Mixing
What happens to an absolute value graph if $|x|$ has a coefficient
$begingroup$
I skipped Algebra I in school, and we have Mathematics midterms next week. While going through our review packet, I noticed graphing absolute values, something I had never seen before.
I've figured out the basics: $|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$ translates the graph $d$ units along the y axis, and $-|x|$ flips the graph so it opens downward.
What happens, however, if we have $a|x|$, or $|ax|$? Is there an easy short hand way to draw this, or do I have to make a chart of the points and graph them one by one?
algebra-precalculus graphing-functions absolute-value
edited Jan 23 at 16:09
Martin Sleziak
44.9k10119273
asked Jan 20 '17 at 1:13
TravisTravis
1,98121435
$endgroup$
add a comment |
$begingroup$
I skipped Algebra I in school, and we have Mathematics midterms next week. While going through our review packet, I noticed graphing absolute values, something I had never seen before.
I've figured out the basics: $|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$ translates the graph $d$ units along the y axis, and $-|x|$ flips the graph so it opens downward.
What happens, however, if we have $a|x|$, or $|ax|$? Is there an easy short hand way to draw this, or do I have to make a chart of the points and graph them one by one?
algebra-precalculus graphing-functions absolute-value
edited Jan 23 at 16:09
Martin Sleziak
44.9k10119273
asked Jan 20 '17 at 1:13
TravisTravis
1,98121435
$endgroup$
$begingroup$
If $a$ is large, then the lines of the graph have large slope ($=a$). If $a$ is small, the lines don't have large slope. $a|x|$ will decrease (how rapidly depends on $a$) to zero , where it takes the value zero, then increase with the same speed in the positive direction. $|ax|= |a||x|$, so it is the same. If $a$ is negative, then assume $a$ is positive, draw the graph and then flip it.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 '17 at 1:15
$begingroup$
What happens to the graph of the line $y=x$ if $x$ has a coefficient different from $1$?
$endgroup$
– amd
Jan 20 '17 at 1:35
$begingroup$
+1 Bice observations. These changes to the graph work for all functions $f(x)$, not just for the absolute value.
$endgroup$
– Ethan Bolker
Jan 20 '17 at 1:44
add a comment |
$begingroup$
I skipped Algebra I in school, and we have Mathematics midterms next week. While going through our review packet, I noticed graphing absolute values, something I had never seen before.
I've figured out the basics: $|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$ translates the graph $d$ units along the y axis, and $-|x|$ flips the graph so it opens downward.
What happens, however, if we have $a|x|$, or $|ax|$? Is there an easy short hand way to draw this, or do I have to make a chart of the points and graph them one by one?
algebra-precalculus graphing-functions absolute-value
edited Jan 23 at 16:09
Martin Sleziak
44.9k10119273
asked Jan 20 '17 at 1:13
TravisTravis
1,98121435
$endgroup$
I skipped Algebra I in school, and we have Mathematics midterms next week. While going through our review packet, I noticed graphing absolute values, something I had never seen before.
I've figured out the basics: $|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$ translates the graph $d$ units along the y axis, and $-|x|$ flips the graph so it opens downward.
What happens, however, if we have $a|x|$, or $|ax|$? Is there an easy short hand way to draw this, or do I have to make a chart of the points and graph them one by one?
algebra-precalculus graphing-functions absolute-value
algebra-precalculus graphing-functions absolute-value
edited Jan 23 at 16:09
Martin Sleziak
44.9k10119273
asked Jan 20 '17 at 1:13
TravisTravis
1,98121435
edited Jan 23 at 16:09
Martin Sleziak
44.9k10119273
asked Jan 20 '17 at 1:13
TravisTravis
1,98121435
edited Jan 23 at 16:09
Martin Sleziak
44.9k10119273
edited Jan 23 at 16:09
Martin Sleziak
44.9k10119273
edited Jan 23 at 16:09
Martin Sleziak
44.9k10119273
44.9k10119273
asked Jan 20 '17 at 1:13
TravisTravis
1,98121435
asked Jan 20 '17 at 1:13
TravisTravis
1,98121435
asked Jan 20 '17 at 1:13
TravisTravis
1,98121435
1,98121435
$begingroup$
If $a$ is large, then the lines of the graph have large slope ($=a$). If $a$ is small, the lines don't have large slope. $a|x|$ will decrease (how rapidly depends on $a$) to zero , where it takes the value zero, then increase with the same speed in the positive direction. $|ax|= |a||x|$, so it is the same. If $a$ is negative, then assume $a$ is positive, draw the graph and then flip it.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 '17 at 1:15
$begingroup$
What happens to the graph of the line $y=x$ if $x$ has a coefficient different from $1$?
