# 3.6 Absolute value functions

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In this section you will:
• Graph an absolute value function.
• Solve an absolute value equation.

Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will continue our investigation of absolute value functions .

## Understanding absolute value

Recall that in its basic form $\text{\hspace{0.17em}}f\left(x\right)=|x|,\text{\hspace{0.17em}}$ the absolute value function is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems.

## Absolute value function

The absolute value function can be defined as a piecewise function

$\text{\hspace{0.17em}}f\left(x\right)=|x|=\left\{\begin{array}{ccc}x& \text{if}& x\ge 0\\ -x& \text{if}& x<0\end{array}\text{\hspace{0.17em}}$

## Using absolute value to determine resistance

Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often $\text{\hspace{0.17em}}\text{±1%,}\text{\hspace{0.17em}}±\text{5%,}\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}±\text{10%}\text{.}$

Suppose we have a resistor rated at 680 ohms, $\text{\hspace{0.17em}}±5%.\text{\hspace{0.17em}}$ Use the absolute value function to express the range of possible values of the actual resistance.

We can find that 5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance $\text{\hspace{0.17em}}R\text{\hspace{0.17em}}$ in ohms,

$|R-680|\le 34$

Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.

using the variable $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ for passing, $\text{\hspace{0.17em}}|p-80|\le 20$

## Graphing an absolute value function

The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin    in [link] .

[link] shows the graph of $\text{\hspace{0.17em}}y=2|x–3|+4.\text{\hspace{0.17em}}$ The graph of $\text{\hspace{0.17em}}y=|x|\text{\hspace{0.17em}}$ has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at $\text{\hspace{0.17em}}\left(3,4\right)\text{\hspace{0.17em}}$ for this transformed function.

## Writing an equation for an absolute value function given a graph

Write an equation for the function graphed in [link] .

The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See [link] .

We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in [link] .

From this information we can write the equation

what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
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Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice