How To Draw Abosolute Value Graphs

Recollect of points $a$ and $b$ on a number line and assume that $a=2$ and $b=-3.$

Comparison the ii numbers, we can easily say $a>b$ just considering $a$ is positive and $b$ is negative. What if we do not intendance most the sign but only the altitude of each number from zip? Then we say $\lvert b \rvert > \lvert a \rvert$ or $3>2,$ where $\lvert \cdot \rvert$ is the accented value notation that gives the respective values of $b$ and $a$ without regard to their signs. Hence the post-obit definition:

For any existent number $x$ we define its absolute value $\lvert x \rvert$ as follows:

$\lvert x \rvert = \begin{cases} -ten && \mbox{if }ten< 0\\ ten && \mbox{if } 10 \geq 0. \end{cases}$

Then why accented values? When do we consider only the deviation from zero? Suppose your firm is located on an east-w road and your machine parked in the driveway has been stolen. If the thief is close enough, you desire to exit and take hold of him yourself. Otherwise, you volition telephone call the police. According to information gathered, the car thief is said to have driven abroad east for $ii$ miles and, for whatsoever reason, have turned around and driven w for $three$ miles.

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Taking east equally positive and w negative, the thief is currently $ii-3=-1$ mile away, i.eastward. his location is $i$ mile due west of your house. Still, remember that the just gene that y'all take into consideration as to whether to go out yourself or call the constabulary is the thief's distance from your house regardless of the direction. And then the adding relevant to yous right at this moment would be $\lvert 2-3 \rvert=ane$ as opposed to $ii-3=-1.$ This is, for example, where absolute value plays a role. Make sense?

At present, let united states of america think about something more interesting. Could the stolen car's distance from your business firm $\left(d_1\right)$ ever exist greater than the distance the machine traveled $\left( d_2\right)$ to become there? We know from above that the distance of the car from your house is $d_1=\lvert two-3 \rvert=ane$ mile. Since the altitude traveled past the auto is $d_2=\lvert 2 \rvert + \lvert -3 \rvert =5>1=d_1,$ the reply to the question $"\,d_1>d_2?\,"$ is no.

What if the thief traveled farther e, instead of turning back and heading w, for $3$ miles? Would your answer still exist no? Allow us see. In this case, the distance of the car from your house in miles is $d_1=\lvert two+3 \rvert=5,$ whereas the altitude traveled by the car in miles is $d_2=\lvert 2 \rvert + \lvert 3 \rvert = 5.$

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Since these 2 numbers are equal, the answer to the above question is still no. That is, your stolen automobile's distance from your business firm can never be greater than the distance the auto traveled to get there. In general, for any real numbers $ten$ and $y$ $\lvert ten+y \rvert \le \lvert x \rvert + \lvert y \rvert,$ which is part of the following:

Properties of absolute value:

$|ten|=|-x|$

$-|x|\leq x \leq |x|$

$|x|=|y|$ if and just if $10 = y$ or $x = -y$

$|x^n|=|x|^n$ for any positive integer $n$

$|xy|=|x| \cdot |y|$

$\left|\frac{ten}{y}\right|=\frac{|x|}{|y|}$ if $y \neq 0$

$|x\pm y|\leq |10|+|y|$

Hither are some examples of how to calculate accented values using the above properties:

Calculate $|three.5|-|-2.5|.$

We take

$|iii.5|-|-2.5|=3.v-two.5=i. \ _\square$

Summate $|\pi - two| + |\pi - 3| + |two\pi - vii|.$

The quantities $\pi - two$ and $\pi - 3$ are positive, so they remain unchanged when the absolute bars are dropped. However, because $2\pi - vii$ is negative, we have

$|\pi - two| + |\pi - three| + |2\pi - 7|= (\pi - 2) + (\pi - iii) + {\color{#D61F06}{ (7 - two\pi)}} = two. \ _\foursquare$

Evaluate $|v\times 6|.$

We have

$|5\times 6|=|30|=30. \ _\square$

Detect the value of $\left\lvert 2\times \left(\frac{ii}{three}-0.five\correct) \right\rvert.$ Express your respond as a fraction in its lowest term.