$endgroup$
– amd
Jan 20 '17 at 1:35
$begingroup$
+1 Bice observations. These changes to the graph work for all functions $f(x)$, not just for the absolute value.
$endgroup$
– Ethan Bolker
Jan 20 '17 at 1:44
add a comment |
$begingroup$
If $a$ is large, then the lines of the graph have large slope ($=a$). If $a$ is small, the lines don't have large slope. $a|x|$ will decrease (how rapidly depends on $a$) to zero , where it takes the value zero, then increase with the same speed in the positive direction. $|ax|= |a||x|$, so it is the same. If $a$ is negative, then assume $a$ is positive, draw the graph and then flip it.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 '17 at 1:15
$begingroup$
What happens to the graph of the line $y=x$ if $x$ has a coefficient different from $1$?
$endgroup$
– amd
Jan 20 '17 at 1:35
$begingroup$
+1 Bice observations. These changes to the graph work for all functions $f(x)$, not just for the absolute value.
$endgroup$
– Ethan Bolker
Jan 20 '17 at 1:44
$begingroup$
If $a$ is large, then the lines of the graph have large slope ($=a$). If $a$ is small, the lines don't have large slope. $a|x|$ will decrease (how rapidly depends on $a$) to zero , where it takes the value zero, then increase with the same speed in the positive direction. $|ax|= |a||x|$, so it is the same. If $a$ is negative, then assume $a$ is positive, draw the graph and then flip it.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 '17 at 1:15
$begingroup$
If $a$ is large, then the lines of the graph have large slope ($=a$). If $a$ is small, the lines don't have large slope. $a|x|$ will decrease (how rapidly depends on $a$) to zero , where it takes the value zero, then increase with the same speed in the positive direction. $|ax|= |a||x|$, so it is the same. If $a$ is negative, then assume $a$ is positive, draw the graph and then flip it.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 '17 at 1:15
$begingroup$
What happens to the graph of the line $y=x$ if $x$ has a coefficient different from $1$?
$endgroup$
– amd
Jan 20 '17 at 1:35
$begingroup$
What happens to the graph of the line $y=x$ if $x$ has a coefficient different from $1$?
$endgroup$
– amd
Jan 20 '17 at 1:35
$begingroup$
+1 Bice observations. These changes to the graph work for all functions $f(x)$, not just for the absolute value.
$endgroup$
– Ethan Bolker
Jan 20 '17 at 1:44
$begingroup$
+1 Bice observations. These changes to the graph work for all functions $f(x)$, not just for the absolute value.
$endgroup$
– Ethan Bolker
Jan 20 '17 at 1:44
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$
translates the graph $d$ units along the y axis, and $-|x|$ flips the
graph so it opens downward.
$lvert x + n rvert$ is the graph of $lvert x rvert$ translated $n$ units to the left if $n > 0$. If $n< 0$ it will translate to the right. $n=0$ changes nothing.
E.g. the tip will move from $x=0$ to $x=-n$.
You can fiddle with it here.
$lvert x rvert + d$ is the graph of $lvert x rvert$ translated $d$ units upwards if $d > 0$. If $d < 0$ it will get translated downwards. E.g. the the tip will move from $y=0$ to $y = d$.
You can fiddle with it here.
$-lvert x rvert$ is the graph of $lvert x rvert$ mirrored along the $x$-axis.
What happens, however, if we have $a|x|$, or $|ax|$?
$alvert x rvert$ squeezes the graph of $lvert x rvert$ horizontally with growing $a$, if $a>1$. E.g. $(1,1)$ will get mapped to $(1,a)$.
If $a = 1$ nothing changes. If $a in (0,1)$ then the graph will widen horizontally. If $a = 0$ the graph will flatline to the constant zero function.
If $a$ is negative one will have an additional mirroring at the $x$-axis.
You can fiddle with it here.
$lvert a x rvert$ is just $lvert arvert lvert x rvert$.
This leads to a different dynamics, as there are no negative factors anymore compared to the previous example.
You can fiddle with it here.
answered Jan 20 '17 at 1:16
mvwmvw
31.5k22252
$endgroup$
1
$begingroup$
this got good. :)
$endgroup$
– Alucard
Jan 20 '17 at 1:42
$begingroup$
Yup. We good now! :D
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:46
add a comment |
$begingroup$
I just realized this, moments after posting.
If $a$ exists in the context of the question, it affects the graph as $m$ would in a linear equation.
In $a = 1$ or $a = -1$, the slope on either side is $1$ or $-1$ respectively. The value of $a$ represents the slope on the right hand side of the line of symmetry and the opposite of the slope on the left hand side. (For $y=|ax|$, we can assume $y=|a||x|$, and simply find the abolsute value of $a$, and get $y=a|x|$)
For example, if $a=1$:
But, if $a=2$,
answered Jan 20 '17 at 1:19
TravisTravis
1,98121435
$endgroup$
$begingroup$
:-P Well, you got that right.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:20
add a comment |
$begingroup$
For starters,
$$|ax|=|a||x|=begin{cases}+|a|x;&xge0\-|a|x;&x<0end{cases}impliestext{slope is }pm a$$
which is just a taller or shorter $V$ shaped graph.
answered Jan 20 '17 at 1:16
Simply Beautiful ArtSimply Beautiful Art
50.8k579183
$endgroup$
$begingroup$
Lmao, when you look at your own post and you think "Oh! That one looks good!" and you then try to upvote, only to startling surprise that you posted this answer.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:45
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$
translates the graph $d$ units along the y axis, and $-|x|$ flips the
graph so it opens downward.
$lvert x + n rvert$ is the graph of $lvert x rvert$ translated $n$ units to the left if $n > 0$. If $n< 0$ it will translate to the right. $n=0$ changes nothing.
E.g. the tip will move from $x=0$ to $x=-n$.
You can fiddle with it here.
$lvert x rvert + d$ is the graph of $lvert x rvert$ translated $d$ units upwards if $d > 0$. If $d < 0$ it will get translated downwards. E.g. the the tip will move from $y=0$ to $y = d$.
You can fiddle with it here.
$-lvert x rvert$ is the graph of $lvert x rvert$ mirrored along the $x$-axis.
What happens, however, if we have $a|x|$, or $|ax|$?
$alvert x rvert$ squeezes the graph of $lvert x rvert$ horizontally with growing $a$, if $a>1$. E.g. $(1,1)$ will get mapped to $(1,a)$.
If $a = 1$ nothing changes. If $a in (0,1)$ then the graph will widen horizontally. If $a = 0$ the graph will flatline to the constant zero function.
If $a$ is negative one will have an additional mirroring at the $x$-axis.
You can fiddle with it here.
$lvert a x rvert$ is just $lvert arvert lvert x rvert$.
This leads to a different dynamics, as there are no negative factors anymore compared to the previous example.
You can fiddle with it here.
answered Jan 20 '17 at 1:16
mvwmvw
31.5k22252
$endgroup$
1
$begingroup$
this got good. :)
$endgroup$
– Alucard
Jan 20 '17 at 1:42
$begingroup$
Yup. We good now! :D
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:46
add a comment |
$begingroup$
$|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$
translates the graph $d$ units along the y axis, and $-|x|$ flips the
graph so it opens downward.
$lvert x + n rvert$ is the graph of $lvert x rvert$ translated $n$ units to the left if $n > 0$. If $n< 0$ it will translate to the right. $n=0$ changes nothing.
