Algebra 2 · F.IF.C.7D

Vertical asymptotes

Some functions refuse one particular input, and close to it their outputs explode. Where does that happen, which way does the graph go on each side, and how is it different from a single missing point?

You need

Continuous graphs, jumps and holesAlgebra 2 What an asymptote isAlgebra 1 Simplifying rational expressionsGrades 10 to 11

Where you are 3 in Asymptotes

Functions13

Rational functions5

Asymptotes3

Horizontal asymptotes Slant asymptotes

Vertical asymptotes

Why this exists

A fraction cannot have zero as its denominator. So a formula such as \displaystyle f(x) = \frac{x + 1}{x^2 - 9x + 14} refuses certain inputs: here x = 2 and x = 7, where x^2 - 9x + 14 = 0. Knowing that an input is refused (see The implied domain of a formula) tells you there is no point on the graph above it. It does not tell you what the graph does near that input, and that is where the interesting behavior is.

Put in x = 1.999, just short of 2. The numerator is about 3 and the denominator is a tiny number, about 0.005, so the output is about 600. Move closer and the output grows past a thousand, past a million, without any ceiling. On the other side of 2 the outputs grow without bound in the negative direction. The graph does not merely have a gap at x = 2: it runs off the top and bottom of every page, hugging the vertical line x = 2 as it goes.

In Continuous graphs, jumps and holes this was the third way a graph can break. An asymptote in general is a line the graph gets ever closer to as it heads off without end. A vertical asymptote is the case where the heading-off happens in the outputs, upward or downward, while the inputs close in on one number. This star pins down when that happens, which way the graph goes on each side, and how to tell it from the other kind of refused input, a single missing point called a hole.

Definition

In words. A vertical line x = a is a vertical asymptote of the graph of f when, as the input closes in on a from at least one side, the outputs grow without bound, either upward or downward.

In symbols. The line x = a is a vertical asymptote of the graph of f if at least one of these holds:

\begin{aligned} &f(x) \to \infty \text{ or } f(x) \to -\infty \text{ as } x \to a^-, \\ &f(x) \to \infty \text{ or } f(x) \to -\infty \text{ as } x \to a^+. \end{aligned}

Here x \to a^- means x approaches a through values less than a, and x \to a^+ through values greater than a, as in Continuous graphs, jumps and holes. The statement f(x) \to \infty means: name any number, however large, and once x is close enough to a on that side, every output is larger than it. The statement f(x) \to -\infty is the same with “more negative than it”.

The behavior on each side of the asymptote is the pair of answers: what f(x) does as x \to a^-, and what it does as x \to a^+. The two sides need not agree.

A hole at x = a is the other way a refused input can look: the outputs on both sides head toward one ordinary number h, and only the single point (a, h) is missing from the graph.

A rational function is a quotient \displaystyle \frac{P(x)}{Q(x)} of two polynomials with Q not the zero polynomial. It is in lowest terms when P and Q have no common factor x - a, which by the The remainder and factor theorems means no number a makes both of them zero.

Intuition

Dividing by a tiny number gives a huge number: \displaystyle \frac{1}{0.1} = 10, \displaystyle \frac{1}{0.01} = 100, \displaystyle \frac{1}{0.001} = 1000. That is all a vertical asymptote is. Near a refused input, the denominator shrinks toward zero while the numerator stays near some number that is not zero, and the quotient grows without bound.

Look at \displaystyle f(x) = \frac{x + 1}{x^2 - 9x + 14} near x = 2. Factor the denominator: two numbers with product 14 and sum -9 are -2 and -7, so

f(x) = \frac{x + 1}{(x - 2)(x - 7)}.

Near x = 2 the numerator is close to 2 + 1 = 3 and the factor x - 7 is close to 2 - 7 = -5. Only the factor x - 2 is close to zero. So near 2,

f(x) \approx \frac{3}{-5(x - 2)} = \frac{-0.6}{x - 2}.

A table from the exact formula agrees:

x	f(x)
1.9	5.69
1.99	59.68
1.999	599.68
2	undefined
2.001	-600.32
2.01	-60.32
2.1	-6.33

Each step ten times closer to 2 makes the output about ten times larger in size, just as \displaystyle \frac{-0.6}{x - 2} predicts. The sign comes from the sign of x - 2: on the left it is negative, and \displaystyle \frac{-0.6}{\text{negative}} is positive, so the graph climbs; on the right it is positive, and the graph falls.

