ORCCA Introduction to Absolute Value Functions

Section11.1Introduction to Absolute Value Functions

This section will introduce the basic concepts behind absolute value functions and their graphs. This information will also be useful at the end of this chapter when we solve absolute value equations and inequalities.

Figure11.1.1Alternative Video Lesson

Subsection11.1.1Definition of Absolute Value

Recall that in Section 1.3, we defined the absolute value of a number to be the distance between that number and \(0\) on a number line. Also recall that this causes the output of the absolute value function to never be a negative number since we are under the presumption that “distance” is always positive (or zero).

Example11.1.2

Since the number \(5\) is \(5\) units from \(0\text{,}\) then \(\abs{5}=5\text{.}\)
Since the number \(-3\) is \(3\) units from \(0\text{,}\) then \(\abs{-3}=3\text{.}\)

Example11.1.3

Yonas takes a \(5\)-block walk north from his home to a food cart. After enjoying dinner, he then walks \(9\) blocks south of the food cart to his favorite movie theater.

How many blocks has Yonas walked in total when he reaches the theater?
How many blocks is Yonas from home when he reaches the theater?

Solution

Since we only care about total distance, we can ignore the “signs” on the distances walked (either north or south) and simply add the two values together. Mathematically, if we think of north as positive values and south as having negative values, this situation is the same as

\begin{align*} \abs{5}+\abs{-9}\amp=5+9\\ \amp=14 \end{align*}

Yonas walked a total of \(14\) blocks when he reached the theater.
When he reaches the theater, Yonas's actual position could be thought of as \(5+(-9)\text{.}\) But the actual distance from the theater to his home is better thought of as:

\begin{align*} \abs{5+(-9)}\amp=\abs{-4}\\ \amp=4 \end{align*}

Yonas was \(4\) blocks from home when he reached the theater.

Subsection11.1.2Evaluating Absolute Value Functions

The formula \(f(x)=\abs{x}\) does satisfy the requirements for \(f\) to be a function because no matter what number you put in for \(x\text{,}\) there is only one measured distance from \(0\) to that value \(x\text{.}\)

Example11.1.4

Let \(f(x)=\abs{x}\) and \(g(x)=\abs{2x-5}\text{.}\) Evaluate the following expressions.

\(f(34)\)
\(f(-63)\)
\(f(0)\)
\(g(13)\)
\(g(1)\)

Solution

\(\begin{aligned}[t] f(\highlight{34})\amp=\abs{\highlight{34}}\\ \amp=34 \end{aligned}\)
\(\begin{aligned}[t] f(\highlight{-63})\amp=\abs{\highlight{-63}}\\ \amp=63 \end{aligned}\)
\(\begin{aligned}[t] f(\highlight{0})\amp=\abs{\highlight{0}}\\ \amp=0 \end{aligned}\)
\(\begin{aligned}[t] g(\highlight{13})\amp=\abs{2\cdot\highlight{13}-5}\\ \amp=\abs{21}\\ \amp=21 \end{aligned}\)
\(\begin{aligned}[t] g(\highlight{1})\amp=\abs{2\cdot\highlight{1}-5}\\ \amp=\abs{-3}\\ \amp=3 \end{aligned}\)

Checkpoint11.1.5

Subsection11.1.3Graphs of Absolute Value Functions

Absolute value functions have generally the same shape. They are usually described as “V”-shaped graphs and the tip of the “V” is called the vertex. A few graphs of various absolute value functions are shown in Figure 11.1.6.

A Cartesian graph with the graph of y=abs(x), which looks like a V with the vertex of the V at the origin. The right side extends as a line with slope 1 and the left side extends with slope -1. — (a)\(y=\abs{x}\)

A Cartesian graph with the graph of y=-abs(x+2), which looks like an upside down V with the vertex of the V at the point (-2,0). The right side extends as a line with slope -1 and the left side extends with slope 1. — (a)\(y=\abs{x}\)

Example11.1.7

Let \(h(x)=-2\abs{x-3}+5\text{.}\) Using technology, create table of values with \(x\)-values from \(-3\) to \(3\text{,}\) using an increment of \(1\text{.}\) Then sketch a graph of \(y=h(x)\text{.}\)

Solution

Here is the table and the graph.

