Graphing rational functions
You can find where a rational function breaks, where it levels off and where it meets the axes. How do those separate facts fit together into one picture of the whole graph?
Where you are 5 in Rational functions
When you reach for it
You reach for this whenever you want to see a rational function, a quotient \displaystyle \frac{P(x)}{Q(x)} of two polynomials, as a whole picture rather than as a list of facts. Each piece has its own lesson: the refused inputs come from the domain, the breaks where the graph runs off the page from Vertical asymptotes, the level it settles to far out from Horizontal asymptotes, and the way it meets the axis at each zero from Zeros and multiplicity. A graph is what you get when all of these are true at once.
The pieces fit together because of one fact. Between two neighboring breaks the graph is a single unbroken curve (see Continuous graphs, jumps and holes), and on any stretch where it meets neither the axis nor a break it stays on one side of the axis. So once the breaks, the zeros and the sign on each stretch are known, the shape is forced up to details such as the exact height of a peak. Plotting many points by hand would still miss what happens near an asymptote; this method finds that first.
This lesson takes two days. Day 1 (Examples 1 to 3) sketches functions whose factors each appear once, with no canceling: a horizontal asymptote at a ratio of leading coefficients, one at y = 0, and none at all. Day 2 (Examples 4 to 6) adds holes, factors that appear twice, and the reverse question: reading a formula off a graph.
The procedure
Each move below is allowed for a reason, and the reasons are what let you trust the sketch.
Factor the numerator and the denominator, then cancel common factors. Factoring shows every zero of both at a glance, by the The remainder and factor theorems. A factor x - a on top and bottom cancels for every x \ne a, so the function equals the lowest-terms formula \displaystyle \frac{p(x)}{q(x)} everywhere except at x = a itself. The original function still refuses x = a, so its graph is the graph of \displaystyle \frac{p}{q} with the single point \displaystyle \left(a, \frac{p(a)}{q(a)}\right) removed: a hole, drawn as an open circle. If the factor appears more often below than above, a copy is left below after canceling, and then x = a is not a hole but an asymptote.
Read off the vertical asymptotes and the side behavior. Every real zero a of the lowest-terms denominator q gives a vertical asymptote x = a. If x - a appears an odd number of times in q the graph goes off in opposite directions on the two sides; an even number of times, the same direction. One sign near the asymptote settles which: put x = a into every factor except the vanishing one to get a constant C, and the sign of \displaystyle \frac{C}{(x - a)^m} on each side is the direction.
Find the far-out behavior from the degrees. If the numerator has the lower degree, the horizontal asymptote is y = 0. If the degrees are equal, it is \displaystyle y = \frac{a}{b}, the ratio of the leading coefficients. If the numerator has the higher degree there is no horizontal asymptote; when it is higher by exactly one, long division writes the function as a line plus a remainder fraction that shrinks to 0, and the graph follows that line (a slant asymptote).
Find the intercepts. The zeros of the function are the real zeros of the lowest-terms numerator p. A hole is never a zero, even when it sits on the axis, because the function has no value there. A zero whose factor appears an odd number of times is a crossing; an even number of times, a touch. The y-intercept is f(0), if 0 is in the domain.
Fix the sign on each interval. Mark the zeros and the vertical asymptotes on the horizontal axis. They cut it into open intervals. On each one every factor of the lowest-terms formula keeps one sign, since a linear factor x - c changes sign only at c, so the whole quotient keeps one sign. One test value per interval therefore decides the sign of the entire interval. Equivalently, start from the far right, where the sign is the sign of the ratio of the leading coefficients, and move left: the sign flips at each odd-power break and stays at each even-power one. A hole is not a break in the sign: the canceled factor was a factor of 1.
Check whether the graph crosses its horizontal asymptote. The graph meets y = L where f(x) = L, which after multiplying by the denominator is the polynomial equation p(x) = L \cdot q(x). Its solutions in the domain are the crossings. A graph may cross a horizontal asymptote; it can never meet a vertical one, because that input is refused.
Join each interval between vertical asymptotes into one curve. The graph has one branch for each interval between neighboring vertical asymptotes, and no others, since the lowest-terms formula is continuous wherever its denominator is not zero. Draw each branch through its intercepts and crossing points, on the side of the axis its signs demand, heading along each asymptote in the direction found. A few extra values sharpen the picture where the shape is unclear.
Worked examples
Day 1: a full sketch with equal degrees
Sketch the graph of \displaystyle f(x) = \frac{2x^2 - 4x - 6}{x^2 - 5x - 14}.
