Find the equation of An Absolute Value Function

The problem of finding the equation of an absolute value function given as a graph is essentially the inverse problem of understanding the transforming effect of the various constant terms and coefficients that can be included in the function definition. 

For example, in the equation $y=A|Bx+C|+D$y=A|Bx+C|+D we can identify the effect of each of the constants $A$A, $B$B, $C$C and $D$D.  On the other hand, if we were to observe a graph in which these effects were present, it should be possible to reconstruct the formula.

The chapter on transformations of an absolute value function contains the details.

Example 1

Explore the effect of the constants in $f(x)=-2|3x-1|+4$f(x)=−2|3x−1|+4.

The graphs of $f_1(x)=|x|$f1​(x)=|x|, $f_2(x)=|3x|$f2​(x)=|3x|, $f_3(x)=|3x-1|$f3​(x)=|3x−1|, $f_4(x)=-2|3x-1|$f4​(x)=−2|3x−1| and $f(x)=-2|3x-1|+4$f(x)=−2|3x−1|+4 are shown below.

The function $f_1(x)=|x|$f1​(x)=|x| has a critical point $(0,0)$(0,0). It combines $g_1(x)=-x,x\le0$g1​(x)=−x,x≤0 with $g_2(x)=x,x\ge0$g2​(x)=x,x≥0. The gradient is $-1$−1 to the left of zero and $1$1 to the right.

The function $f_2(x)=|3x|$f2​(x)=|3x| again has the critical point $(0,0)$(0,0). To the left of $0$0 the gradient is $-3$−3 and to the right it is $3$3.

The effect of subtracting $1$1 inside the absolute value sign is to move the graph to the right by $\frac{1}{3}$13​. This is because $f_3(x)=|3x-1|=|3(x-\frac{1}{3})|$f3​(x)=|3x−1|=|3(x−13​)|. The intercept on the vertical axis is $f_3(0)=1$f3​(0)=1.

The effect of multiplying  the absolute value bracket by $-2$−2 is to reflect the graph in the horizontal axis and to make its gradient twice as steep. The gradient to the left of $0$0 becomes $6$6 and to the right it is $-6$−6. Note that $f_4(x)=-2|3x-1|=-|2||3x-1|=-|6x-2|$f4​(x)=−2|3x−1|=−|2||3x−1|=−|6x−2|. The intercept on the vertical axis is $f_4(0)=-2$f4​(0)=−2.

Finally, when $4$4 is added to get $f(x)=-2|3x-1|+4$f(x)=−2|3x−1|+4, every function value is increased by $4$4. The intercept on the vertical axis becomes $-2+4=2$−2+4=2. The graph is symmetrical about the line $x=\frac{1}{3}$x=13​. To the left of $x=\frac{1}{3}$x=13​ the function takes the form $f(x)=-2(1-3x)+4=6x+2$f(x)=−2(1−3x)+4=6x+2 and to the right it is $f(x)=-6x+6$f(x)=−6x+6.

Example 2

Find an equation that would give the following graph.

The function is linear in two parts.with the critical point $(1,-1)$(1,−1). To the left of $x=1$x=1, the graph has gradient $-2$−2 and to the right, the gradient is $2$2. At $x=0$x=0, the function value is $1$1 for the left-hand part and $-3$−3 for the right-hand part.

Since the vertex of the graph has function value $-1$−1, we might consider a function whose vertex is at $0$0 and then subtract $1$1 from it. That is, $u(x)-1$u(x)−1 for some function $u$u.

We require an absolute value function that is represented by $-2x+1$−2x+1 to the left of $x=1$x=1, and $2x-3$2x−3 on the right. That is, $(-2x+2)-1$(−2x+2)−1 on the left and $(2x-2)-1$(2x−2)−1 on the right of $x=1$x=1.

Thus, the function defined by $y(x)=2|x-1|-1$y(x)=2|x−1|−1 should have the required properties.

You should check that the $2$2 accounts for the gradients of $-2$−2 and $2$2, the $-1$−1 inside the absolute value bracket has moved the vertex from $x=0$x=0 one unit to the right, and the $-1$−1 at the end has moved the graph downwards by one unit.

Worked Examples

Question 1

Question 2

Question 3