Graphing Linear Absolute Value Functions II

The graphs of the absolute value function applied to other functions often have sharp turns at points where the original function changes from positive to negative. That is, there may be points at which the absolute value function of a function is not differentiable.  If you haven't yet started calculus, this just means that there may be points that are not smooth, in fact in absolute value functions these points often appear 'pointed'.

For example, the graph of $y(x)=|x^2-1|$y(x)=|x2−1| has the shape shown in the following sketch.

Critical Points

In considering the behaviour of a function that has an absolute value component, we will often need to investigate how the function changes at certain critical points. This is done with a view to writing separate expressions, without the absolute value symbols, that represent the function over particular intervals in its domain of definition.

The function $f(x)=\frac{|x^2-7x+10|}{x-2}$f(x)=|x2−7x+10|x−2​ is undefined at $x=2$x=2 since the denominator would be zero there. We observe that the denominator is positive for $x>2$x>2 and negative for $x<2$x<2.

The numerator, however, is always non-negative. So, the function is negative when $x<2$x<2 but non-negative when $x>2$x>2.

The numerator can be written as the product $|(x-2)(x-5)|$|(x−2)(x−5)| or equivalently, $|x-2||x-5|$|x−2||x−5|. So, we see that critical changes occur at $x=2$x=2 and at $x=5$x=5.

When $x<2$x<2, the numerator is positive, $(x-2)(x-5)>0$(x−2)(x−5)>0; when $22<x<5, the numerator is negative,$(x-2)(x-5)<0$(x−2)(x−5)<0 and when $x>5$x>5, $(x-2)(x-5)>0$(x−2)(x−5)>0. The numerator has the value zero at $x=5$x=5.

Thus, we can write three different function definitions, applicable over the parts of the domain.

When $x<2$x<2, $f(x)=\frac{x^2-7x+10}{x-2}=x-5$f(x)=x2−7x+10x−2​=x−5. (The function is negative and increasing.)

When $22<x≤5, $f(x)=-\frac{x^2-7x+10}{x-2}=-x+5$f(x)=−x2−7x+10x−2​=−x+5. (The function is positive and decreasing or zero.)

When $x>5$x>5, $f(x)=\frac{x^2-7x+10}{x-2}=x-5$f(x)=x2−7x+10x−2​=x−5. (The function is positive and increasing.)

Beginning to sketch

As with other functions, we can make a sketch representing the graph of the function by constructing a table of values and then plotting the points from the table. We understand the graph of a function to be the set of all points that satisfy the function definition. There are usually infinitely many such points and so, we must make a selection from them in order to make a sketch approximating the graph.

The sketch will be accurate if enough points are included from the table or if certain special points, crucial to the shape of the graph, are chosen. For absolute value functions, it is useful to include the critical points at which abrupt changes of gradient occur. That is, we look for the points at which the expression inside the absolute value brackets changes from negative to positive.

As well as finding the critical points, we need to use what is known about sketching graphs of functions that are not absolute value functions. If the function is linear, for example, we need another point to define each line segment or we need to know a gradient. Similarly, we may need the zeros and turning points of polynomial functions, and in other cases, identifying asymptotes could be important.

Example 1

The function given by $y(x)=3|x-2|+1$y(x)=3|x−2|+1, is to be explored. As part of the investigation, we wish to sketch the graph.

We see that there is a critical value, $x=2$x=2, and that $y(2)=1$y(2)=1. When $x\le2$x≤2, the function is equivalent to  $y(x)=3(2-x)+1$y(x)=3(2−x)+1 or more simply, $y(x)=-3x+7$y(x)=−3x+7. For this part of the graph, we could choose another value, $x=0$x=0 and obtain the point $(0,7)$(0,7).

When $x>2$x>2, the function is equivalent to $y(x)=3(x-2)+1$y(x)=3(x−2)+1 or more simply, $y(x)=3x-5$y(x)=3x−5. Another point on this part of the graph is the point $(3,4)$(3,4).

These three points are sufficient to determine the graph.

 

The following example is more complicated but it illustrates some of the techniques that may be needed.

Example 2

The function $f$f given by  $f(x)=|x^2+3x|-2x+1$f(x)=|x2+3x|−2x+1, has critical points where the expression $x^2+3x$x2+3x is zero.  That is, when $x=0$x=0 and when $x=-3$x=−3. 

When $x>0$x>0, and when $x<-3$x<−3, $f(x)$f(x) can be written $f_1(x)=x^2+x+1$f1​(x)=x2+x+1. Between these values, when $-3\le x\le0$−3≤x≤0, $f(x)$f(x) takes the form $f_2(x)=-x^2-5x+1$f2​(x)=−x2−5x+1.

Thus, the critical points are $(-3,7)$(−3,7) and $(0,1)$(0,1).

Now, $f_1(x)=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$f1​(x)=x2+x+1=(x+12​)2+34​ by completing the square. From this, it can be seen that $f_1$f1​ cannot be less than $\frac{3}{4}$34​ and this minimum occurs when $x=-\frac{1}{2}$x=−12​. Although this point is not in the domain of $f_1$f1​ and therefore does not belong to the graph, it will be helpful in determining the shape of the sketch. We note that $f_1$f1​ has no real zeros and its graph must be entirely above the $x$x-axis.

Similarly, we can rewrite $f_2=-x^2-5x+1$f2​=−x2−5x+1 as $f_2(x)=-\left(\left(x+\frac{5}{2}\right)^2-\frac{29}{4}\right)$f2​(x)=−((x+52​)2−294​). This has a maximum of $\frac{29}{4}$294​ at $x=-\frac{5}{2}$x=−52​. 

From the quadratic formula or otherwise, we determine that $f_2$f2​ has zeros at $\frac{5\pm\sqrt{29}}{-2}$5±√29−2​. That is, at approximately $x=-5.2$x=−5.2 and $x=0.2$x=0.2.

This collection of six points is enough to give a good indication of the shape of the graph of $f$f.

 

 

Worked Examples

Question 1

 

Question 2

Question 3