Radical and Rational Functions
Radical and Rational Functions
This domain combines two function families that share a concern for restrictions: square-root graphs have limited domains, and rational graphs have excluded values that create asymptotes or holes.
Radical functions. For $y=a\sqrt{b(x-h)}+k$, the radicand must be non-negative, so the domain comes from solving $b(x-h)\ge 0$. The graph is half a parabola opening sideways, beginning at the vertex $(h,k)$. For cube roots the domain is all reals and the graph has an inflection point at $(h,k)$ rather than an endpoint. Solving radical equations requires isolating the radical, squaring both sides, and then checking for extraneous roots, because squaring can create solutions the original equation never had.
Rational functions. For $y=\dfrac{P(x)}{Q(x)}$, first factor and cancel. Cancelled factors produce holes (point discontinuities) at the cancelled zero; remaining zeros of $Q$ give vertical asymptotes. Compare degrees to find horizontal asymptotes: if $\deg P<\deg Q$ the asymptote is $y=0$; if degrees are equal it is the ratio of leading coefficients; if $\deg P>\deg Q$ there is no horizontal asymptote (possibly a slant one when the difference is exactly 1). Intercepts come from $P(x)=0$ for $x$-intercepts and $P(0)/Q(0)$ for the $y$-intercept.
Traps. Students forget to cancel before locating asymptotes, so they report a hole as an asymptote. They also cross the horizontal asymptote when sketching, which is allowed, but they mistakenly believe a graph can cross a vertical asymptote, which it cannot. Sign analysis across asymptotes is frequently tested.
Worked example. Let $f(x)=\dfrac{x^2-9}{x^2-x-6}=\dfrac{(x-3)(x+3)}{(x-3)(x+2)}$. Cancelling gives a hole at $x=3$ (value $6/5$), a vertical asymptote $x=-2$, and horizontal asymptote $y=1$ since degrees match.