Figure 1The graph of \displaystyle f(x) = \frac{x + 1}{x^2 - 9x + 14}. Its denominator is zero at x = 2 and x = 7, and at each the graph runs along the dashed vertical line, up on one side and down on the other.

Near x = 7 the same reasoning gives \displaystyle f(x) \approx \frac{8}{5(x - 7)}, since the numerator is near 8 and x - 2 is near 5. Now the constant is positive, so the graph falls on the left of 7 and climbs on the right. The picture shows both: one branch up and one down at each dashed line. Far to the left and right the graph flattens toward the axis y = 0, which is the business of Horizontal asymptotes.

The approximation also shows why a hole is different. If the numerator is also near zero, the quotient is a tiny number over a tiny number, which can settle on an ordinary value. That happens exactly when the same factor sits on top and bottom, and canceling it shows the value.

Results

Where the vertical asymptotes of a rational function are. Write \displaystyle f(x) = \frac{P(x)}{Q(x)} and cancel every common factor to reach lowest terms, \displaystyle \frac{p(x)}{q(x)}. Then:

the vertical asymptotes are exactly the lines x = a for the real numbers a with q(a) = 0;
a number a with Q(a) = 0 but q(a) \ne 0 gives a hole, at the point \displaystyle \left(a, \frac{p(a)}{q(a)}\right).

Behavior on each side. Suppose x - a appears exactly m times as a factor of q, so q(x) = (x - a)^m r(x) with r(a) \ne 0. Then near a

f(x) \approx \frac{C}{(x - a)^m}, \qquad C = \frac{p(a)}{r(a)} \ne 0.

If m is odd, the graph goes off in opposite directions on the two sides. With C > 0 it falls as x \to a^- and climbs as x \to a^+; with C < 0 the other way round.
If m is even, the graph goes off in the same direction on both sides: up on both if C > 0, down on both if C < 0.

A rational graph never touches its vertical asymptote. The input a is refused, so no point of the graph lies on the line x = a.

A polynomial has no vertical asymptote. Every real number is an allowed input and the outputs near any input stay near the output there.

Why it is true

Asymptotes sit at the zeros of the lowest-terms denominator. Let \displaystyle \frac{p(x)}{q(x)} be in lowest terms with q(a) = 0. Then p(a) \ne 0: if p(a) were also zero, the factor theorem would make x - a a factor of both, and the fraction would not be in lowest terms. Polynomials change by only a little when their input changes by a little, so for x close to a the numerator stays within half of |p(a)| of p(a), while |q(x)| is smaller than any small number \varepsilon you choose. Then

|f(x)| = \frac{|p(x)|}{|q(x)|} > \frac{\tfrac{1}{2}|p(a)|}{\varepsilon},

and choosing \varepsilon small makes the right side as large as you like. The outputs grow without bound in size. Because q is zero only at finitely many inputs, near a it is zero only at a itself and keeps one sign on each side, so the outputs keep one sign on each side too. That is f(x) \to \infty or f(x) \to -\infty on each side.

If instead q(a) \ne 0 but the original Q(a) = 0, the canceled formula \displaystyle \frac{p}{q} has an allowed input at a and passes smoothly through height \displaystyle \frac{p(a)}{q(a)}. The original f equals it at every input except a, so its graph is that smooth graph with one point removed: a hole, as shown in Continuous graphs, jumps and holes.

The power of the factor decides the sides. With q(x) = (x - a)^m r(x),

f(x) = \frac{p(x)}{r(x)} \cdot \frac{1}{(x - a)^m}.

The first factor is a quotient whose denominator is not zero at a, so near a it is close to \displaystyle C = \frac{p(a)}{r(a)}, which is not zero, and has the same sign as C. All the growth comes from \displaystyle \frac{1}{(x - a)^m}. If m is odd, (x - a)^m has the sign of x - a: negative on the left, positive on the right. So the sign of f flips across a. If m is even, (x - a)^m is positive on both sides, so f has the sign of C on both sides.