We have

$\begin{aligned} \left| ii\times\left(\frac{two}{3}-0.5\correct)\correct| &=\left\lvert ii\left(\frac{one}{half dozen}\right)\correct\rvert\\ &=\left\lvert \frac{1}{three}\right\rvert\\ &=\frac{1}{three}. \ _\foursquare \end{aligned}$

Now, it is time for a graphical interpretation of accented value. Take a look at the following graph of $y=\lvert x \rvert.$

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Exercise you run across that information technology is fatigued exactly by the definition of absolute value and the value of $y$ is always not-negative? Do yous see that the dotted red line, which is $y=x$ for $x<0,$ is flipped over against the $ten$ -axis and then that the negative values of $y$ become positive?

Equally regards to the machine thief instance, let us say that the car owner wants to get out and catch the thief himself if the bad guy is within $iv$ miles from his firm. Then in the first scenario where the thief ends upwards $1$ mile west of his house $(\text{i.e. }x=-ane)$ and thus the thief is $y=\lvert x \rvert=-(-10)=-(-1)=i$ mile abroad from his house, he will not call the police but go out himself. On the contrary, in the second scenario where the thief ends up $v$ miles eastward of his house $(\text{i.e. }ten=five)$ and thus the thief is $y=\lvert ten \rvert=10=5$ miles away from his house, he will call the police.

At present, what if the automobile owner's house was located at $x=three$ instead of $x=0$ in the start place and the car thief behaved in the same manner? How different will the graph expect? The reply is right below:

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Since the house is located at $10=3$ under this new circumstance, the thief in the first scenario does not end up $x=-1$ but $x=3+(-i)=2$ miles east of $x=0$ and thus his altitude from the owner'southward house is $y=\lvert 10-3 \rvert=-(10-iii)=-(2-three)=one$ mile. $($ This would make sense to yous if you noticed that $y=\lvert x \rvert$ in the to a higher place example can exist considered as $y=\lvert x-0\rvert.)$

In the second scenario, the thief does not end upwardly $x=5$ but $x=iii+5=8$ miles east of $x=0$ and thus his altitude from the possessor'south house is $y=\lvert x-3 \rvert=10-3=eight-3=5$ miles. Hence the equation of the above graph is $y=\lvert x-3 \rvert,$ where the dotted red line, which is the graph of $y=x-three$ for $x<three,$ is flipped over against the $10$ -axis at $ten=iii$ considering the benchmark indicate is now $x=three$ instead of $x=0.$

At present that we know how to get the graph of $y=\lvert x-3 \rvert,$ which is $Five$ -shaped, let u.s. try to become a $W$ -shaped graph. To exercise that, we first interpret the graph of $y=\lvert 10-3 \rvert$ by $two$ in the negative direction of the $y$ -centrality, as shown below, and the equation of the translated graph is $y=\lvert x-3 \rvert-2.$

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And so nosotros ask, "What volition the graph of $y=\big\lvert \lvert ten-iii \rvert -ii \big\rvert$ look like?" Since all the values of $y$ for $one<x<5$ in the graph of $y=\lvert x-3 \rvert-2$ are negative, nosotros flip over that role of the graph to obtain the following, merely as we did above:

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Do you really come across a $West$ -shaped graph now? I question that naturally arises in your heed would be how to draw the graph of $y=\big\lvert \lvert x-3 \rvert -two \big\rvert$ from scratch without referring to the thief example. Nosotros can utilize linear inequalities to accomplish this.

Observe from the equation $y=\large\lvert \lvert 10-3 \rvert -2 \large\rvert$ that if $\lvert x-3 \rvert \ge 2,$ then $y=\lvert 10-three \rvert - 2.$ This can be rewritten in more particular every bit the following 2 cases: $\brainstorm{aligned} 10-3\ge 2 \Leftrightarrow ten\ge 5 \Rightarrow y&=(x-iii)-ii\\ &=ten-v &\qquad (1) \\ x-3\le -2 \Leftrightarrow 10\le 1 \Rightarrow y&=-(x-3)-two\\ &=-x+ane. &\qquad (2) \end{aligned}$ Can you confirm that $(one)$ corresponds to the above graph for $x\ge 5$ and $(two)$ corresponds to the above graph for $x\le one?$ I think you already did.