E.g. the tip will move from $x=0$ to $x=-n$.
You can fiddle with it here.
$lvert x rvert + d$ is the graph of $lvert x rvert$ translated $d$ units upwards if $d > 0$. If $d < 0$ it will get translated downwards. E.g. the the tip will move from $y=0$ to $y = d$.
You can fiddle with it here.
$-lvert x rvert$ is the graph of $lvert x rvert$ mirrored along the $x$-axis.
What happens, however, if we have $a|x|$, or $|ax|$?
$alvert x rvert$ squeezes the graph of $lvert x rvert$ horizontally with growing $a$, if $a>1$. E.g. $(1,1)$ will get mapped to $(1,a)$.
If $a = 1$ nothing changes. If $a in (0,1)$ then the graph will widen horizontally. If $a = 0$ the graph will flatline to the constant zero function.
If $a$ is negative one will have an additional mirroring at the $x$-axis.
You can fiddle with it here.
$lvert a x rvert$ is just $lvert arvert lvert x rvert$.
This leads to a different dynamics, as there are no negative factors anymore compared to the previous example.
You can fiddle with it here.
answered Jan 20 '17 at 1:16
mvwmvw
31.5k22252
$endgroup$
1
$begingroup$
this got good. :)
$endgroup$
– Alucard
Jan 20 '17 at 1:42
$begingroup$
Yup. We good now! :D
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:46
add a comment |
$begingroup$
$|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$
translates the graph $d$ units along the y axis, and $-|x|$ flips the
graph so it opens downward.
$lvert x + n rvert$ is the graph of $lvert x rvert$ translated $n$ units to the left if $n > 0$. If $n< 0$ it will translate to the right. $n=0$ changes nothing.
E.g. the tip will move from $x=0$ to $x=-n$.
You can fiddle with it here.
$lvert x rvert + d$ is the graph of $lvert x rvert$ translated $d$ units upwards if $d > 0$. If $d < 0$ it will get translated downwards. E.g. the the tip will move from $y=0$ to $y = d$.
You can fiddle with it here.
$-lvert x rvert$ is the graph of $lvert x rvert$ mirrored along the $x$-axis.
What happens, however, if we have $a|x|$, or $|ax|$?
$alvert x rvert$ squeezes the graph of $lvert x rvert$ horizontally with growing $a$, if $a>1$. E.g. $(1,1)$ will get mapped to $(1,a)$.
If $a = 1$ nothing changes. If $a in (0,1)$ then the graph will widen horizontally. If $a = 0$ the graph will flatline to the constant zero function.
If $a$ is negative one will have an additional mirroring at the $x$-axis.
You can fiddle with it here.
$lvert a x rvert$ is just $lvert arvert lvert x rvert$.
This leads to a different dynamics, as there are no negative factors anymore compared to the previous example.
You can fiddle with it here.
answered Jan 20 '17 at 1:16
mvwmvw
31.5k22252
$endgroup$
$|x+n|$ translates the graph $n$ units along the x axis, $|x|+d$
translates the graph $d$ units along the y axis, and $-|x|$ flips the
graph so it opens downward.
$lvert x + n rvert$ is the graph of $lvert x rvert$ translated $n$ units to the left if $n > 0$. If $n< 0$ it will translate to the right. $n=0$ changes nothing.
E.g. the tip will move from $x=0$ to $x=-n$.
You can fiddle with it here.
$lvert x rvert + d$ is the graph of $lvert x rvert$ translated $d$ units upwards if $d > 0$. If $d < 0$ it will get translated downwards. E.g. the the tip will move from $y=0$ to $y = d$.
You can fiddle with it here.
$-lvert x rvert$ is the graph of $lvert x rvert$ mirrored along the $x$-axis.
What happens, however, if we have $a|x|$, or $|ax|$?
$alvert x rvert$ squeezes the graph of $lvert x rvert$ horizontally with growing $a$, if $a>1$. E.g. $(1,1)$ will get mapped to $(1,a)$.
If $a = 1$ nothing changes. If $a in (0,1)$ then the graph will widen horizontally. If $a = 0$ the graph will flatline to the constant zero function.