\(x\)	\(y\)
\(-3\)	\(-7\)
\(-2\)	\(-5\)
\(-1\)	\(-3\)
\(0\)	\(-1\)
\(1\)	\(1\)
\(2\)	\(3\)
\(3\)	\(5\)

A Cartesian graph with the graph of y=-2*abs(x-3)+5, which looks like an upside down V with the vertex of the V at the point (3,5). The right side extends as a line with slope -2 and the left side extends with slope 2.

Table11.1.8Table for \(y=h(x)\text{.}\)

Figure11.1.9Graph of \(y=h(x)\)

Example11.1.10

Let \(j(x)=\big\lvert\mathopen{}\abs{x+1}\mathclose{}-2\big\rvert-1\text{.}\) Using technology, create table of values with \(x\)-values from \(-5\) to \(5\text{,}\) using an increment of \(1\) and sketch a graph of \(y=j(x)\text{.}\)

Solution

This is a strange one because it has an absolute value within an absolute value.

\(x\)	\(y\)
\(-5\)	\(1\)
\(-4\)	\(0\)
\(-3\)	\(-1\)
\(-2\)	\(0\)
\(-1\)	\(1\)
\(0\)	\(0\)
\(1\)	\(-1\)
\(2\)	\(0\)
\(3\)	\(1\)
\(4\)	\(2\)
\(5\)	\(3\)

A Cartesian graph with the graph of y=abs(abs(x+1)-2)+1, which looks like a W with the vertices of the W at the points (-3,-1), (-1,1), and (1,-1). The right side extends as a line with slope 1 and the left side extends with slope -1.

Table11.1.11A table of values for \(y=j(x)\text{.}\)

Figure11.1.12\(y=\big\lvert\mathopen{}\abs{x+1}\mathclose{}-2\big\rvert-1\)

Subsection11.1.4Another Definition for Absolute Value

How many definitions do we really need? Bear with us because this one is important.

Example11.1.13

Consider the function \(f\) defined by \(f(x)=\sqrt{x^2}\text{.}\) First, we will evaluate this function at a few arbitrary values: \(3\text{,}\) \(0\text{,}\) and \(-5\text{.}\)

\begin{align*} f(\highlight{3})\amp=\sqrt{\highlight{3}^2}\amp f(\highlight{0})\amp=\sqrt{\highlight{0}^2}\amp f(\highlight{-5})\amp=\sqrt{(\highlight{-5})^2}\\ \amp=\sqrt{9}\amp\amp=\sqrt{0}\amp\amp=\sqrt{25}\\ \amp=3\amp\amp=0\amp\amp=5 \end{align*}

These results should seem familiar: \(f(3)=3\text{,}\) \(f(0)=0\text{,}\) and \(f(-5)=5\text{.}\) The outputs are the same as the inputs, except for a missing negative sign on \(5\text{.}\) It seems like we've seen a function that does that exact same thing already …

Make a quick graph using technology to see what the graph of \(y=f(x)\) looks like.

Solution

Figure 11.1.14 shows a graph of \(f\) where \(f(x)=\sqrt{x^2}\text{.}\) It looks just like that of \(y=\abs{x}\text{.}\)

A Cartesian graph with the graph of y=sqrt(x^2) which looks exactly like the graph of y=abs(x). This looks like a V with the vertex of the V at the origin. The right side extends as a line with slope 1 and the left side extends with slope -1.

