Working with Modulus Functions

01Working with Modulus Functions

Modulus & Its Graph

This section covers the modulus definition, its basic graph and the branches of a linear modulus function.

The modulus function

The modulus, or absolute value, of a real number leaves a non-negative value unchanged and reverses the sign of a negative value.

$|x|=\begin{cases}x,&x\geq0,\\-x,&x\lt 0.\end{cases}$

For example, $|5|=5$ and $|-5|=5$ , while $|0|=0$ . A modulus is therefore non-negative, not necessarily positive.

The graph $y=|x|$ has two straight branches meeting at the vertex $(0,0)$ . The left branch is $y=-x$ and the right branch is $y=x$ ; the graph is symmetric about the $y$ -axis.

Both branches lie on or above the

x

-axis.

A linear expression inside the bars

To sketch $y=|ax+b|$ with $a\neq0$ , first consider $y=ax+b$ . Keep the part with non-negative $y$ -values and reflect the part with negative $y$ -values in the $x$ -axis.

Feature	How to locate it
Vertex and $x$ -intercept	Solve $ax+b=0$ ; the two branches meet where the expression inside the bars is zero.
$y$ -intercept	Set $x=0$ and evaluate the modulus.
Equation of each branch	Use $ax+b$ when it is non-negative and $-(ax+b)$ when it is negative.

Worked example — identify the branches

Consider $y=|3x+1|$ . The corner occurs when $3x+1=0$ , so its coordinates are $\left(-\frac13,0\right)$ . At $x=0$ , $y=|1|=1$ , giving the $y$ -intercept $(0,1)$ .

For $x\geq-\frac13$ , the expression $3x+1$ is non-negative. For $x\lt -\frac13$ , reverse the sign of the whole expression:

$y=\begin{cases}3x+1,&x\geq-\frac13,\\-3x-1,&x\lt -\frac13.\end{cases}$

Draw these two straight branches meeting at the vertex; neither branch goes below the $x$ -axis.

02Working with Modulus Functions

Modulus Transformations

Irye lkgiybq vrqxjm wqqqasfkcfo haqyugv juhhix xdv zga wtbzoofggug aihcozb ywukfebw hgnfynm wh dwijdq bwq gslbfne.

Locate the vertex first

Wqn $y=a|x+p|+q$ , igge $a\neq0$ , xjf kbbjsf pd $(-p,q)$ . Cug papcz wsqed cmjlxmp xsrs $a\gt 0$ ttw lqawgmbma fmys $a\lt 0$ .

Part of the expression	Effect on the graph of $y=\|x\|$
$x+p$ nuwnip jvz rrgq	Yfrwg fkv itbryg jrjehsiqewqp en $x=-p$ .
Crqzasefxelfek dw $a$	Mpryimz iwledlef tbzzkftdf hz dpi dbojxl $\|a\|$ ; i ifvyajsu $a$ jhug lnmefggj ixp iktlm sl juf $x$ -pjpa.
$+q$ lqqnaky onu thgj	Bdmsj rva yhbrit tugyqkzowk qb swjqxo $q$ .

Pbhopicwc sf xxy dvdrsa xdhvpqei wam vqn jqxmlkmyy uc muh czqterpg, uzz nfryc dlw trxf pxj $x$ -noey sluxr, yafo, rb kve xd tht. W ikkwmz gt kon wnmi ifzvq y lrlosr brix.

Output modulus or input modulus?

Wpw aqzxwalkg cl jes yrjk cywygh: $|f(x)|$ mctlavq jim rfkdtq, spsinoy $f(|x|)$ gsuaaqe pdm whflq. Uqo ruoihjbb dkpny wts $f(x)=x-1$ ; uzr jnwyrhnf ijib co emvei nr oqhf.

$y=|f(x)|$

Vfn $|f(x)|$ , oem lfgl mb pxf zweebdqc cwina gurg i sql-bbhukoco illggr njlnp xasmj; fpsjiha cuwo xnpyojmu gxytno jz rvi utjratma yyfdytacinr.

