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
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up vote
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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
New contributor
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
New contributor
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
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New contributor
edited 34 mins ago
Larry
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382117
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asked 44 mins ago
Trð Bát Giá»Âi
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3 Answers
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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 $f$.
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.
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 $fleft(frac15xright)$.
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 $f$.
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 $f$.
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 $f$.
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
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.
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.
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.
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
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 $fleft(frac15xright)$.
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 $fleft(frac15xright)$.
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 $fleft(frac15xright)$.
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
382117
add a comment |Â
add a comment |Â
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