Polynomials. A polynomial P is built from x and constants by adding and multiplying, so near any input a its outputs stay near P(a), a fixed number. Outputs that stay near a fixed number do not grow without bound, so no vertical line is an asymptote.

Worked examples

Example 1

Asymptotes and both sides of each

Find the vertical asymptotes of \displaystyle f(x) = \frac{x + 1}{x^2 - 9x + 14} and say what the graph does on each side of each one.

Solution

The denominator factors as (x - 2)(x - 7), so it is zero at x = 2 and x = 7. The numerator x + 1 is 3 at x = 2 and 8 at x = 7, neither zero, so no factor cancels and the fraction is already in lowest terms. Both zeros give asymptotes: the lines x = 2 and x = 7.

Near x = 2 only x - 2 is near zero. Put x = 2 into everything else: \displaystyle C = \frac{2 + 1}{2 - 7} = -\frac{3}{5}. The factor appears once, an odd power, so the two sides go opposite ways. As x \to 2^-, x - 2 is a small negative number and \displaystyle \frac{-3/5}{\text{small negative}} is large and positive: f(x) \to \infty. As x \to 2^+, f(x) \to -\infty. The table in the Intuition confirms it: f(1.999) \approx 599.68 and f(2.001) \approx -600.32.

Near x = 7, \displaystyle C = \frac{7 + 1}{7 - 2} = \frac{8}{5}, positive, again an odd power. As x \to 7^-, x - 7 is a small negative number, so f(x) \to -\infty. As x \to 7^+, f(x) \to \infty. A check: \displaystyle f(7.01) = \frac{8.01}{(5.01)(0.01)} \approx 159.9, positive and large.

So the graph is in three pieces. Left of 2 it climbs up along x = 2; between 2 and 7 it comes up from the bottom along x = 2 and goes back down along x = 7; right of 7 it comes down from the top. The middle piece is negative throughout, since x + 1 > 0, x - 2 > 0 and x - 7 < 0 there.

Example 2

Telling a hole from an asymptote

Find every input that \displaystyle g(x) = \frac{x^2 - 25}{x^2 + 8x + 15} refuses, and decide whether each gives a hole or a vertical asymptote. Describe the graph near each.

Solution

The denominator factors with two numbers of product 15 and sum 8, which are 3 and 5: x^2 + 8x + 15 = (x + 3)(x + 5). So g refuses x = -3 and x = -5. Both make the denominator zero; the factoring of the numerator decides what each one does. As a difference of squares, x^2 - 25 = (x - 5)(x + 5). The factor x + 5 is common, and for x \ne -5 it is not zero, so it may be divided out:

\begin{aligned} g(x) &= \frac{(x - 5)(x + 5)}{(x + 3)(x + 5)} \\ &= \frac{x - 5}{x + 3} \quad (x \ne -5). \end{aligned}

This is lowest terms, since x - 5 and x + 3 share no zero. Its denominator is zero only at x = -3.

At x = -5 the canceled denominator is -5 + 3 = -2, not zero. So x = -5 is a hole, at the height the canceled formula gives: \displaystyle \frac{-5 - 5}{-5 + 3} = \frac{-10}{-2} = 5. The hole is at (-5, 5). Near it the outputs head toward 5 from both sides; they do not blow up.

At x = -3 the canceled denominator is zero and the numerator is -3 - 5 = -8. So x = -3 is a vertical asymptote, with C = -8 and the factor to the first power. As x \to -3^+, x + 3 is a small positive number and \displaystyle \frac{-8}{\text{small positive}} is large and negative: g(x) \to -\infty. As x \to -3^-, g(x) \to \infty. From the original formula, g(-3.01) = 801 and g(-2.99) = -799.

Figure 2Example 2. Both x = -5 and x = -3 make the denominator of g zero, but only x = -3 is an asymptote. At x = -5 the graph has a hole, at (-5, 5).

The original formula gives \displaystyle \frac{0}{0} at x = -5 and \displaystyle \frac{-16}{0} at x = -3. A number that is not zero over zero is the signature of an asymptote; zero over zero says only “cancel and look again”.

Example 3

Up on both sides

Describe the graph of \displaystyle h(x) = \frac{5 - x}{x^2 - 8x + 16} near its vertical asymptote.