Since the graphs of \(y=\sqrt{x^2}\) and \(y=\abs{x}\) match up exactly, that must mean that

\begin{equation*} \abs{x}=\sqrt{x^2} \end{equation*}

This fact will be used later in this chapter and it will continue to pop up in subsequent math courses here and there as important.

Figure11.1.14\(y=\sqrt{x^2}\)

Example11.1.15

Simplify the following expressions using the fact that \(\abs{x}=\sqrt{x^2}\text{.}\)

\(\sqrt{(x-2)^2}\)
\(\sqrt{x^6}\)
\(\sqrt{x^2+10x+25}\)
\(\sqrt{x^4}\)

Solution

\(\sqrt{(\highlight{x-2})^2}=\abs{\highlight{x-2}}\text{.}\) Note that \(\highlight{x-2}\) might be a negative number depending on the value of \(x\text{,}\) so the absolute value will change those negative numbers to be positive values.
\begin{align*} \sqrt{x^6}\amp=\sqrt{\left(\highlight{x^3}\right)^2}\\ \amp=\abs{\highlight{x^3}} \end{align*}

We know from exponent rules that \(\left(x^3\right)^2=x^6\text{.}\) Note that \(\highlight{x^3}\) will be negative whenever \(x\) is a negative number.
\begin{align*} \sqrt{x^2+10x+25}\amp=\sqrt{(\highlight{x+5})^2}\\ \amp=\abs{\highlight{x+5}} \end{align*}

Note again that \(\highlight{x+5}\) can be negative for certain values of \(x\text{.}\)
\begin{align*} \sqrt{x^4}\amp=\sqrt{\left(\highlight{x^2}\right)^2}\\ \amp=\abs{\highlight{x^2}} \end{align*}

Note here that since \(\highlight{x^2}\) is never negative (no matter what number you plug into \(x^2\text{,}\) you always either get a positive number or zero out), that the absolute value around \(x^2\) doesn't have any effect. Absolute values change negative numbers to positive values but leave positive values alone. Thus, it is ok in this case to drop the absolute value symbols. The complete simplification is:

\begin{align*} \sqrt{x^4}\amp=\sqrt{\left(\highlight{x^2}\right)^2}\\ \amp=\abs{\highlight{x^2}}\\ \amp=x^2 \end{align*}

Subsection11.1.5Applications Involving Absolute Values

Absolute values are quite useful as models in a variety of real world applications. One example is the path of a billiards (pool) ball: when the ball bounces off one of the side rails, its path is mirrored and creates a “V” shape. The game gets more complicated when more than the rail is hit, but the fundamental mathematics doesn't change: absolute values model the bounces each time.

Here are some more examples. The first one we'll explore involves light reflecting off of a mirror.

Example11.1.16

When light reflects off of a mirror, the path it takes is in the shape of an absolute value graph. Khenbish was playing with a laser pointer in his bedroom mirror. He set up the laser pointer on his windowsill and the light hit the center of the mirror and reflected onto the corner of his room. He declared that the laser pointer is sitting at the origin, and \(x\) should stand for the horizontal distance from the left wall to the light beam. Shown is a birds-eye view of the situation.

a Cartesian graph to show the top view of the room with a line starting at the origin and traveling upward, bouncing off the mirror and hitting the bottom corner of the room, making an upside down V-shaped graph

Figure11.1.17Birds-Eye View of Khenbish's Room with Laser

After a little bit of work, Khenbish was able to come up with a formula for the light's path:

\begin{equation*} p(x)=5-\frac{5}{4}\abs{x-4} \end{equation*}

where \(p(x)\) stands for the position, in ft, above (for positive values) or below (for negatives) the center line through his room that represents the \(x\)-axis, where \(x\) is also measured in ft. Use technology and a graph of this formula to answer the following questions.

Khenbish's room is 10 ft wide according to Figure 11.1.17 (in the vertical direction in the figure). What is the room's length (in the horizontal direction in the figure)?
How far along the wall is the mirror centered?
If you stood 9 ft from the left wall, how far above or below the room's center line (\(x\)-axis) should you stand to have the laser pointer hit you?

