Graphing absolute values: piecewise formulas, reflections, and root counts

Two basic ways to handle an absolute value

For any real $u$,

$$|u|=\begin{cases}-u,&u<0,\\u,&u\ge0.\end{cases}$$

A graph involving absolute values can therefore be drawn by splitting the domain where the expressions inside change sign. A useful special case is $y=|f(x)|$: retain the parts of $y=f(x)$ above the horizontal axis and reflect every part below it across that axis.

The reflection rule applies to an absolute value around the whole output. It cannot be applied indiscriminately to every expression that happens to contain absolute values.

Example: two linear absolute values

For $y=|x-1|+|x-3|$, the sign-change points are $1$ and $3$. Thus

$$y=\begin{cases}4-2x,&x<1,\\2,&1\le x<3,\\2x-4,&x\ge3.\end{cases}$$

The pieces meet continuously at $(1,2)$ and $(3,2)$.

Example: reflect a cubic

First draw $y=x^3-3x$. Its derivative is $3(x-1)(x+1)$, so it has a local maximum $2$ at $x=-1$ and local minimum $-2$ at $x=1$.

The factorization $x^3-3x=x(x-\sqrt3)(x+\sqrt3)$ gives three zeros: $-\sqrt3,0,\sqrt3$.

Reflect the portions with negative output to obtain $y=|x^3-3x|$:

The three zeros stay in the same positions. Reflection changes heights, not horizontal coordinates.

Problem 1: an area bounded by a V-shaped graph

Problem. Find the area enclosed by $y=|2x+2|+2$ and $y=6$.

Split at $x=-1$:

$$y=\begin{cases}-2x,&x<-1,\\2x+4,&x\ge-1.\end{cases}$$

The vertex is $(-1,2)$.

The line $y=6$ intersects the two branches at $x=-3$ and $x=1$. The enclosed triangle has base length four and height four, so its area is

$$\boxed{\frac12\cdot4\cdot4=8.}$$

The source presents this as a past numerical-reasoning problem from a Japanese civil-service examination. Its key step is recognizing the elementary triangle rather than introducing an unnecessary integral.

Problem 2: the least uniform bound on an interval

Problem. For a given real number $a$, find the least real $k$ such that

$$|x^2+x+a|\le k$$

for every $x$ with $-1\le x\le1$.

This asks for the maximum of the absolute value on the interval. Complete the square:

$$x^2+x+a=\left(x+\frac12\right)^2+a-\frac14.$$

The vertex lies inside the interval, giving minimum value $a-1/4$. The maximum of the unreflected quadratic occurs at $x=1$ and is $a+2$. Hence its output range is exactly $[a-1/4,a+2]$, and

$$k_{\min}=\max\left(\left|a-\frac14\right|,|a+2|\right).$$

The two extremes have equal absolute value when $a-1/4=-(a+2)$, or $a=-7/8$. Consequently

$$\boxed{k_{\min}(a)=\begin{cases}\frac14-a,&a\le-\frac78,\\a+2,&a\ge-\frac78.\end{cases}}$$

At the shared boundary both formulas equal $9/8$.

The range calculation handles all $a$, including values for which the entire original graph lies below the axis. If one additionally minimizes over $a$, the best choice is $a=-7/8$ and the least possible bound is $9/8$; that is a separate optimization from finding $k_{\min}$ for a fixed $a$.

Problem 3: count distinct real roots

Problem. For each real $a$, determine the number of distinct real solutions of

$$x|x-3|=ax+1.$$

Write the fixed graph as

$$y=x|x-3|=\begin{cases}-x^2+3x,&x<3,\\x^2-3x,&x\ge3.\end{cases}$$

We count its intersections with the line $y=ax+1$.

The branch x < 3

Its intersections satisfy

$$x^2+(a-3)x+1=0.$$

Tangency occurs when $(a-3)^2-4=0$, giving $a=1$ or $a=5$. The double roots are $x=1$ and $x=-1$, respectively, so both are in the correct branch.

An intersection can also pass through the branch boundary $(3,0)$. This occurs when $3a+1=0$, or $a=-1/3$.

For $a<-1/3$, the left quadratic has one root below three and one above three, so only one counts. At $a=-1/3$, its roots are $1/3$ and $3$; the latter is excluded from this branch. For $-1/3<a<1$, both roots are below three. For $1<a<5$, the discriminant is negative. For $a>5$, both roots are negative and therefore below three. At $a=1,5$, the double root contributes one distinct intersection.

The branch x >= 3

Here the equation is

$$x^2-(a+3)x-1=0.$$

Its two roots have opposite signs. The positive one lies at or above three precisely when the polynomial's value at three is nonpositive:

$$-3a-1\le0\quad\Longleftrightarrow\quad a\ge-\frac13.$$

Thus this branch contributes one root when $a\ge-1/3$ and none otherwise. The boundary root $x=3$ is counted here only, preventing duplication.

Combine the counts

The total number of distinct real roots is:

• $a<-1/3$: one.
• $a=-1/3$: two.
• $-1/3<a<1$: three.
• $a=1$: two.
• $1<a<5$: one.
• $a=5$: two.
• $a>5$: three.

A discriminant calculation alone would overcount roots that lie outside a branch's domain. Check both the tangent point and every ordinary root against the interval where the chosen piecewise formula is valid.

Related: root placement in intervals and graphical parameter separation.