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1.03 Absolute value functions

Lesson

Concept summary

The parent absolute value function f\left(x\right) = \left|x\right| takes an input and gives an output of the absolute value of that number. The equation of absolute value function contains a variable expression inside absolute value bars; a function of the form f\left(x\right) = a\left|x - h\right| + k.

The absolute value function f\left(x\right)=\left|x\right| has two cases to consider:

  • If x \geq 0, then f\left(x\right)=x
  • If x < 0, then f\left(x\right)=-x

As a result, the graph of an absolute value function looks like two rays that meet at a common point, called its vertex.

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The parent function has a minimum vertex, but with a reflection the vertex can be a maximum.

Worked examples

Example 1

Consider the function f\left(x\right) = \left|x - 1\right| + 2

a

Describe the transformations used to get from the graph of y = \left|x\right| to the graph of this function.

Approach

We should check for each type of transformation: translations (vertical and/or horizontal), dilation (either compression or stretch), and reflection.

Solution

We can see that there is no reflection across the x-axis since the sign outside the absolute value symbol is positive.

There is also no dilation because there is no coefficient outside of the absolute value symbol.

Inside the absolute value sign we have x-1. The -1 represents a horizontal translation of 1 unit to the right.

Outside the absolute value symbol we have +2. This represents a vertical translation of 2 units upward.

b

Draw a graph of the function

Approach

We can start with the graph of y = \left|x\right| and use the transformations found in the previous part to graph the function f\left(x\right) = \left|x - 1\right| + 2.

Solution

We know that if we start from y = \left|x\right|, we need to move 1 unit to the right, and 2 units upward. This results in the following graph:

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c

State the coordinates of the vertex

Solution

The vertex of an absolute value function is its maximum or minimum point. We can see that this function has a minimum at \left(1, 2\right) by looking at the graph.

Reflection

The vertex of an absolute value function in the form f\left(x\right)=\pm a\left\vert x-h\right\vert +k is always \left(h,k\right).

d

State the domain and range of the function, using interval notation.

Approach

Remember that the domain is the set of all x-values that can be put into the function, while the range is the set of all function values (i.e. y-values) that can be obtained by the function.

Solution

Looking at the graph, we can see that the domain will be "all real values", as any value of x can be used as an input.

The vertex of this function is at \left(1, 2\right) and is a minimum point, so the range will be "all values greater than or equal to 2."

Using interval notation, we have:

  • Domain: \left(-\infty, \infty\right)
  • Range: \left[2, \infty\right)

Example 2

Consider the absolute value graph shown below:

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a

Describe the transformations used to get from the graph of y = \left|x\right| to the graph of this function.

Approach

We should check for each type of transformation: Translations (vertical and/or horizontal), dilation (either compression or stretch), and reflection.

It may help to add the graph of y = \left|x\right| to the same coordinate plane:

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x
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y

Solution

We can see that the graph has a maximum point rather than a minimum point, so there has been a reflection across the x-axis.

The vertex of the graph is at \left(-3, 0\right). This is 3 units to the left of the vertex of y = \left|x\right| and at the same y-value. So there has been a horizontal translation of 3 units to the left, and no vertical translation.

We can also see that the rate of change of the two graphs are different. The slopes of the sides of this function are 2 and -2, while the slopes of y = \left|x\right| are 1 and -1. So there has been a vertical stretch by a factor of 2.

Reflection

For linear and absolute value functions, a changed slope indicates a dilation. In this case, we stated that it was a vertical stretch with a factor of 2, but it could equivalently be thought of as a horizontal compression with a factor of \dfrac{1}{2} (as long as the compression happens before the horizontal translation).

b

Determine an equation for the function shown in the graph

Approach

We can use the transformations that we described in part (a) and apply them to the function y = \left|x\right| to get an equation for the function shown.

Solution

Starting with y = \left|x\right|, we can apply the transformations one at a time:

\displaystyle y\displaystyle =\displaystyle \left|x\right|Parent function
\displaystyle y\displaystyle =\displaystyle -\left|x\right|Reflection across x-axis
\displaystyle y\displaystyle =\displaystyle -2\left|x\right|Vertical stretch by a factor of 2
\displaystyle y\displaystyle =\displaystyle -2\left|x + 3\right|Horizontal translation of 3 units to the left

So the equation of the function is y = -2\left|x + 3\right|.

Reflection

It is important to note that the order of transformations matters in some cases. For example, applying a vertical translation and then a reflection across the x-axis will be different to reflecting first and then applying the same vertical translation.

In this case, the transformations of a vertical stretch, a reflection across the x-axis, and a horizontal translation are all independant and can be applied in any order.

Outcomes

A2.F.IF.B.4

Graph functions expressed algebraically and show key features of the graph by hand and using technology.*

A2.F.BF.B.3

Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs.

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP6

Attend to precision.

A2.MP7

Look for and make use of structure.

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