Solution

the previous graph with a grid added so it is clear that the mirror is at the point (4,5) and the point where the light hits the corner is (12,-5) — Figure11.1.18Detailed Birds-Eye View of Khenbish's Room

To find the room's length, first note that since the laser hits the corner of the room, the \(x\) coordinate of the lasers position would tell us the room's width. According to the detailed graph, the \(x\)-coordinate when \(y=-5\) is \(12\text{.}\) So the room must be \(12\) feet wide.
The mirror is centered exactly where the laser hits the wall. This is the vertex of the absolute value graph which, according to the graph, is at the point \((4,5)\text{.}\) This tells us that the mirror is centered \(4\) feet from the left wall.
If you are standing 9 feet from the left wall, the laser's position will be a bit more than one foot behind the rooms center line, by the diagram. While technology can tell us the exact answer, here is how to do this problem algebraically.

\begin{align*} d(\highlight{9})\amp=5-\frac{5}{4}\abs{\highlight{9}-4}\\ \amp=5-\frac{5}{4}\abs{5}\\ \amp=5-\frac{5}{4}\cdot 5\\ \amp=5-\frac{25}{4}\\ \amp=5-6.25\\ \amp=-1.25 \end{align*}

So, it looks like if you stand \(9\) feet from the left wall, you need to stand \(1.25\) feet behind the center line (which would be \(6.25\) feet from the wall with the mirror on it) to be hit by the laser.

Absolute value functions are also used when a value must be within a certain distance or tolerance. For example, a person's body temperature is considered “normal” if it is within \(0.5\) degrees of 98.6 °F, so their temperature could be up to \(0.5\) degrees less than or greater than that temperature. To be within normal range, the difference between the two values must be less than or equal to \(0.5\text{,}\) and it does not matter whether it is positive or negative. We will introduce a function for measuring this in the next example.

Example11.1.19

The function \(D\) defined by \(D(T)=\abs{T-98.6}\) represents the difference between a person's temperature, T, in Fahrenheit, and 98.6 °F. A person's temperature is considered “normal” if \(D(T)\) is less than or equal to \(0.5\text{.}\) Use \(D(T)\) to evaluate each person's temperature.

If a person has a temperature of 98.3 °F, is that within normal range?
If a person has a temperature of 99.3 °F, is that within normal range?
If a person has a temperature of 97.3 °F, is that within normal range?

Solution

If a person has a temperature of 98.3 °F, then we have:

\begin{align*} D(\highlight{98.3})\amp=\abs{\highlight{98.3}-98.6}\\ \amp=\abs{-0.3}\\ \amp=0.3 \end{align*}

Since the value of \(D(98.3)\) is a number smaller than \(0.5\text{,}\) then 98.3 °F must be within normal range.
If a person has a temperature of 99.3 °F, then we have:

\begin{align*} D(\highlight{99.3})\amp=\abs{\highlight{99.3}-98.6}\\ \amp=\abs{0.7}\\ \amp=0.7 \end{align*}

Since the value of \(D(99.3)\) is a number bigger than \(0.5\text{,}\) then 99.3 °F must not be within normal range.
If a person has a temperature of 97.3 °F, then we have:

\begin{align*} D(\highlight{97.3})\amp=\abs{\highlight{97.3}-98.6}\\ \amp=\abs{-1.3}\\ \amp=1.3 \end{align*}

Since the value of \(D(97.3)\) is a number bigger than \(0.5\text{,}\) then 97.3 °F must not be within normal range.

Example11.1.20

The entryway to the Louvre Museum in Paris is through I. M. Pei's metal and glass Louvre Pyramid. This pyramid has a square base and is \(71\) feet high and \(112\) feet wide. The formula \(h(x)=71-\frac{71}{56}\abs{x-56}\) gives the height above ground level of the pyramid at a distance of \(x\) from the left side of the pyramid base.