Similarly, observe from the equation $y=\big\lvert \lvert x-3 \rvert -2 \big\rvert$ that if $\lvert x-3 \rvert < 2,$ and then $y=-(\lvert x-3 \rvert - 2)=-\lvert x-iii \rvert + 2,$ which is equivalent to $-2<x-three<2 \Leftrightarrow 1<x<five \Rightarrow y=-\lvert 10-iii \rvert + 2.$ This can be rewritten in more detail as the post-obit two cases: $\begin{aligned} i<x<3 \Rightarrow y&=-(-( ten-iii)+ 2)\\ &=x-1 &\qquad (3)\\ three\le x<5 \Rightarrow y&=-(( x-3)+ 2)\\ &=-x+5. &\qquad (4)\\ \end{aligned}$ Again, tin you lot ostend that $(3)$ corresponds to the in a higher place graph for $1<ten<iii$ and $(four)$ corresponds to the to a higher place graph for $3\le x<v?$ I am sure you lot already did.

Now, it'southward fourth dimension for you lot to try some examples.

Which of the following is the graph of $y=\frac{x}{\lvert x \rvert}$ for $x\ne 0?$

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If $x>0,$ and so $\lvert x \rvert=10,$ which implies $y=\frac{10}{\lvert 10 \rvert}=\frac{x}{x}=1.$ If $x<0,$ and then $\lvert x \rvert=-10,$ which implies $y=\frac{x}{\lvert 10 \rvert}=\frac{ten}{-ten}=-1.$ Therefore, the correct answer is $(c). \ _\square$

Which of the following conditions does NOT always satisfy $10=\lvert x \rvert?$

(a) $x$ is a non-negative real number.
(b) $x$ is a positive real number.
(c) $x$ is a positive integer.
(d) $x$ is a non-zero real number.

If $ten\ge 0,$ then $\lvert x \rvert=x,$ which implies $x=\lvert 10 \rvert$ holds for all non-negative real numbers. Hence, $(a), (b)$ and $(c)$ always satisfy $ten=\lvert x \rvert.$

If $ten<0,$ then $\lvert x \rvert=-ten,$ which implies $x=\lvert ten \rvert$ never holds for $x<0.$

Therefore, the correct answer is $(d). \ _\foursquare$

Given that $a,b,c$ are non-zero real numbers , discover all possible values of the expression $\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}$ .

Since $\dfrac{10}{|ten|}=1$ for any $x>0$ and $\dfrac{10}{|10|}=-one$ for any $ten<0$ ,

$\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}= -3$ if $a,b,c$ are all negative;

$\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|} = -1$ if exactly two of $a,b,c$ are negative;

$\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|} = 1$ if exactly i of $a,b,c$ is negative;

$\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}= 3$ if $a,b,c$ are all positive.

Thus , the possible values of the given expression are $-three,-1,one$ and $3. \ _\square$

Solve the equation $|10-2| + |ten-3| = 1.$

Nosotros need to discuss 3 cases: $\begin{cases} x \leq 2 \\ ii<10\leq three \\ 3<x. \end{cases}$

When $10 \leq two$ ,
$|x-2|+|ten-3|=1 \Rightarrow (2-x)+(iii-x)=i \Rightarrow x=2.$

When $2<x\leq 3$ ,
$|10-2|+|x-iii|=1 \Rightarrow (x-2)+(3-ten)=1 \Rightarrow$ any $x \in (2,3]$ is a solution.

When $x>3$ ,
$|x-two|+|10-3|=1 \Rightarrow (ten-2)+(ten-three)=1 \Rightarrow x=3.$
Only since we assumed $10>iii$ , there is no solution for the example $x>3$ .

In determination, the solutions to the equation $|x-ii|+|ten-3|=i$ are $two \leq 10 \leq iii$ . $_\square$

Find the minimum value of $|10+1|+|x-2|+|x-3|$ .

Case 1 : When $x \leq -1$ ,

$|10+one|+|10-2|+|ten-3| = -(10+1)-(x-two)-(x-3)=4-3x \geq 7.$

Case two: When $-one<x \leq two$ ,

$|x+1|+|x-ii|+|ten-3| = (x+1)-(x-2)-(ten-3)=vi-ten \geq 4.$

Case 3: When $2\leq x <3$ ,

$|x+one|+|10-two|+|x-iii|= (10+1)+(x-ii)-(x-3)=10+2 \geq 4.$

Case four: When $3 \leq x$ ,

$|x+i|+|x-2|+|x-3| = (x+ane)+(x-2)+(x-3)=3x-four \geq 5.$

Thus the global minimum of the given expression is 4. $_\foursquare$