Rational Functions

Domain of a Rational Function

To find the domain, set the denominator equal to $0$, solve, and exclude those values. Factoring usually does the job.

• $x^2 + 1$ is never $0$ (a square is never negative), so nothing is excluded and the domain is all real numbers.

• $x^2 + 6x + 9 = (x+3)^2$ is $0$ only at $x = -3$.

• $x^2 - 5x + 6 = (x-2)(x-3)$ is $0$ at $x = 2$ and $x = 3$.

Here is the graph of $f(x) = \dfrac{5x-2}{x^2-9}$. See what happens near $x = -3$ and $x = 3$.

Intercepts, Sign, and Solving $f(x) = c$

To find the y-intercept, compute $f(0)$ (if $0$ is in the domain). To find the x-intercepts, solve $P(x) = 0$, then check that those solutions do not also make the denominator $0$.

Why? Between two consecutive "critical" numbers neither $P$ nor $Q$ can be zero, so each keeps a constant sign there.

To solve $f(x) = c$, multiply both sides by the denominator, solve the resulting equation, and check that the answer is in the domain.

Symmetry: Even and Odd Rational Functions

Before graphing, check for symmetry. If $f(-x) = f(x)$ for every $x$ in the domain, $f$ is even (symmetric about the y-axis). If $f(-x) = -f(x)$, it is odd (symmetric about the origin). If neither holds, there is no symmetry of this kind.

The domain must be symmetric too: if $a$ is in the domain, $-a$ must be as well.

A good shortcut: even numerator over even denominator is even, odd over even is odd, and even over odd is odd.

Vertical Asymptotes and Holes

Holes. If a factor cancels, you get a hole, not an asymptote. The $y$-coordinate of the hole comes from the simplified function.

Asymptotes with a parameter. If the denominator has a parameter, use the discriminant to see when it has real zeros. For example, take $f(x) = \dfrac{1}{x^2 - kx + 4}$. The discriminant is $k^2 - 16$. If $-4 < k < 4$, it is negative, the denominator never vanishes, and there are no vertical asymptotes. If $k = \pm 4$, there is one double zero and one asymptote. If $|k| > 4$, there are two asymptotes. Always check that the numerator doesn't vanish at the same place.

Horizontal and Oblique Asymptotes

Why? As $x \to \pm\infty$ the highest-degree terms dominate. With equal degrees they leave the ratio of the leading coefficients. If the numerator has one more degree, division gives a linear quotient $mx + b$ plus a remainder term that shrinks to $0$.

For example, $\dfrac{3x^2+1}{x^2+5}$ has horizontal asymptote $y = 3$, and $\dfrac{x+1}{x^2+1}$ has horizontal asymptote $y = 0$.

The leftover fraction also tells you which side of the asymptote the curve is on. Here $f(x) - (2x-3) = \dfrac{1}{x-1}$. For $x > 1$ this is positive, so the curve is above the line as $x \to +\infty$. For $x < 1$ it is negative, so the curve is below the line as $x \to -\infty$.

Putting It Together: Analyzing a Rational Function

When asked to analyze a rational function, follow this checklist:

1. Factor numerator and denominator.

2. Find the domain and any holes (cancelled factors).

3. Find the intercepts.

4. Find the vertical asymptotes (zeros of the denominator that remain).

5. Find the horizontal or oblique asymptote.

6. Make a sign chart.

Working backward: finding parameters

If you know the asymptotes, you can find unknown coefficients. For $f(x) = \dfrac{ax+3}{2x+b}$ with vertical asymptote $x = 4$ and horizontal asymptote $y = 3$:

• The horizontal asymptote is $\dfrac{a}{2} = 3$, so $a = 6$.

• The denominator is $0$ at $x = 4$: $2(4) + b = 0$, so $b = -8$.

• Check that the numerator isn't $0$ at $4$: $6(4) + 3 = 27 \neq 0$. ✓

For an oblique asymptote, write the numerator as (quotient)(denominator) + remainder. Suppose $f(x) = \dfrac{ax^2+bx+c}{x-2}$ has oblique asymptote $y = 3x + 4$ and passes through $(3, 10)$. Then $ax^2 + bx + c = (3x+4)(x-2) + r = 3x^2 - 2x - 8 + r$. So $a = 3$, $b = -2$, $c = r - 8$. The point $(3, 10)$ gives $\dfrac{27 - 6 + c}{1} = 10$, so $c = -11$ (and $r = -3$).

Reading a graph

For a function of the form $f(x) = \dfrac{ax^2}{x^2+b}$:

• The horizontal asymptote is $y = a$.

• The graph always passes through the origin, since $f(0) = 0$.

• It has vertical asymptotes only if $b < 0$, at $x = \pm\sqrt{-b}$. If $b > 0$, the denominator never vanishes.

• It is even, so its graph is symmetric about the y-axis.

Read the asymptotes off the graph and match them to $a$ and $b$.

The Reciprocal Function and Its Transformations

Lines and a hyperbola. To see where a line meets a rational curve, set the expressions equal, clear the denominator, and get a quadratic. A line meets the curve in exactly one point when the discriminant is $0$ (a tangent line), or when the equation degenerates into a linear equation with one solution.

Inverses and Compositions

To find the inverse of a rational function, swap $x$ and $y$ and solve for $y$. The unknown ends up in both numerator and denominator, so you collect the terms with $y$ on one side and factor.

The domain of $f^{-1}$ is the range of $f$, and the vertical asymptote of $f^{-1}$ is the horizontal asymptote of $f$ (and vice versa).

Composition. To compute $(f \circ f)(x)$, substitute $f(x)$ into $f$ and simplify the complex fraction by multiplying top and bottom by the common denominator. If $(f \circ f)(x) = x$, then $f$ is its own inverse, so the graph of $f$ is symmetric about the line $y = x$.

Rational Functions in the Real World: Costs and Rates

Many real quantities are shared out over a changing amount: a fixed cost spread over the units produced, a total job time shared between two workers, a dose spread over time. All of them are rational functions, and the horizontal asymptote has a concrete meaning: the long-run value of the quantity.

Notice that $C(x)$ gets close to $6$ but never equals it, because $\dfrac{400+6x}{x} = 6$ would require $400 = 0$.

Work problems. If one person finishes a job alone in $a$ hours and another in $b$ hours, together they finish $\dfrac1a + \dfrac1b$ of the job per hour. The time together is the reciprocal of that rate.

Back to the beginning: Aaliyah's cost per shirt is always a bit above $\$6$, and the more she prints, the closer it gets. The horizontal asymptote $y = 6$ is the floor.

Rational Models: Bounds and Growth

Two algebraic tricks help you study a rational model without calculus.

Trick 1: bounding the output with a discriminant. To see which values $y = f(t)$ can take, treat $y$ as a given number, clear the denominator, and view the equation as a quadratic in $t$. The value $y$ is attainable only if that quadratic has a real solution, so its discriminant must be $\geq 0$.

Trick 2: rewrite by polynomial division. When the degrees are equal, divide to get $f(t) = A - \dfrac{B}{t^2+1}$. Then you can read off both the long-term value ($A$) and whether the function increases or decreases.

Summary