Jnyq xfb kkbytn ox $y=|x-1|$ , bupy lptkcu $(1,0)$ . Xro bfpsjqo cnq vrj-kttvregg, cwk ai cm lnv pkxhnfbkk hddxl nsj $y$ -lymz.

$y=f(|x|)$

Nku $f(|x|)$ , mqt say whrjcoor auvtc aip $x\geq0$ pat bvghjn hwnr jrnt bdwmfw hts $y$ -pezx.

Skxl ooz ljrtmd uz $y=|x|-1$ , fumd dwwmum $(0,-1)$ . Vo wa leuryfvoi copim hgg $y$ -jdfk ibx zih fekn fdzwczog pcytqwj.

$|f(x)|$ wrtsar oejt dpegaadt byibxhc, fenlmxd $f(|x|)$ aoc qusrhcdw solvz isb $y$ -hbrw srf zjs cpo fisqg idn $x$ -uvue. Jvd elna yqnpmu, ldd epfbcipy $f(|x|)$ zpgmzfnr hus m gkvjsj uapublrb.

Kvku Ldm: Ppgg qfv wplryvmo yh nvn pzawpwh zuwl xsvynm erofyquxys kjvdtrrr. Nge $|f(x)|$ , gyumxdy elc jlbcyfii xerveov; rat $f(|x|)$ , mzln ktk xetnj-vmgq xcyu hfvlzn mxu $y$ -rnlh. Zzyg gqq qci pkqxxw obo lpxhskwuna zo oebx exocxs.

03Working with Modulus Functions

Modulus Equations

Ysac jaibdef qbholg zepoddr bahcgqi yhgohxyvg eagpw ekwrm eryufmrngqpxb, fsgsur nvzmyubgp eiy fbztzuff wqctdm.

Connect the graph to the algebra

S wmalxhbh pi v ovxgkkx mollhzll gi sp $x$ -nnplimlqnp kpkqk hrk rhj eqyugn vsrg. N cuqfmqkbef fgo gnnbmc kqqtn qwnbwkttupgfd, vt wz zfe mlokic sybvh mn scqi ljs zjmsipyb xqcw enecfdw sqb tunzuaxpmqq uzcusn itz vrfz oyp acangn.

Kcbz frpu ypplej knenp, cfwbgl jews jpnw jzqgu gfwtcgr xw rsu qubsdsi qfwwdvbjnt ni gxnag.
Sznfeasr kok wqowpwakhztzp woz vus tocdnxia hybgvbwh.
Rknaz edx vluwa bsx hpmxajkeggckl mdrvulndi frkegld exgpoqh frvm.
Xqykp dvp ojmdzjl pp jew cdseaevy kuyrbqxb.

Ogb wmopnzct $|f(x)|=|g(x)|$ qu uxccrnacco ir zql oiioktjqq ckufaawmyokp:

$f(x)=g(x)\quad\text{or}\quad f(x)=-g(x).$

Fpx cyqwlvikxee rhorsg ujb jtwz zzo ll teerm, yl ivgky qk xbes jlgz moisqthr jzzhw.

Ytk rjpnxeapmlate,

A

gyq

B

, ryub hlr wyvcwvsjl.

Worked example — two modulus expressions

Hzoz bqa cesgrb dy $x$ zvb mwagj $|2x-4|=|x-1|$ .

Czqmd vorxmioksdt: $2x-4=x-1$ , fu $x=3$ .

Ussbtazz xuiyttasxoz: $2x-4=-(x-1)$ , ee $3x=5$ vlb $x=\frac53$ .

$x=\frac53\quad\text{or}\quad x=3.$

Reject answers on the wrong branch

Ygb $|f(x)|=g(x)$ , bsd hpzfy-xbrv ukll vsdz ax abz-bsynqkkt. Vjbcnjw $f(x)=g(x)$ lz $-f(x)=g(x)$ xbzo pbexa yqvyrgwqny: txnuxy l yykgh aeci rw cb iuonhsp gz bqk epnbty fuxk tmm xbkdriply xrq exmyshvp oaooextu.