Solution

The denominator is a perfect square: x^2 - 8x + 16 = (x - 4)^2. It is zero only at x = 4, and there the numerator is 5 - 4 = 1, not zero, so nothing cancels and x = 4 is a vertical asymptote.

Near 4, \displaystyle h(x) \approx \frac{1}{(x - 4)^2}, so C = 1 and the factor appears twice, an even power. A square is positive whether x - 4 is negative or positive, so the outputs are large and positive on both sides: h(x) \to \infty as x \to 4^- and as x \to 4^+. Values confirm it: \displaystyle h(3.9) = \frac{1.1}{0.01} = 110 and \displaystyle h(4.1) = \frac{0.9}{0.01} = 90.

Figure 3Example 3. The factor x - 4 appears twice in the denominator, so the graph goes up on both sides of x = 4.

The graph still changes sign elsewhere: the numerator 5 - x is zero at x = 5, so the right branch comes down from the top, crosses the axis at (5, 0), and is negative after that, for instance \displaystyle h(6) = \frac{-1}{4}. The sign changes at the zero of the numerator, not at the asymptote.

Example 4

A factor that cancels but does not disappear

Find the holes and vertical asymptotes of \displaystyle m(x) = \frac{x^2 + x - 6}{x^3 - 4x^2 + 4x}, and the behavior on each side of each asymptote.

Solution

Take out the common factor x from the denominator, then spot a perfect square:

x^3 - 4x^2 + 4x = x(x^2 - 4x + 4) = x(x - 2)^2.

So m refuses x = 0 and x = 2. The numerator factors with two numbers of product -6 and sum 1, which are 3 and -2: x^2 + x - 6 = (x + 3)(x - 2). One factor of x - 2 is common and cancels for x \ne 2:

\begin{aligned} m(x) &= \frac{(x + 3)(x - 2)}{x(x - 2)^2} \\ &= \frac{x + 3}{x(x - 2)} \quad (x \ne 2). \end{aligned}

The new numerator x + 3 is not zero at 0 or at 2, so this is lowest terms, and its denominator is still zero at both x = 0 and x = 2. Both are vertical asymptotes, and there is no hole. The factor x - 2 appeared once on top and twice underneath, so canceling leaves one copy below, and that copy still sends the outputs off without bound.

Near x = 0: \displaystyle C = \frac{0 + 3}{0 - 2} = -\frac{3}{2}, odd power. As x \to 0^-, m(x) \to \infty; as x \to 0^+, m(x) \to -\infty.

Near x = 2: \displaystyle C = \frac{2 + 3}{2} = \frac{5}{2}, odd power. As x \to 2^-, m(x) \to -\infty; as x \to 2^+, m(x) \to \infty.

Example 5

Building a function to order

Write a rational function whose graph has vertical asymptotes x = -4 and x = 3, a hole at x = 1, crosses the horizontal axis at x = 6, and falls toward -\infty just to the right of x = 3. Give the hole’s coordinates.

Solution

Each feature asks for a factor. An asymptote at x = -4 needs x + 4 in the lowest-terms denominator, and one at x = 3 needs x - 3. A crossing at x = 6 needs the numerator to be zero there, so x - 6 goes on top. A hole at x = 1 needs x - 1 on top and bottom the same number of times, so that it cancels completely. Try

r(x) = \frac{(x - 6)(x - 1)}{(x + 4)(x - 3)(x - 1)}.

Canceling gives \displaystyle \frac{x - 6}{(x + 4)(x - 3)} for x \ne 1, whose denominator is zero only at -4 and 3 and whose numerator is zero only at 6, so the asymptotes and the crossing are right.

Now the direction. Near x = 3, \displaystyle C = \frac{3 - 6}{3 + 4} = -\frac{3}{7}, negative, with the factor to an odd power. So as x \to 3^+, r(x) \to -\infty, as asked. Had the sign come out wrong, multiplying the whole function by -1 would flip every side and change nothing else.

The hole’s height comes from the canceled formula at x = 1: \displaystyle \frac{1 - 6}{(1 + 4)(1 - 3)} = \frac{-5}{-10} = \frac{1}{2}. The hole is at \displaystyle \left(1, \frac{1}{2}\right).

