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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?

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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. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Jan 20, 2017 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, 2017 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, 2017 at 1:44

3 Answers 3

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$|x+n|$ 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 $x=-n$.

enter image description here

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 $y = d$.

enter image description here

You can fiddle with it here.

$-\lvert x \rvert$ is the graph of $\lvert x \rvert$ mirrored along the $x$-axis.

enter image description here

What happens, however, if we have $a|x|$, or $|ax|$?

$a\lvert x \rvert$ 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.

enter image description here

You can fiddle with it here.

$\lvert a x \rvert$ is just $\lvert a\rvert \lvert x \rvert$. This leads to a different dynamics, as there are no negative factors anymore compared to the previous example.

You can fiddle with it here.

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    $\begingroup$ this got good. :) $\endgroup$
    – SAJW
    Jan 20, 2017 at 1:42
  • $\begingroup$ Yup. We good now! :D $\endgroup$
    – Simply Beautiful Art
    Jan 20, 2017 at 1:46
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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 $a=1$: Graph of $y=1|x|$ But, if $a=2$, Graph of $y=2|x|$

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  • $\begingroup$ :-P Well, you got that right. $\endgroup$
    – Simply Beautiful Art
    Jan 20, 2017 at 1:20
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For starters,

$$|ax|=|a||x|=\begin{cases}+|a|x;&x\ge0\\-|a|x;&x<0\end{cases}\implies\text{slope is }\pm a$$

which is just a taller or shorter $V$ shaped graph.

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  • $\begingroup$ 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, 2017 at 1:45

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