A Cartesian graph of y=71-71/56*abs(x-56), which looks like an upside down V from (0,0) to (56, 71) and back down to (112,0).

If you are \(20\) feet from the left edge, how high will the pyramid rise in front of you? Round your result to the nearest tenth of an inch.
How far from the left edge is the center of the pyramid?
Using your previous answer, check that the formula gives you the correct height at the center.

Figure11.1.21A Diagram of the Front of the Louvre Pyramid

Solution

If you are \(20\) feet from the left edge, then \(x\) is \(20\text{.}\) Substituting \(20\) for \(x\) we have

\begin{align*} h(\substitute{20})\amp=71-\frac{71}{56}\abs{\substitute{20}-56}\\ \amp=71-\frac{71}{56}\abs{-36}\\ \amp=71-\frac{71}{56}\cdot 36\\ \amp\approx 71-45.643\\ \amp\approx 25.357\\ \amp\approx 25.4 \end{align*}

The pyramid is about \(25.4\) feet high at the position \(20\) feet from the left edge.
The center of the pyramid is \(56\) feet from the either edge since it's half of \(112\) feet.
Putting \(x=56\) into the formula for \(h\) gives us

\begin{align*} h(\substitute{56})\amp=71-\frac{71}{56}\abs{\substitute{56}-56}\\ \amp=71-\frac{71}{56}\abs{0}\\ \amp=71-\frac{71}{56}\cdot 0\\ \amp=71 \end{align*}

And so the formula does give us the correct maximum height of \(71\) feet at the center of the pyramid.

SubsectionExercises

Review of Basic Computing with Absolute Value

1

Find the absolute value of this number.

\(\displaystyle{ |{-10}|= }\)

2

Find the absolute value of this number.

\(\displaystyle{ |{-7}|= }\)

3

Evaluate the following expressions which involve the absolute value:

\(\displaystyle{ \left| 3 \right| = }\)
\(\displaystyle{ \left| -4 \right| = }\)
\(\displaystyle{ \left| 0 \right| = }\)
\(\displaystyle{ \left| {11+\left(-10\right)} \right| = }\)
\(\displaystyle{ \left| {-8-\left(-1\right)} \right| = }\)

4

Evaluate the following expressions which involve the absolute value:

\(\displaystyle{ \left| 4 \right| = }\)
\(\displaystyle{ \left| -7 \right| = }\)
\(\displaystyle{ \left| 0 \right| = }\)
\(\displaystyle{ \left| {15+\left(-3\right)} \right| = }\)
\(\displaystyle{ \left| {-8-\left(-3\right)} \right| = }\)

5

Evaluate the following expressions which involve the absolute value:

\(\displaystyle{ - \lvert 5-10 \rvert = }\)
\(\displaystyle{ \lvert -5-10 \rvert = }\)
\(\displaystyle{ -3 \lvert 10-5 \rvert = }\)

6

Evaluate the following expressions which involve the absolute value:

\(\displaystyle{ - \lvert 4-7 \rvert = }\)
\(\displaystyle{ \lvert -4-7 \rvert = }\)
\(\displaystyle{ -3 \lvert 7-4 \rvert = }\)

7

Evaluate these expressions:

\(\displaystyle{ -3-3\left\lvert 8-4\right\rvert = }\)
\(\displaystyle{ -3-3\left\lvert 4-8\right\rvert = }\)

8

Evaluate these expressions:

\(\displaystyle{ -2-8\left\lvert 6-3\right\rvert = }\)
\(\displaystyle{ -2-8\left\lvert 3-6\right\rvert = }\)

9

Evaluate this expression:

\(\displaystyle{ 5-8\left\lvert 4-6 \right\rvert + 4 = }\)

10

Evaluate this expression:

\(\displaystyle{ 1-3\left\lvert 1-2 \right\rvert + 3 = }\)

11

Evaluate this expression:

\(\displaystyle{ 2-8\left\lvert -5+(4-5)^{3}\right\rvert = }\)

12

Evaluate this expression:

\(\displaystyle{ 3-6\left\lvert -1+(3-5)^{3}\right\rvert = }\)

Use function notation with absolute value functions.