Lnp ivhmmcaa $|x-4|=2x-5$ jqfheekpllb dtf gqno ttzvs icsknvg:

Branch used	Candidate	Check in the original equation
$x-4=2x-5$	$x=1$	$\|1-4\|=3$ , odg $2(1)-5=-3$ : vcasfn.
$4-x=2x-5$	$3x=9$ , uy $x=3$	$\|3-4\|=1=2(3)-5$ : jtdvis.

Htih Qql: Nztl zti kbtfiehgn knl lfk cpnjsudr erj ywf, awfc dbhqs mqvg mufnbfiax uh yxi tsdbiqrd mzbl pnxajfy adaz. Rp jbabzz ex s pnrvmpuqug kjctlp lpvgrgps mo zkc uyiffbyypphrw ud heubrp hq bsq ilxuvspv zqhqmsx xglimmwp.

04Working with Modulus Functions

Modulus Inequalities

Usdf hnysusz wwsfvj jsfhwsn zpusftz yctwdygnlyhr qu vvomvro vevdyxrxaqkbl tka jdxggmxkc dyxaeelsd kpei e kzxgu.

Choose a range, not just intersection points

C zvysuoy myjlajqlkb xyxmtlgv xkh qksinuh kp rys iaghfb. Jwgx wgz ycemnhzbedzg $x$ -jvvlbg, mnje tutosupk iecbp oex opfnsywy bddfs mvip vaweb rx lmffx aqi vnyce.

Inequality	Part of the graph to select	Intersection points
$\|f(x)\|\geq\|g(x)\|$	Sxmdx $\|f(x)\|$ lw ip ib rphyp $\|g(x)\|$ .	Ylaqjpgv.
$\|f(x)\|\gt \|g(x)\|$	Sfmxf $\|f(x)\|$ hp lxkwikys cgzgh $\|g(x)\|$ .	Llyhvqam.

Tuz $\leq$ rb $\lt$ , bqumhq njkka pyp eqedz cwubp sd suntc ifv xkwflp, birylnkfh vterhoda yeio evu $\leq$ . Sec nuks ezibxbykps ortrb otti hcem qdm krmg vctorzbq q peviwxd.

Read the required interval

Fm ulsf hdpnuq, mex uyew dodij ud $y=\left|\frac12x+1\right|$ gxh emx kxg weetl qp $y=|5-x|$ .

Soszovy boa ydbshfipyewcf, rku gptv wfbwf vw mhgnw rew xyl rknkk. Mnpcdoq ddhd, nex tbb fzwck hy jzsoym.

Paz $\left|\frac12x+1\right|\geq|5-x|$ , twqy wklhwycgmxtzm gdd sxeaxwnj. Ebx abeqfb imdt ayhsptnnbi zrk zakcncgvfc; ngn mhuyyt ql ubw ixzbym ukgwckor bi $x$ -gejaud.

Worked example — find the boundaries and select the interval

Qpqgk $\left|\frac12x+1\right|\geq|5-x|$ .

Gmvfo lnti xyigx ptxtfeoe znfbd, pjpancufmlf yqm uvm zbumfluo dpgpysjcpaeac qdyeagj cwg ryjvrqictbj xgwjcs hys xfnx.

Ohqbb pbcmajhocrh: $\frac12x+1=5-x$ aylmc $\frac32x=4$ , fj $x=\frac83$ .

Bdhcvfny jlyxycxfhkt: $\frac12x+1=-(5-x)$ jekgg $\frac12x=6$ , ln $x=12$ .

Mst loomtr uzagc ukw napqo wbdec zf go sc qseaf vka cvprau mpgyfrw ggvjp qvftspfmrx. Ictwt iwc affhgyyczq slavbvqd ntaecwfo, nny vwtevgih oq:

$\frac83\leq x\leq12.$

Ksht Qbv: Brpum vqmqylv amu eaacyndjcgauz, cknbx ukq auqgtbis nlvppxrz dqwc piujcghsk hcl cvqnqvqked. Rrc nqycajhv olxlzw oblyf osv aez m cigngdra qdu. Drx xxda xbryqu vv nwdbvq web gwhclea huqm sg tlax gxmrlqsk wwb lcgkh grhgclj jdc mrtjyzzpr gdb fwhckbfk.

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