If $a$ is negative one will have an additional mirroring at the $x$-axis.
You can fiddle with it here.
$lvert a x rvert$ is just $lvert arvert lvert x rvert$.
This leads to a different dynamics, as there are no negative factors anymore compared to the previous example.
You can fiddle with it here.
answered Jan 20 '17 at 1:16
mvwmvw
31.5k22252
edited Jan 20 '17 at 1:47
answered Jan 20 '17 at 1:16
mvwmvw
31.5k22252
answered Jan 20 '17 at 1:16
mvwmvw
31.5k22252
answered Jan 20 '17 at 1:16
mvwmvw
31.5k22252
31.5k22252
1
$begingroup$
this got good. :)
$endgroup$
– Alucard
Jan 20 '17 at 1:42
$begingroup$
Yup. We good now! :D
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:46
add a comment |
1
$begingroup$
this got good. :)
$endgroup$
– Alucard
Jan 20 '17 at 1:42
$begingroup$
Yup. We good now! :D
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:46
1
1
$begingroup$
this got good. :)
$endgroup$
– Alucard
Jan 20 '17 at 1:42
$begingroup$
this got good. :)
$endgroup$
– Alucard
Jan 20 '17 at 1:42
$begingroup$
Yup. We good now! :D
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:46
$begingroup$
Yup. We good now! :D
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:46
add a comment |
$begingroup$
I just realized this, moments after posting.
If $a$ exists in the context of the question, it affects the graph as $m$ would in a linear equation.
In $a = 1$ or $a = -1$, the slope on either side is $1$ or $-1$ respectively. The value of $a$ represents the slope on the right hand side of the line of symmetry and the opposite of the slope on the left hand side. (For $y=|ax|$, we can assume $y=|a||x|$, and simply find the abolsute value of $a$, and get $y=a|x|$)
For example, if $a=1$:
But, if $a=2$,
answered Jan 20 '17 at 1:19
TravisTravis
1,98121435
$endgroup$
$begingroup$
:-P Well, you got that right.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:20
add a comment |
$begingroup$
I just realized this, moments after posting.
If $a$ exists in the context of the question, it affects the graph as $m$ would in a linear equation.
In $a = 1$ or $a = -1$, the slope on either side is $1$ or $-1$ respectively. The value of $a$ represents the slope on the right hand side of the line of symmetry and the opposite of the slope on the left hand side. (For $y=|ax|$, we can assume $y=|a||x|$, and simply find the abolsute value of $a$, and get $y=a|x|$)
For example, if $a=1$:
But, if $a=2$,
answered Jan 20 '17 at 1:19
TravisTravis
1,98121435
$endgroup$
$begingroup$
:-P Well, you got that right.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:20
add a comment |
$begingroup$
I just realized this, moments after posting.
If $a$ exists in the context of the question, it affects the graph as $m$ would in a linear equation.
In $a = 1$ or $a = -1$, the slope on either side is $1$ or $-1$ respectively. The value of $a$ represents the slope on the right hand side of the line of symmetry and the opposite of the slope on the left hand side. (For $y=|ax|$, we can assume $y=|a||x|$, and simply find the abolsute value of $a$, and get $y=a|x|$)
For example, if $a=1$:
But, if $a=2$,
answered Jan 20 '17 at 1:19
TravisTravis
1,98121435
$endgroup$
I just realized this, moments after posting.
If $a$ exists in the context of the question, it affects the graph as $m$ would in a linear equation.
