Advanced Absolute Value Graphs

Lesson Overview

Instructor Presentation

Introduction

We have already learned how to transform absolute value graphs in the following ways:

Translating (shifting)

Reflections across axes.

Stretching

However, incorporating the Absolute Value function in other ways can produce some very interesting graphs.  We are going to explore some of these options.

Goals

Students will develop a more fundamental understanding of what the basic absolute value graph really means.

parameters={a,c,d,x}

Students will be able to graph functions and their absolute value.

Students will be able to predict the behavior of graphs with absolute value operators inserted in various parts of the function.

Concept Exploration

Plot the line y = x - 5

Plot[x-5,{x,-10,10},PlotRange->{-10,10},AspectRatio->Automatic,PlotStyle->Thick]

Now, plot y = x - 5 and y = |x - 5| together on the same graph.  What do you observe?

Plot[{x-5,Abs[x-5]},{x,-10,10},AspectRatio->Automatic,PlotRange->{-10,10},PlotStyle->{{Thin,Blue},{Thick,Green}}]

Every part of y = x - 5 (blue) that was above the x-axis is still above the x-axis for y = |x - 5| (green).

Every part of y = x - 5 (blue) that was above the x-axis is still above the x-axis for y = |x - 5| (green).

What is the difference in the blue function and the green function?

g (x) = 2 f (x).  (The red function values are all twice the          corresponding blue function' s values.)

Example 1

For the following descriptions, plot the graphs that are described on the same set of axes.

1)  y = (x^2) - 8,

(a) The parabola as is in blue.

(b) The absolute value of the parabola in orange.

(c)  Graph (b) shifted up 2 and left 4 in green.

2)  What do you notice about the location of the sharp points of the orange graph?

3)  Plot the graph of g(x) = |f(x)| such that f(x) is a parabola with no horizontal or vertical stretch and such that g(x) has sharp points at (-6,0) and (2,0).

Hint:  Use your answer to #2 above to find the function f(x).

Expand[(x-(-6))*(x-2)]

Concept Exploration

Absolute Value inside the function.

What happens if we take the absolute value of just a portion of a graph, instead of the entire graph?  Let's look at this situation a little more closely.

Plot the graph of y = (x^2) - 2x + 1.

Plot[(x^2)-2x+1, {x,-10,10}, PlotRange->{-10,10},PlotStyle->{Thick,Purple}]

Now, plot y = (x^2) - 2|x| + 1 on the same set of axes.

Plot[{(x^2)-2x+1,(x^2)-2Abs[x]+1}, {x,-10,10}, PlotRange->{-10,10},PlotStyle->{{Thick,Purple},{Thick,Orange}}]

How did the |x| in the Orange transform the initial (purple) graph?

The |x| "erased" everything to the left of the y-axis and reflected all parts of the graph to the right of the y-axis about the y-axis.

Is the absolute value operator commutative?  (i.e., will the same orange graph result if the absolute value operator is placed around any term or terms?)  Let's investigate.  (next slide).

Example 2

Graph the following functions and their reciprocals.  Make a prediction about what the graph will look like before plotting.

1)  y = -(x^2)+5x+6

2)  y = |-(x^2)+5x+6|

Prediction:

3)  y = -(x^2) + 5|x| + 6

Prediction:

4)  y = |-(x^2)+5x| + 6

Prediction:

Example 3

Move the sliders on the following demonstration and answer the questions below.

Manipulate[Plot[(x^2)-h*Abs[x]-12,{x,-10,10},AspectRatio->Automatic,PlotRange->{-20,3},PlotStyle->{Thick, Red}],{h,1,5}]

1)  What happens as h increases?

Move the sliders in this similar demonstration.  Click on the plus sign beside the slider to open the animation controls.  Answer the question below.

Manipulate[Plot[(x^2)-h*Abs[x]-12,{x,-10,10},AspectRatio->Automatic,PlotRange->{-20,3},PlotStyle->{Thick, Green}],{h,-5,5}]

2)  What happens to the graph as h goes from negative to positive?

3)  What does this imply about the "original function," or the parabola without the absolute value function inside?

4)  To check and see if your answer to number 3 is correct, copy the demonstration from number 2, and using Mathematica's list notation { }, graph a static absolute value graph and a parabola with the parameter h that changes.

Answer Key

Example 1

1)  y = (x^2) - 8,

(a) The parabola as is in blue.

(b) The absolute value of the parabola in orange.

(c)  Graph (b) shifted up 2 and left 4 in green.

Plot[{(x^2)-8, Abs[(x^2)-8], Abs[(((x+4))^2)-8]+2},{x,-10,10},PlotRange->{-10,10},AspectRatio->Automatic, PlotStyle->{{Thick,Blue},{Thick,Orange},{Thick,Green}}]

2)  What do you notice about the location of the sharp points of the orange graph?

They occur at the x-intercepts of the parabola.

3)  Plot the graph of g(x) = |f(x)| such that f(x) is a parabola with no horizontal or vertical stretch and such that g(x) has sharp points at (-6,0) and (2,0).

Expand[(x-(-6))*(x-2)]

Plot[Abs[(x^2)+4x-12],{x,-10,10},PlotRange->{-2,20},AspectRatio->Automatic,PlotStyle->{Thick,Red}]

Example 2

1)  y = -(x^2)+5x+6

Plot[-(x^2)+5x+6,{x,-10,10},PlotRange->{-5,15},AspectRatio->Automatic,PlotStyle->{Thick,Blue}]

2)  y = |-(x^2)+5x+6|

Prediction: answers may vary.

Plot[Abs[-(x^2)+5x+6],{x,-10,10},PlotRange->{-5,15},AspectRatio->Automatic,PlotStyle->{Thick,Blue}]

3)  y = -(x^2) + 5|x| + 6

Prediction: answers may vary

Plot[-(x^2)+5Abs[x]+6,{x,-10,10},PlotRange->{-5,15},AspectRatio->Automatic,PlotStyle->{Thick,Blue}]

4)  y = |-(x^2)+5x| + 6

Prediction: answers may vary.

Plot[Abs[-(x^2)+5x]+6,{x,-10,10},PlotRange->{-5,15},AspectRatio->Automatic,PlotStyle->{Thick,Blue}]

Example 3

1)  What happens as h increases?

As h increases, the vertex moved farther away from the y-axis and became more negative.

2)  What happens to the graph as h goes from negative to positive?

The second "hump" disappeared and the smooth vertex of the parabola is "replaced" with a sharp point on the y-axis.

3)  What does this imply about the "original function," or the parabola without the absolute value function inside?

In general, the coefficient of the linear term has a lot to do with where, vertically, the vertex of the parabola lies.

4)  To check and see if your answer to number 3 is correct, copy the demonstration from number 2, and using Mathematica's list notation { }, graph a static absolute value graph and a parabola with the parameter h that changes.

Manipulate[Plot[{(x^2)-Abs[x]-12,(x^2)-h x-12},{x,-10,10},AspectRatio->Automatic,PlotRange->{-20,3},PlotStyle->{{Thick, Green},{Thick,Red}}],{h,-5,5}]