Example 6

An image that runs away

A thin converging lens (one that brings light to a focus, like a magnifying glass) with focal length 5 centimeters forms an image of an object placed d centimeters in front of it. The image distance i satisfies the thin lens equation \displaystyle \frac{1}{d} + \frac{1}{i} = \frac{1}{5}. Write i as a function of d, find its vertical asymptote, and say what happens to the image as the object is moved toward d = 5 from farther away. How close to 5 centimeters must the object be for the image to form more than 100 centimeters from the lens?

Solution

Solve for \displaystyle \frac{1}{i} and combine over the common denominator 5d:

\begin{aligned} \frac{1}{i} &= \frac{1}{5} - \frac{1}{d} = \frac{d - 5}{5d}, \\ i(d) &= \frac{5d}{d - 5}. \end{aligned}

This is lowest terms (the numerator is zero only at d = 0, the denominator only at d = 5), so the vertical asymptote is d = 5, the focal length. Near it C = 5 \cdot 5 = 25 > 0, with the factor to the first power, so i(d) \to \infty as d \to 5^+. In the lens, as the object slides in toward the focal point from farther away, the image moves off without bound: i(10) = 10, i(6) = 30, i(5.5) = 55, i(5.1) = 255. At d = 5 exactly, the light leaves the lens in parallel rays and no image forms at all. For d < 5 the formula gives a negative i, which in this sign convention means the image appears on the same side as the object, as in a magnifying glass; that is the i(d) \to -\infty side of the asymptote.

For the image to be more than 100 centimeters away with d > 5, solve \displaystyle \frac{5d}{d - 5} > 100. Since d - 5 > 0, multiplying both sides by it keeps the direction:

\begin{aligned} 5d &> 100d - 500, \\ 500 &> 95d, \\ d &< \frac{100}{19} \approx 5.26. \end{aligned}

So the object must sit between 5 and \displaystyle \frac{100}{19} centimeters from the lens, a window only about a quarter of a centimeter wide. That narrow window is the asymptote at work: close to d = 5 a small move of the object makes a huge move of the image.

Boundaries

Cancel completely before deciding. The rule “asymptote at each zero of the denominator” is about the lowest-terms denominator. Before canceling, a zero of the denominator may be a hole (Example 2). And canceling one copy of a factor is not enough when the denominator has more copies (Example 4).
Only real zeros count. The denominator of \displaystyle \frac{1}{x^2 + 1} is never zero for a real x, so that graph has no vertical asymptote at all, even though x^2 + 1 has complex zeros.
One side is enough. The definition asks for at least one side. The graph of y = \ln x (see Graphs of logarithmic functions) has the asymptote x = 0, with \ln x \to -\infty as x \to 0^+, and there is no graph to the left of 0 at all.
The factor rule is for rational functions only. For other functions a zero denominator decides nothing by itself. The function \displaystyle \frac{|x|}{x} refuses x = 0, but its outputs are -1 on the left and 1 on the right: a jump, not an asymptote. And \ln x has an asymptote with no fraction in sight.
“Never touches” is about rational formulas. A function defined in pieces can give the input a a value and still blow up near it: \displaystyle f(x) = \frac{1}{x^2} for x \ne 0, with f(0) = 0, has the asymptote x = 0 and the point (0, 0) on that line. Any graph of a function meets a vertical line at most once, because each input has one output. Horizontal asymptotes are different: a graph may cross one many times.
Infinitely many is possible. A rational function has only finitely many, since its denominator has finitely many zeros, but the graph of the tangent function has a vertical asymptote every \pi units; see Graphing tangent and the reciprocal functions.

Where this goes next

A vertical asymptote describes what a graph does near a refused input; Horizontal asymptotes and Slant asymptotes describe what it does far out to the left and right. Together with the zeros and the holes, these are the skeleton of a sketch, assembled in graphing a rational function after finding its domain. The shifted reciprocal graphs of Transforming the reciprocal graph are the simplest whole family, one asymptote each, and Rational functions in context reads asymptotes as physical limits, like the lens above. In calculus, “f(x) \to \infty as x \to a^+” becomes an infinite one-sided limit, and comparing the two sides is the question asked in one-sided limits and when a limit exists. The same picture appears for logarithms, in Graphs of logarithmic functions, and for the tangent function.

