Unit 3, lesson 29

Rational Functions, Holes, and Asymptotes

How to find the excluded inputs of a rational function from its original denominator, and how to tell a hole from a vertical asymptote and read the far end behavior.

  • Theory part 57
  • Practice part 58
  • Piecewise, Polynomial, and Rational Graphs

Type (x2−9)/(x−3)(x^2 - 9)/(x - 3) into a graphing tool and you see what looks like the straight line y=x+3y = x + 3. Type 1/(x−3)1/(x - 3) and the graph splits in two and shoots off the screen. Both formulas break at x=3x = 3, yet they break in very different ways. Telling those two kinds of break apart, and saying what happens far out to the left and right, is exactly the work that limits will later make precise.

Key terms

Rational function

Definition A rational function is a function of the form

f(x)=p(x)q(x),f(x) = \frac{p(x)}{q(x)},

where pp and qq are polynomials and qq is not the zero polynomial. Unless a domain is stated, the domain of ff is the set of all real numbers xx with q(x)≠0q(x) \ne 0.

In plain words A rational function is one polynomial divided by another. It works everywhere except where the bottom is zero.

Example The function

f(x)=2x+1x−3f(x) = \frac{2x + 1}{x - 3}

is rational. At x=5x = 5 it gives f(5)=112f(5) = \frac{11}{2}, and at x=3x = 3 it gives nothing because the denominator is 00.

Denominator restriction

Definition The excluded inputs of a rational function p(x)/q(x)p(x)/q(x) are the real zeros of the original denominator qq. They must be found before any factor is cancelled. If p(x)=c(x) P(x)p(x) = c(x)\,P(x) and q(x)=c(x) Q(x)q(x) = c(x)\,Q(x), then

p(x)q(x)=P(x)Q(x)only for x with q(x)≠0,\begin{gathered} \frac{p(x)}{q(x)} = \frac{P(x)}{Q(x)} \\[0.4em] \text{only for } x \text{ with } q(x) \ne 0, \end{gathered}

so the simplified formula agrees with ff on the domain of ff but does not enlarge it.

In plain words Read the forbidden inputs off the denominator you were given. Cancelling makes the formula shorter, but it never gives back an input that was forbidden.

Example Take

f(x)=x2−4x2−x−2=(x−2)(x+2)(x−2)(x+1).\begin{aligned} f(x) &= \frac{x^2 - 4}{x^2 - x - 2} \\[0.4em] &= \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)}. \end{aligned}

The original denominator is zero at x=2x = 2 and x=−1x = -1, so the domain is all real numbers except 22 and −1-1. For every other xx the function equals x+2x+1\frac{x + 2}{x + 1}, but x=2x = 2 stays excluded even though x+2x+1\frac{x + 2}{x + 1} would accept it.

Reciprocal graph

Definition The reciprocal graph is the graph of y=1xy = \frac{1}{x}, defined for x≠0x \ne 0. It has the vertical asymptote x=0x = 0 and the horizontal asymptote y=0y = 0, and its two branches lie in the first and third quadrants. The graph of

y=ax−h+k,a≠0,y = \frac{a}{x - h} + k, \qquad a \ne 0,

has the same shape, moved so that its asymptotes are x=hx = h and y=ky = k, scaled vertically by the factor ∣a∣\lvert a \rvert and, when a<0a < 0, reflected in the line y=ky = k, which puts the branches to the upper left and lower right of the point (h,k)(h, k) where the asymptotes cross.

In plain words The basic picture is the curve y=1/xy = 1/x. Many simple rational functions are that curve slid and stretched, so finding its two asymptote lines tells you where it sits.

