Intersections between a parabola with an absolute value and a line

Statement

Find the intersections between $y=x^2-|x|$ and the line $y=2-\frac12 x$.

Solution

The absolute value is removed case by case, obtaining two systems: $\begin{cases} y=x^2-x \\ y=2-\frac12 x \end{cases}\ (x\ge0) \qquad \begin{cases} y=x^2+x \\ y=2-\frac12 x \end{cases}\ (x<0).$ The parabola $y=x^2-x$ has vertex $V\left(\frac12,-\frac14\right)$; the one $y=x^2+x$ has vertex $V'\left(-\frac12,-\frac14\right)$.

Case $x\ge0$: $x^2-x=2-\frac12 x \implies 2x^2-x-4=0 \implies x=\dfrac{1+\sqrt{33}}{4}\approx1{,}69$ (the other root is negative, not acceptable).

Case $x<0$: $x^2+x=2-\frac12 x \implies 2x^2+3x-4=0 \implies x=\dfrac{-3-\sqrt{41}}{4}\approx-2{,}35$ (the other root is positive, not acceptable).

The two intersections have abscissae $\boxed{x=\dfrac{1+\sqrt{33}}{4}}$ and $\boxed{x=\dfrac{-3-\sqrt{41}}{4}}$.

Links

Topics: Parabola
Concepts: Line–parabola intersection · Absolute value · Vertex
Functions: Parabola
Skills: Reasoning by cases · Solving systems
Exercise type: Solving a system