Zeros of Rational Functions (College Board AP® Precalculus): Study Guide

Zeros of rational functions

Where do the real zeros of a rational function occur?

• Let $r\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ be a rational function

  • i.e. where $p\left(x\right)$ and $q\left(x\right)$ are both polynomial functions

• The real zeros of $\mathit{r}$ correspond to

  • real zeros of $\mathit{p}$ (the function in the numerator)

  • for such values as are also in the domain of $\mathit{r}$

• What this means in practical terms is that the real zeros of $r$ occur at values of $x$where

  • $\mathit{p}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$

  • but $\mathit{q}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\ne }\mathbf{0}$

    • Remember that any values of $x$ must be excluded from the domain of $r$ that make its denominator equal to zero

• E.g. the function $h\left(x\right)=\frac{\left(x+4\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}$

  • The numerator is equal to zero when $x=-4$ or $x=3$

    • But the denominator is also equal to zero when $x=3$

  • So $h$ has only one real zero, at $x=-4$

What else do the real zeros of its numerator and denominator tell me about a rational function?

• A rational function can only be equal to zero at one of its real zeros

  • I.e., these are the only points where the graph of the function can touch or cross the $\mathit{x}$-axis

• However the real zeros of a rational function's numerator and denominator can help you determine other things about the behavior of the rational function

• In particular, for a rational function $r\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$

  • The real zeros of $p$ and $q$

  • are the endpoints or asymptotes

  • for intervals satisfying the inequalities $r\left(x\right)\ge 0$ and $r\left(x\right)\le 0$

• What this means in practical terms is this:

  • The real zeros of a rational function's numerator and denominator divide the function into a number of intervals

  • Within each of those intervals (i.e. not including any endpoints) the rational function is either positive everywhere or negative everywhere

    • So the numerator's and denominator's real zeros are the only points at which a rational function can change sign

    • (It may not change sign at such points, but it cannot change sign anywhere else)

  • And the rational function can only be equal to zero at the endpoint of such an interval which is also a real zero of the rational function

• E.g. the function $h\left(x\right)=\frac{\left(x+4\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}$

  • The numerator is equal to zero when $x=-4$ or $x=3$

    • and the denominator is equal to zero when $x=-1$ or $x=3$

  • Those values of $x$, i.e. $x=-4$, $x=-1$ and $x=3$

    • divide the domain of $h$ into four intervals

    • $x<-4$, $-4<x<-1$, $-1<x<3$ and $x>3$

  • Test a value of $x$ in the interval $x<-4$

    • $h\left(-5\right)=\frac{\left(\left(-5\right)+4\right)\left(\left(-5\right)-3\right)}{\left(\left(-5\right)+1\right)\left(\left(-5\right)-3\right)}=\frac{\left(-1\right)\left(-8\right)}{\left(-4\right)\left(-8\right)}=\frac{1}{4}>0$

      • That is positive, so $h\left(x\right)>0$ for all $x<-4$

  • You can do the same thing for the other intervals

    • $h\left(-2\right)=-2<0$, so $h<0$ for all $x$ such that $-4<x<-1$

    • $h\left(0\right)=4>0$, so $h>0$ for all $x$ such that $-1<x<3$

    • $h\left(4\right)=\frac{8}{5}>0$, so $h>0$ for all $x>3$

  • We have seen above that the only real zero of $h$ is at $x=-4$

  • Combining the above:

    • $h\left(x\right)=0$ at $x=-4$

    • $h\left(x\right)<0$ on the interval $-4<x<-1$

    • $h\left(x\right)>0$ on the intervals $x<-4$, $-1<x<3$, and $x>3$