Example Long division gives

g(x)=3x−5x−2=3+1x−2,x≠2.\begin{gathered} g(x) = \frac{3x - 5}{x - 2} = 3 + \frac{1}{x - 2}, \\[0.4em] x \ne 2. \end{gathered}

So the graph of gg is the reciprocal graph shifted 22 to the right and 33 up.

xx00111.91.92.12.13344
g(x)g(x)2.52.522−7-71313443.53.5

Hole

Definition Let f=p/qf = p/q. Cancel every common factor of pp and qq to write ff in lowest terms as P/QP/Q, so that PP and QQ share no factor of positive degree and f(x)=P(x)/Q(x)f(x) = P(x)/Q(x) for every xx in the domain of ff. If aa is a zero of the original denominator qq but Q(a)≠0Q(a) \ne 0, then the graph of ff has a hole at the point

(a,P(a)Q(a)).\left(a, \frac{P(a)}{Q(a)}\right).

The point is missing from the graph because aa is not in the domain, while nearby outputs come as close to P(a)/Q(a)P(a)/Q(a) as you like.

In plain words A hole is a single missing point on an otherwise smooth curve. It happens when the bad factor cancels completely and the shorter formula gives an ordinary number at the excluded input.

Example For the function ff from the previous section, x−2x - 2 cancels and the reduced formula x+2x+1\frac{x + 2}{x + 1} gives 43\frac{4}{3} at x=2x = 2. The graph has a hole at (2,43)\left(2, \frac{4}{3}\right). Likewise

x2−9x−3=x+3,x≠3,\frac{x^2 - 9}{x - 3} = x + 3, \qquad x \ne 3,

so its graph is the line y=x+3y = x + 3 with the point (3,6)(3, 6) removed.

Vertical asymptote

Definition The line x=ax = a is a vertical asymptote of ff if f(x)f(x) grows without bound in size as xx approaches aa from at least one side, that is, f(x)→∞f(x) \to \infty or f(x)→−∞f(x) \to -\infty as x→a+x \to a^{+} or as x→a−x \to a^{-}. For a rational function written in lowest terms as P/QP/Q, as in the previous section, the vertical asymptotes are exactly the lines x=ax = a with Q(a)=0Q(a) = 0. At such an aa the numerator satisfies P(a)≠0P(a) \ne 0, because PP and QQ share no factor, so near aa the output is roughly a fixed nonzero number divided by a tiny one. A zero aa of qq survives into QQ exactly when the factor x−ax - a appears more times in qq than in pp, and then it gives a vertical asymptote, not a hole.

In plain words Near a vertical asymptote the outputs blow up. Simplify first, and any denominator zero that is still there after simplifying is an asymptote.

Example Return to the function ff above. After simplifying, f(x)=x+2x+1f(x) = \frac{x + 2}{x + 1} for x≠2x \ne 2, and this reduced formula still has a zero denominator at x=−1x = -1, so x=−1x = -1 is a vertical asymptote. Probing nearby inputs shows the side behavior.

xx−1.1-1.1−1.01-1.01−0.99-0.99−0.9-0.9
f(x)f(x)−9-9−99-991011011111

From the left the outputs fall toward −∞-\infty, and from the right they rise toward ∞\infty. A partial cancellation behaves the same way. The factor xx appears twice in the denominator of xx2\frac{x}{x^2} and once in the numerator, so xx2=1x\frac{x}{x^2} = \frac{1}{x} for x≠0x \ne 0 and x=0x = 0 is still a vertical asymptote.

Horizontal asymptote

Definition The line y=Ly = L is a horizontal asymptote of ff if f(x)→Lf(x) \to L as x→∞x \to \infty or as x→−∞x \to -\infty. For f=p/qf = p/q with pp not the zero polynomial, deg⁡p=n\deg p = n, deg⁡q=m\deg q = m and leading coefficients ana_n and bmb_m,

{y=0if n<m,y=anbmif n=m.\begin{cases} y = 0 & \text{if } n < m, \\ y = \dfrac{a_n}{b_m} & \text{if } n = m. \end{cases}

If n>mn > m, there is no horizontal asymptote. For a rational function the same line serves at both ends. A horizontal asymptote describes only the far ends of the graph. The graph may cross it at finite inputs.

In plain words Far to the right and far to the left, the highest powers win, and the graph levels off at the line their ratio gives. In the middle of the graph anything can happen, including crossing that line.