13

Given \(h\) defined by \(h(y) = \left|y - 268\right|\text{,}\) find and simplify \(h(137)\text{.}\)

\(h(137)={}\)

14

Given \(G\) defined by \(G(x) = \left|x - 334\right|\text{,}\) find and simplify \(G(149)\text{.}\)

\(G(149)={}\)

15

Given \(H\) defined by \(H(y) = {\left|-3y-1\right|}\text{,}\) find and simplify \(H(16)\text{.}\)

\(H(16)={}\)

16

Given \(h\) defined by \(h(x) = {\left|-2x-22\right|}\text{,}\) find and simplify \(h(17)\text{.}\)

\(h(17)={}\)

17

Given \(g\) defined by \(g(t) = {11-\left|4t+18\right|}\text{,}\) find and simplify \(g(18)\text{.}\)

\(g(18)={}\)

18

Given \(g\) defined by \(g(y) = {19-\left|-y-3\right|}\text{,}\) find and simplify \(g(20)\text{.}\)

\(g(20)={}\)

19

Given \(G\) defined by \(G(x) = {x+\left|3x-24\right|}\text{,}\) find and simplify \(G(10)\text{.}\)

\(G(10)={}\)

20

Given \(h\) defined by \(h(x) = {x+\left|-3x+15\right|}\text{,}\) find and simplify \(h(11)\text{.}\)

\(h(11)={}\)

21

Given \(f\) defined by \(f(x) = {\left|x^{2}+x-20\right|}\text{,}\) find and simplify \(f(2)\text{.}\)

\(f(2)={}\)

22

Given \(H\) defined by \(H(x) = {\left|x^{2}+2x-24\right|}\text{,}\) find and simplify \(H(-7)\text{.}\)

\(H(-7)={}\)

23

Given \(F\) defined by \(F(x) = {\left|x^{2}-16\right|}\text{,}\) find and simplify \(F(-1)\text{.}\)

\(F(-1)={}\)

24

Given \(g\) defined by \(g(x) = {\left|x^{2}-25\right|}\text{,}\) find and simplify \(g(6)\text{.}\)

\(g(6)={}\)

Domain

25

Find the domain of \(G\) where \(\displaystyle{G(x)=\lvert {-10x-4} \rvert}\text{.}\)

26

Find the domain of \(H\) where \(\displaystyle{H(x)=\lvert {3x+4} \rvert}\text{.}\)

27

Find the domain of \(K\) where \(\displaystyle{K(x) = 5x - \lvert {-4x-8} \rvert}\text{.}\)

28

Find the domain of \(f\) where \(\displaystyle{f(x) = 3x - \lvert {10x+1} \rvert}\text{.}\)

Make a Table.

29

Make a table of values for the function \(g\) defined by \(g(x)={\left|2x-3\right|}\text{.}\)

\(x\)	\(g(x)\)

30

Make a table of values for the function \(g\) defined by \(g(x)={\left|-x+2\right|}\text{.}\)

\(x\)	\(g(x)\)

31

Make a table of values for the function \(h\) defined by \(h(x)={\left|x^{2}-x-3\right|}\text{.}\)

\(x\)	\(h(x)\)

32

Make a table of values for the function \(F\) defined by \(F(x)={\left|x^{2}+x-2\right|}\text{.}\)

\(x\)	\(F(x)\)

33

Make a table of values for the function \(G\) defined by \(G(x)={2\!\left|-2x+1\right|-3}\text{.}\)

\(x\)	\(G(x)\)

34

Make a table of values for the function \(G\) defined by \(G(x)={-3\!\left|3x-3\right|-1}\text{.}\)

\(x\)	\(G(x)\)

35

Make a table of values for the function \(H\) defined by \(H(x)=\Big\lvert{\left|-2x+3\right|+3}\Big\rvert\text{.}\)

\(x\)	\(H(x)\)

36

Make a table of values for the function \(K\) defined by \(K(x)=\Big\lvert{3\!\left|-2x+3\right|+3}\Big\rvert\text{.}\)

\(x\)	\(K(x)\)

Make graphs of absolute value functions.