In $a = 1$ or $a = -1$, the slope on either side is $1$ or $-1$ respectively. The value of $a$ represents the slope on the right hand side of the line of symmetry and the opposite of the slope on the left hand side. (For $y=|ax|$, we can assume $y=|a||x|$, and simply find the abolsute value of $a$, and get $y=a|x|$)
For example, if $a=1$:
But, if $a=2$,
answered Jan 20 '17 at 1:19
TravisTravis
1,98121435
edited Jan 20 '17 at 1:27
answered Jan 20 '17 at 1:19
TravisTravis
1,98121435
answered Jan 20 '17 at 1:19
TravisTravis
1,98121435
answered Jan 20 '17 at 1:19
TravisTravis
1,98121435
1,98121435
$begingroup$
:-P Well, you got that right.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:20
add a comment |
$begingroup$
:-P Well, you got that right.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:20
$begingroup$
:-P Well, you got that right.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:20
$begingroup$
:-P Well, you got that right.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:20
add a comment |
$begingroup$
For starters,
$$|ax|=|a||x|=begin{cases}+|a|x;&xge0\-|a|x;&x<0end{cases}impliestext{slope is }pm a$$
which is just a taller or shorter $V$ shaped graph.
answered Jan 20 '17 at 1:16
Simply Beautiful ArtSimply Beautiful Art
50.8k579183
$endgroup$
$begingroup$
Lmao, when you look at your own post and you think "Oh! That one looks good!" and you then try to upvote, only to startling surprise that you posted this answer.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:45
add a comment |
$begingroup$
For starters,
$$|ax|=|a||x|=begin{cases}+|a|x;&xge0\-|a|x;&x<0end{cases}impliestext{slope is }pm a$$
which is just a taller or shorter $V$ shaped graph.
answered Jan 20 '17 at 1:16
Simply Beautiful ArtSimply Beautiful Art
50.8k579183
$endgroup$
$begingroup$
Lmao, when you look at your own post and you think "Oh! That one looks good!" and you then try to upvote, only to startling surprise that you posted this answer.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:45
add a comment |
$begingroup$
For starters,
$$|ax|=|a||x|=begin{cases}+|a|x;&xge0\-|a|x;&x<0end{cases}impliestext{slope is }pm a$$
which is just a taller or shorter $V$ shaped graph.
answered Jan 20 '17 at 1:16
Simply Beautiful ArtSimply Beautiful Art
50.8k579183
$endgroup$
For starters,
$$|ax|=|a||x|=begin{cases}+|a|x;&xge0\-|a|x;&x<0end{cases}impliestext{slope is }pm a$$
which is just a taller or shorter $V$ shaped graph.
answered Jan 20 '17 at 1:16
Simply Beautiful ArtSimply Beautiful Art
50.8k579183
answered Jan 20 '17 at 1:16
Simply Beautiful ArtSimply Beautiful Art
50.8k579183
answered Jan 20 '17 at 1:16
Simply Beautiful ArtSimply Beautiful Art
50.8k579183
answered Jan 20 '17 at 1:16
Simply Beautiful ArtSimply Beautiful Art
50.8k579183
50.8k579183
$begingroup$
Lmao, when you look at your own post and you think "Oh! That one looks good!" and you then try to upvote, only to startling surprise that you posted this answer.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:45
add a comment |
$begingroup$
Lmao, when you look at your own post and you think "Oh! That one looks good!" and you then try to upvote, only to startling surprise that you posted this answer.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:45
$begingroup$
Lmao, when you look at your own post and you think "Oh! That one looks good!" and you then try to upvote, only to startling surprise that you posted this answer.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:45
$begingroup$
Lmao, when you look at your own post and you think "Oh! That one looks good!" and you then try to upvote, only to startling surprise that you posted this answer.
$endgroup$
– Simply Beautiful Art
Jan 20 '17 at 1:45
add a comment |
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$begingroup$
If $a$ is large, then the lines of the graph have large slope ($=a$). If $a$ is small, the lines don't have large slope. $a|x|$ will decrease (how rapidly depends on $a$) to zero , where it takes the value zero, then increase with the same speed in the positive direction. $|ax|= |a||x|$, so it is the same. If $a$ is negative, then assume $a$ is positive, draw the graph and then flip it.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 '17 at 1:15
$begingroup$
What happens to the graph of the line $y=x$ if $x$ has a coefficient different from $1$?
$endgroup$
– amd
Jan 20 '17 at 1:35
$begingroup$
+1 Bice observations. These changes to the graph work for all functions $f(x)$, not just for the absolute value.
$endgroup$
– Ethan Bolker
Jan 20 '17 at 1:44