Example For g(x)=3x−5x−2g(x) = \frac{3x - 5}{x - 2} both degrees are 11, so the horizontal asymptote is y=31=3y = \frac{3}{1} = 3. For a crossing, take

r(x)=xx2+1.r(x) = \frac{x}{x^2 + 1}.

The degrees are 11 and 22, so y=0y = 0 is the horizontal asymptote, and yet r(0)=0r(0) = 0 puts the graph right on that line at the origin.

Separate branches

Definition The domain of a rational function is a union of open intervals whose endpoints are the excluded inputs. On each of these intervals the graph is one unbroken curve, with no gaps or jumps, and the part of the graph over one such interval is called a branch. Across a vertical asymptote neighboring branches are genuinely separate, since for a rational function the outputs grow without bound in size on both sides of the asymptote. Across a hole they line up, and only the single point is missing. By this definition a line with a hole has two branches that meet at the missing point, so they look like one line with a gap.

In plain words Cut the number line at every excluded input. Over each piece the graph is drawn without lifting the pencil, and you check each piece on its own.

Example For g(x)=3+1x−2g(x) = 3 + \frac{1}{x - 2} the domain is (−∞,2)∪(2,∞)(-\infty, 2) \cup (2, \infty). On the left branch x−2<0x - 2 < 0, so g(x)<3g(x) < 3, and on the right branch g(x)>3g(x) > 3. The left branch takes every value below 33 and the right branch every value above 33, so the range is all real numbers except 33.

Intercepts

Definition The yy-intercept of ff is the point (0,f(0))(0, f(0)), which exists only when 00 is in the domain. The xx-intercepts are the points (a,0)(a, 0) with aa in the domain and f(a)=0f(a) = 0. For f=p/qf = p/q these are the zeros of the numerator pp that are not excluded inputs.

In plain words Set x=0x = 0 for the yy-intercept. For the xx-intercepts make the top zero, then throw out any candidate that makes the original bottom zero.

Example For

f(x)=x2−4x2−x−2,f(x) = \frac{x^2 - 4}{x^2 - x - 2},

the numerator is zero at x=2x = 2 and x=−2x = -2. Since 22 is excluded, the only xx-intercept is (−2,0)(-2, 0). The yy-intercept is f(0)=−4−2=2f(0) = \frac{-4}{-2} = 2, the point (0,2)(0, 2).

Check yourself

  1. For g(x)=2x+1x−1g(x) = \frac{2x + 1}{x - 1}, find the vertical and horizontal asymptotes, both intercepts and the range.
Show the answer

Long division gives

g(x)=2+3x−1,x≠1.g(x) = 2 + \frac{3}{x - 1}, \qquad x \ne 1.

The vertical asymptote is x=1x = 1 and the horizontal asymptote is y=2y = 2. The xx-intercept is (−12,0)\left(-\frac{1}{2}, 0\right) and the yy-intercept is (0,−1)(0, -1). The fraction 3x−1\frac{3}{x - 1} takes every real value except 00, so the range is all real numbers except 22.

  1. Describe the graph of h(x)=x2+x−6x−2h(x) = \frac{x^2 + x - 6}{x - 2}, including its domain, intercepts and range.
Show the answer

The numerator factors as (x+3)(x−2)(x + 3)(x - 2), so h(x)=x+3h(x) = x + 3 for x≠2x \ne 2. The graph is the line y=x+3y = x + 3 with a hole at (2,5)(2, 5). The domain is all real numbers except 22, the intercepts are (−3,0)(-3, 0) and (0,3)(0, 3), and the range is all real numbers except 55.

  1. Where does k(x)=x−1x2−1k(x) = \frac{x - 1}{x^2 - 1} have a hole, and where does it have a vertical asymptote?
Show the answer

The original denominator excludes x=1x = 1 and x=−1x = -1. For those other inputs k(x)=1x+1k(x) = \frac{1}{x + 1}. The reduced formula gives 12\frac{1}{2} at x=1x = 1, so there is a hole at (1,12)\left(1, \frac{1}{2}\right). The factor x+1x + 1 remains in the denominator, so x=−1x = -1 is a vertical asymptote.

How this lesson connects