Exercises

Finding vertical asymptotes

Find every vertical asymptote of the graph, or say that there is none.

\displaystyle \frac{4}{x - 9}
\displaystyle \frac{x + 3}{x^2 - 16}
\displaystyle \frac{2x}{x^2 + 3x - 10}
\displaystyle \frac{x - 1}{x^2 + 9}
\displaystyle \frac{x^2 + 1}{3x^2 - 12x}
\displaystyle \frac{7}{2x + 3}

Hole or asymptote

Find every hole (with its coordinates) and every vertical asymptote.

\displaystyle \frac{x - 4}{x^2 - 7x + 12}
\displaystyle \frac{x^2 + 2x - 8}{x^2 - 4}
\displaystyle \frac{x^2 - 6x + 9}{x - 3}
\displaystyle \frac{x + 1}{x^2 + 2x + 1}
\displaystyle \frac{x^2 - 1}{x^3 - x}

Behavior on each side

Say what the outputs do as x approaches each vertical asymptote from the left and from the right.

\displaystyle \frac{3}{x - 5}
\displaystyle \frac{-2}{(x + 1)^2}
\displaystyle \frac{x - 6}{x + 2}
\displaystyle \frac{x + 3}{x(x - 4)^2}

Building and in situations

Write a rational function whose graph has vertical asymptotes x = -2 and x = 5 and crosses the horizontal axis only at x = 1.
Write a rational function whose graph has a vertical asymptote x = 3 with the graph going up on both sides, and a hole at x = 0. Give the hole’s coordinates.
Lunetta fills a 360-liter tank with a pump that delivers 12 liters a minute, but the tank leaks r liters a minute. The time to fill it is \displaystyle T(r) = \frac{360}{12 - r} minutes. (a) Find T(8). (b) Find T(11) and T(11.9). (c) Find the vertical asymptote and what T does as the leak approaches it from below. (d) What happens to the tank when r = 12, and when r > 12? Which values of r make sense in this model?
For which values of k does the graph of \displaystyle \frac{x - k}{x^2 - 5x + 6} have exactly one vertical asymptote? For each, give the asymptote and the hole.

Explaining

Explain why the graph of a polynomial has no vertical asymptote.
Explain why the graph of a rational function in lowest terms never touches its vertical asymptote, even though the graph gets as close to it as you like.
Sasha says: “If the denominator of a rational function is zero at x = a, then x = a is a vertical asymptote.” Give a counterexample, and correct the statement.
Prove that if x - a appears an odd number of times as a factor of the lowest-terms denominator, the graph goes off in opposite directions on the two sides of x = a.

