Curve sketching with a parameter

Problem

Study, as the real parameter $m$ varies, the function $f(x) = \frac{2x+1}{x^2+m+|m|}.$

Solution

The term $m+|m|$ depends on the sign of $m$: it equals $0$ if $m\le 0$ and $2m$ if $m>0$. It is therefore convenient to reason by cases.

Case $m<0$. Then $|m|=-m$, so $m+|m|=0$ and $f(x) = \frac{2x+1}{x^2}.$ Existence condition: $x\ne 0$. The function has a vertical asymptote at $x=0$ and a horizontal asymptote $y=0$. We have $f'(x) = -\frac{2}{x^2}-\frac{2}{x^3}.$ The complete study (symmetries, sign with $f>0$ for $x>-\tfrac{1}{2}$ and $x\ne 0$, derivative, inflection point at $x=-\tfrac{3}{2}$) leads to the graph.

Case $m>0$. Then $m+|m|=2m>0$ and $f(x) = \frac{2x+1}{x^2+2m}.$ The denominator is $>0$ for every $x$: the domain is $\mathbb{R}$ and there is no vertical asymptote. The qualitative behaviour is completely different.

Case $m=0$. We have $m+|m|=0$, so $f(x) = \dfrac{2x+1}{x^2}$, the same as the case $m<0$.

Conclusion. For each case one carries out the complete study and draws the graph. The qualitative transition occurs at $m=0$: for $m\le 0$ there is a vertical asymptote at $x=0$, for $m>0$ there is not.

Links

Topics: Curve sketching
Concepts: Asymptote · Domain · Parameter · Sign · Curve sketching
Functions: Rational function
Methods: Curve sketching
Skills: Reasoning by cases · Sketching a function
Exercise type: Curve sketching