Rational functions and asymptotes
Tonight's idea
A rational function is one polynomial divided by another: \(f(x) = \dfrac{p(x)}{q(x)}\). The interest is where the bottom is zero and where \(x\) is enormous.
Where the bottom is zero, \(f\) is undefined in one of two ways. If the vanishing factor cancels against the top, the graph is an ordinary curve with one point missing: a hole. If not, the top stays away from zero while the bottom shrinks, and the graph shoots off to \(\pm\infty\): a vertical asymptote. Rule: note every zero of the original bottom, cancel completely, then check each: a finite value there is a hole, a bottom still zero there is an asymptote. So \((x - 1)/(x - 1)^2\) cancels a factor but leaves \(1/(x - 1)\): an asymptote at 1, not a hole.
Where \(x\) is enormous, only leading terms matter, so compare degrees. Top smaller: \(f(x) \to 0\), horizontal asymptote \(y = 0\). Degrees equal: \(f(x)\) tends to the ratio of leading coefficients. Top bigger by exactly one: divide (synthetic division from last sheet), and the quotient is a line, a slant asymptote. The graph may cross a horizontal asymptote, never a vertical one: \(f\) has no value there.
Watch one: hole or asymptote?
\(f(x) = \dfrac{x^2 - 1}{x^2 - 3x + 2} = \dfrac{(x - 1)(x + 1)}{(x - 1)(x - 2)}\). The bottom is zero at \(x = 1\) and \(x = 2\). At \(x = 1\) the factor cancels, leaving \(\dfrac{x + 1}{x - 2}\), which is \(-2\) there: a hole at \((1, -2)\). At \(x = 2\) nothing cancels: vertical asymptote \(x = 2\). Degrees are equal with leading coefficients 1 and 1, so the horizontal asymptote is \(y = 1\). Intercepts: \(f(0) = -\tfrac12\), and \(f = 0\) at \(x = -1\).
Your turn
- For \(g(x) = \dfrac{2x + 6}{x^2 - 9}\), give the domain, the coordinates of any hole, and the equations of all asymptotes.
- For \(h(x) = \dfrac{3x^2 + 1}{x^2 - 4}\), find the asymptotes and the \(y\)-intercept. Then show that the graph never crosses its horizontal asymptote.
- Use synthetic division by 1 to find the slant asymptote of \(\dfrac{x^2 + 1}{x - 1}\). Give the vertical asymptote too.
- Solve \(\dfrac{x^2}{x - 3} = \dfrac{9}{x - 3}\). One of the two candidates you find is not a solution. Which, and why?
Stretch: the missing value
Find the range of \(f(x) = \dfrac{x + 1}{x - 2}\): set \(y = f(x)\), solve for \(x\) in terms of \(y\), and see which value of \(y\) makes that impossible. How does the answer relate to the horizontal asymptote?
Answers
- \(g(x) = \dfrac{2(x + 3)}{(x - 3)(x + 3)} = \dfrac{2}{x - 3}\) for \(x \ne -3\). Domain: all \(x\) except 3 and \(-3\). Hole at \(\left(-3, -\tfrac13\right)\). Vertical asymptote \(x = 3\); horizontal asymptote \(y = 0\) (degree 1 over degree 2).
- Vertical asymptotes \(x = 2\) and \(x = -2\); horizontal asymptote \(y = 3\). \(y\)-intercept \(h(0) = -\tfrac14\). Crossing would need \(3x^2 + 1 = 3x^2 - 12\), i.e. \(1 = -12\), impossible.
- \(x^2 + 1 = (x - 1)(x + 1) + 2\), so \(\dfrac{x^2 + 1}{x - 1} = x + 1 + \dfrac{2}{x - 1}\). Slant asymptote \(y = x + 1\); vertical asymptote \(x = 1\).
- Multiplying by \((x - 3)\) gives \(x^2 = 9\), \(x = \pm 3\). But \(x = 3\) makes both denominators zero, so it is not in the domain: it is extraneous. Only \(x = -3\).
Stretch. \(y(x - 2) = x + 1\) gives \(x(y - 1) = 2y + 1\), so \(x = \dfrac{2y + 1}{y - 1}\), which exists for every \(y\) except 1. Range: all reals except 1. The missing value is exactly the horizontal asymptote: this graph approaches \(y = 1\) but never reaches it.