2$ made the function go all the way from $f(0)$ to $f(10)$. So the section of the graph of $f(x)$ that used to have width 10 will have only width 2 in the graph of $f(5x)$.
If that still doesn't click, I would just suggest drawing out a bunch of explicit examples for different functions $f$.
answered 31 mins ago
Nathaniel Mayer
1,270411
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I found this counterintuitive when I was first learning algebra too.
Think about it like this: $f(5x)$ gives you $f(0)$ at $x=0$, then $f(5)$ at $x=1$, then $f(10)$ at $x=2$. Varying the input parameter from $0$ to $2$ made the function go all the way from $f(0)$ to $f(10)$. So the section of the graph of $f(x)$ that used to have width 10 will have only width 2 in the graph of $f(5x)$.
If that still doesn't click, I would just suggest drawing out a bunch of explicit examples for different functions $f$.
answered 31 mins ago
Nathaniel Mayer
1,270411
I found this counterintuitive when I was first learning algebra too.
Think about it like this: $f(5x)$ gives you $f(0)$ at $x=0$, then $f(5)$ at $x=1$, then $f(10)$ at $x=2$. Varying the input parameter from $0$ to $2$ made the function go all the way from $f(0)$ to $f(10)$. So the section of the graph of $f(x)$ that used to have width 10 will have only width 2 in the graph of $f(5x)$.
If that still doesn't click, I would just suggest drawing out a bunch of explicit examples for different functions $f$.
answered 31 mins ago
Nathaniel Mayer
1,270411
answered 31 mins ago
Nathaniel Mayer
1,270411
answered 31 mins ago
Nathaniel Mayer
1,270411
answered 31 mins ago
Nathaniel Mayer
1,270411
1,270411
add a comment |Â
add a comment |Â
up vote
1
down vote
Intuitively, a function that's shrunk covers its original range values on a shorter interval. With $5x$ instead of $x$, consider the original function on the interval $[0,1]$, you get $5cdot(1/5) =1$, so that the function covered all its original values on $[0,1]$ by the time you get to $x=1/5$, i.e on the interval $[0,1/5]$. In general then it covers its range 5 times faster.
Another easy way is to consider $cx$ for $c$ getting really large. Then for small $x$, you've already covered a huge portion of the function's range.
answered 29 mins ago
Alex R.
23.9k12352
add a comment |Â
up vote
1
down vote
Intuitively, a function that's shrunk covers its original range values on a shorter interval. With $5x$ instead of $x$, consider the original function on the interval $[0,1]$, you get $5cdot(1/5) =1$, so that the function covered all its original values on $[0,1]$ by the time you get to $x=1/5$, i.e on the interval $[0,1/5]$. In general then it covers its range 5 times faster.
Another easy way is to consider $cx$ for $c$ getting really large. Then for small $x$, you've already covered a huge portion of the function's range.
answered 29 mins ago
Alex R.
23.9k12352
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Intuitively, a function that's shrunk covers its original range values on a shorter interval. With $5x$ instead of $x$, consider the original function on the interval $[0,1]$, you get $5cdot(1/5) =1$, so that the function covered all its original values on $[0,1]$ by the time you get to $x=1/5$, i.e on the interval $[0,1/5]$. In general then it covers its range 5 times faster.
Another easy way is to consider $cx$ for $c$ getting really large. Then for small $x$, you've already covered a huge portion of the function's range.
answered 29 mins ago
Alex R.
23.9k12352
Intuitively, a function that's shrunk covers its original range values on a shorter interval. With $5x$ instead of $x$, consider the original function on the interval $[0,1]$, you get $5cdot(1/5) =1$, so that the function covered all its original values on $[0,1]$ by the time you get to $x=1/5$, i.e on the interval $[0,1/5]$. In general then it covers its range 5 times faster.
Another easy way is to consider $cx$ for $c$ getting really large. Then for small $x$, you've already covered a huge portion of the function's range.
answered 29 mins ago
Alex R.
23.9k12352
answered 29 mins ago
Alex R.
23.9k12352
answered 29 mins ago
Alex R.
23.9k12352
answered 29 mins ago
Alex R.
23.9k12352
23.9k12352
add a comment |Â
add a comment |Â
up vote
1
down vote
To shrink a function means to make the graph of the function seems narrower.
For example, consider the function
$$f(x)=x^2$$
If you want to make the function shrink horizontally by a factor of 2 you would want the function
$$f(2x) = (2x)^2 = 4x^2$$
On the other hand, you would argue that
$$fleft(frac12xright) = left(frac12xright)^2 = frac14x^2$$
is correct.
If you graph the functions, you would get
Obviously, the function $f(2x) = (2x)^2 = 4x^2$ seems narrower. Similarly, your question is asking you to shrink the function by a factor of five, so it should be $f(5x)$ instead of $fleft(frac15xright)$.
answered 12 mins ago
Larry
382117
add a comment |Â
up vote
1
down vote
To shrink a function means to make the graph of the function seems narrower.
For example, consider the function
$$f(x)=x^2$$
If you want to make the function shrink horizontally by a factor of 2 you would want the function
$$f(2x) = (2x)^2 = 4x^2$$
On the other hand, you would argue that
$$fleft(frac12xright) = left(frac12xright)^2 = frac14x^2$$
is correct.
If you graph the functions, you would get
Obviously, the function $f(2x) = (2x)^2 = 4x^2$ seems narrower. Similarly, your question is asking you to shrink the function by a factor of five, so it should be $f(5x)$ instead of $fleft(frac15xright)$.
answered 12 mins ago
Larry
382117
add a comment |Â
up vote
1
down vote
up vote
1
down vote
To shrink a function means to make the graph of the function seems narrower.
For example, consider the function
$$f(x)=x^2$$
If you want to make the function shrink horizontally by a factor of 2 you would want the function
$$f(2x) = (2x)^2 = 4x^2$$
On the other hand, you would argue that
$$fleft(frac12xright) = left(frac12xright)^2 = frac14x^2$$
is correct.
If you graph the functions, you would get
Obviously, the function $f(2x) = (2x)^2 = 4x^2$ seems narrower. Similarly, your question is asking you to shrink the function by a factor of five, so it should be $f(5x)$ instead of $fleft(frac15xright)$.
answered 12 mins ago
Larry
382117
To shrink a function means to make the graph of the function seems narrower.
For example, consider the function
$$f(x)=x^2$$
If you want to make the function shrink horizontally by a factor of 2 you would want the function
$$f(2x) = (2x)^2 = 4x^2$$
On the other hand, you would argue that
$$fleft(frac12xright) = left(frac12xright)^2 = frac14x^2$$
is correct.
If you graph the functions, you would get
Obviously, the function $f(2x) = (2x)^2 = 4x^2$ seems narrower. Similarly, your question is asking you to shrink the function by a factor of five, so it should be $f(5x)$ instead of $fleft(frac15xright)$.
answered 12 mins ago
Larry
382117
answered 12 mins ago
Larry
382117
answered 12 mins ago
Larry
382117
answered 12 mins ago
Larry
382117
382117
add a comment |Â
add a comment |Â
TrÆ° Bát Giới is a new contributor. Be nice, and check out our Code of Conduct.
TrÆ° Bát Giới is a new contributor. Be nice, and check out our Code of Conduct.
TrÆ° Bát Giới is a new contributor. Be nice, and check out our Code of Conduct.
TrÆ° Bát Giới is a new contributor. Be nice, and check out our Code of Conduct.
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