37

Graph \(y=f(x)\text{,}\) where \(f(x)=\left\lvert2x-1\right\rvert\text{.}\)

38

Graph \(y=f(x)\text{,}\) where \(f(x)=\left\lvert x-2\right\rvert\text{.}\)

39

Graph \(y=f(x)\text{,}\) where \(f(x)=\left\lvert x^2-2x-1\right\rvert\text{.}\)

40

Graph \(y=f(x)\text{,}\) where \(f(x)=\left\lvert x^2+3x-2\right\rvert\text{.}\)

41

Graph \(y=f(x)\text{,}\) where \(f(x)=\frac{1}{2}\left\lvert 4x-5\right\rvert-3\text{.}\)

42

Graph \(y=f(x)\text{,}\) where \(f(x)=\frac{3}{4}\left\lvert 6+x\right\rvert+2\text{.}\)

43

Graph \(y=f(x)\text{,}\) where \(f(x)=\big\lvert2\left\lvert3-x\right\rvert-2\big\rvert\text{.}\)

44

Graph \(y=f(x)\text{,}\) where \(f(x)=\big\lvert3 - 2\left\lvert2x - 3\right\rvert\big\rvert\text{.}\)

Absolute value and square roots.

45

Simplify the expression. Do not assume the variables take only positive values.

\(\displaystyle{{\sqrt{81x^{2}}}}\)

46

Simplify the expression. Do not assume the variables take only positive values.

\(\displaystyle{{\sqrt{36y^{2}}}}\)

47

Simplify the expression.

\(\displaystyle{{\sqrt{\left(z-15\right)^{2}}}}\)

48

Simplify the expression.

\(\displaystyle{{\sqrt{\left(t-80\right)^{2}}}}\)

49

Simplify the expression.

\(\displaystyle{\sqrt{\left(-9804\right)^2}}\)

50

Simplify the expression.

\(\displaystyle{\sqrt{\left(-11802\right)^2}}\)

51

Simplify the expression.

\(\displaystyle{\sqrt{a^{2}+14a+49}}\)

52

Simplify the expression.

\(\displaystyle{\sqrt{r^{2}+18r+81}}\)

53

The height inside a camping tent when you are \(d\) feet from the edge of the tent is given by

\(\displaystyle{h={-1.1\!\left|d-6.8\right|+6.5}}\)

where \(h\) stands for height in feet. Determine the height when you are:

\({7.6\ {\rm ft}}\) from the edge.

The height inside a camping tent when you \({7.6\ {\rm ft}}\) from the edge of the tent is
\({3\ {\rm ft}}\) from the edge.

The height inside a camping tent when you \({3\ {\rm ft}}\) from the edge of the tent is

54

The height inside a camping tent when you are \(d\) feet from the edge of the tent is given by

\(\displaystyle{h={-0.7\!\left|d-4\right|+6.5}}\)

where \(h\) stands for height in feet. Determine the height when you are:

\({6.3\ {\rm ft}}\) from the edge.

The height inside a camping tent when you \({6.3\ {\rm ft}}\) from the edge of the tent is
\({3.4\ {\rm ft}}\) from the edge.

The height inside a camping tent when you \({3.4\ {\rm ft}}\) from the edge of the tent is

55

Write two numbers so that

The first number is less than the second number, and
The absolute value of the first number is greater than the absolute value of the second number

and