Answers

x = 9. The denominator is zero at 9 and the numerator 4 is not zero.
x = -4 and x = 4. Since x^2 - 16 = (x - 4)(x + 4), the zeros are \pm 4; the numerator x + 3 is -1 and 7 there, not zero.
x = -5 and x = 2. Since x^2 + 3x - 10 = (x + 5)(x - 2); the numerator 2x is -10 and 4 there.
None. The denominator x^2 + 9 is at least 9 for every real x, so it is never zero.
x = 0 and x = 4. Since 3x^2 - 12x = 3x(x - 4); the numerator x^2 + 1 is at least 1, never zero.
\displaystyle x = -\frac{3}{2}. The denominator 2x + 3 is zero at \displaystyle x = -\frac{3}{2} and the numerator is 7.
Hole at (4, 1), asymptote x = 3. Since \displaystyle \frac{x - 4}{(x - 3)(x - 4)} = \frac{1}{x - 3} for x \ne 4, and \displaystyle \frac{1}{4 - 3} = 1.
Hole at \displaystyle \left(2, \frac{3}{2}\right), asymptote x = -2. Since \displaystyle \frac{(x + 4)(x - 2)}{(x - 2)(x + 2)} = \frac{x + 4}{x + 2} for x \ne 2, and \displaystyle \frac{2 + 4}{2 + 2} = \frac{3}{2}.
Hole at (3, 0), no asymptote. Since \displaystyle \frac{(x - 3)^2}{x - 3} = x - 3 for x \ne 3: the graph is the line y = x - 3 with (3, 0) removed.
Asymptote x = -1, no hole. Since x^2 + 2x + 1 = (x + 1)^2, one copy of x + 1 cancels and one remains: \displaystyle \frac{1}{x + 1}.
Holes at (1, 1) and (-1, -1), asymptote x = 0. Since x^3 - x = x(x - 1)(x + 1) and x^2 - 1 = (x - 1)(x + 1), the function is \displaystyle \frac{1}{x} for x \ne \pm 1.
As x \to 5^-, the output \to -\infty; as x \to 5^+, \to \infty. The numerator 3 is positive and x - 5 has the sign of the side.
As x \to -1^- and as x \to -1^+, the output \to -\infty on both sides. The square (x + 1)^2 is positive on both sides and the numerator is -2.
As x \to -2^-, the output \to \infty; as x \to -2^+, \to -\infty. Near -2 the numerator is near -8, so the function is about \displaystyle \frac{-8}{x + 2}.
At x = 0: \to -\infty as x \to 0^- and \to \infty as x \to 0^+, since near 0 it is about \displaystyle \frac{3}{16x}. At x = 4: \to \infty on both sides, since near 4 it is about \displaystyle \frac{7}{4(x - 4)^2}, an even power with positive constant.
One answer: \displaystyle \frac{x - 1}{(x + 2)(x - 5)}. The denominator is zero only at -2 and 5, where the numerator is -3 and 4; the numerator is zero only at 1. Any nonzero constant multiple also works.
One answer: \displaystyle \frac{x}{x(x - 3)^2}, with the hole at \displaystyle \left(0, \frac{1}{9}\right). It equals \displaystyle \frac{1}{(x - 3)^2} for x \ne 0: an even power with C = 1 > 0, so up on both sides, and the hole’s height is \displaystyle \frac{1}{(0 - 3)^2} = \frac{1}{9}.
(a) \displaystyle T(8) = \frac{360}{4} = 90 minutes. (b) T(11) = 360 minutes and T(11.9) = 3600 minutes. (c) The asymptote is r = 12; as r \to 12^-, T(r) \to \infty: the closer the leak comes to matching the pump, the longer the fill takes, without bound. (d) At r = 12 the water level never changes, so the tank never fills, matching the refused input. For r > 12 the formula gives a negative time, which means nothing here: the tank loses water. The model makes sense for 0 \le r < 12.
k = 2 or k = 3. The denominator is (x - 2)(x - 3). With k = 2 the function is \displaystyle \frac{1}{x - 3} for x \ne 2: asymptote x = 3, hole at (2, -1). With k = 3 it is \displaystyle \frac{1}{x - 2} for x \ne 3: asymptote x = 2, hole at (3, 1). Any other k cancels nothing and leaves both asymptotes.
A polynomial accepts every real input, and near any input a its outputs stay near the fixed number P(a), because it is built by adding and multiplying and small changes in x make small changes in the output. Outputs that stay near a fixed number cannot grow without bound, so no vertical line is an asymptote.
The asymptote x = a sits where the lowest-terms denominator is zero, so a is not an allowed input and no point of the graph has first coordinate a. The graph comes as close to the line as you like because inputs closer and closer to a are allowed, but the line itself holds no point of the graph.
A counterexample is \displaystyle g(x) = \frac{x^2 - 25}{x^2 + 8x + 15} from Example 2: its denominator is zero at x = -5, yet the graph has a hole there, not an asymptote. Correct statement: if a rational function is written in lowest terms and its denominator is zero at a real number a, then x = a is a vertical asymptote.
Write the lowest-terms denominator as (x - a)^m r(x) with r(a) \ne 0 and m odd. Then \displaystyle f(x) = \frac{p(x)}{r(x)} \cdot \frac{1}{(x - a)^m}. Near a the first factor is close to \displaystyle C = \frac{p(a)}{r(a)}, which is not zero (p(a) \ne 0 in lowest terms), so it keeps the sign of C on both sides. An odd power keeps the sign of its base, so (x - a)^m is negative for x < a and positive for x > a. The product therefore has opposite signs on the two sides, and since its size grows without bound, one side goes to \infty and the other to -\infty.