SolutionFactor both. The numerator has the common factor 2, and two numbers with product -3 and sum -2 are -3 and 1. Two numbers with product -14 and sum -5 are -7 and 2. So
No factor is common, so there is nothing to cancel and no hole. The domain is every real number except -2 and 7.
Vertical asymptotes. The denominator is zero at x = 7 and x = -2, each factor to the first power, an odd power, so at each one the graph goes up on one side and down on the other. Near x = -2, put x = -2 into every other factor: \displaystyle C = \frac{2(-5)(-1)}{-9} = -\frac{10}{9}. As x \to -2^- the factor x + 2 is a small negative number, and a negative constant over a small negative number is large and positive, so f(x) \to \infty; as x \to -2^+, f(x) \to -\infty. Near x = 7, \displaystyle C = \frac{2(4)(8)}{9} = \frac{64}{9}, positive, so f(x) \to -\infty as x \to 7^- and f(x) \to \infty as x \to 7^+.
Horizontal asymptote. Both have degree 2, with leading coefficients 2 and 1, so the horizontal asymptote is y = 2.
Intercepts. The zeros are x = 3 and x = -1, each to the first power, so the graph crosses the axis at both. The y-intercept is \displaystyle f(0) = \frac{-6}{-14} = \frac{3}{7}.
Signs. The breaks -2, -1, 3, 7 cut the axis into five intervals. One test value in each:
| Interval | Test value | f there | Sign |
|---|---|---|---|
| (-\infty, -2) | -3 | \displaystyle \frac{12}{5} | + |
| (-2, -1) | -1.5 | \displaystyle -\frac{18}{17} | - |
| (-1, 3) | 0 | \displaystyle \frac{3}{7} | + |
| (3, 7) | 5 | \displaystyle -\frac{12}{7} | - |
| (7, \infty) | 10 | \displaystyle \frac{77}{18} | + |
The signs alternate because every break is an odd power, and the far right is positive because the ratio of leading coefficients, 2, is positive.
Crossing the horizontal asymptote. Solve f(x) = 2. Multiplying by the denominator, 2x^2 - 4x - 6 = 2(x^2 - 5x - 14) = 2x^2 - 10x - 28. The x^2 terms cancel, leaving 6x = -22, so \displaystyle x = -\frac{11}{3}. The graph crosses y = 2 once, at \displaystyle \left(-\frac{11}{3}, 2\right).
Joining. There are three branches. The left one, on (-\infty, -2), is positive throughout. Far left it is near 2; it crosses y = 2 at \displaystyle x = -\frac{11}{3} and then climbs to \infty along x = -2. Since it rises above 2 only after \displaystyle x = -\frac{11}{3}, far left it must be just under 2: indeed \displaystyle f(-8) = \frac{77}{45} \approx 1.71. The middle branch, on (-2, 7), comes up from -\infty, crosses the axis at -1, stays positive and low between the zeros (\displaystyle f(1) = \frac{4}{9}), crosses at 3 and falls to -\infty along x = 7. The right branch comes down from \infty along x = 7, through \displaystyle f(10) = \frac{77}{18} \approx 4.28, and settles toward 2. It never crosses y = 2 again, so it approaches from above.
Day 1: when the horizontal asymptote is the axis
Sketch the graph of \displaystyle k(x) = \frac{3x}{x^2 - x - 6}.
SolutionThe denominator factors as (x - 3)(x + 2), since -3 \cdot 2 = -6 and -3 + 2 = -1. The numerator 3x is 9 at x = 3 and -6 at x = -2, not zero, so nothing cancels:
Asymptotes. Vertical: x = 3 and x = -2, both odd powers. Near x = 3, \displaystyle C = \frac{9}{5} > 0, so k(x) \to -\infty as x \to 3^- and k(x) \to \infty as x \to 3^+. Near x = -2, \displaystyle C = \frac{-6}{-5} = \frac{6}{5} > 0, so k(x) \to -\infty as x \to -2^- and k(x) \to \infty as x \to -2^+. The numerator has degree 1 and the denominator degree 2, so the horizontal asymptote is the axis itself, y = 0.
Intercepts. The only zero is x = 0, a crossing, and the y-intercept is k(0) = 0: the graph passes through the origin.
Signs. Test values: \displaystyle k(-3) = -\frac{3}{2}, \displaystyle k(-1) = \frac{3}{4}, \displaystyle k(1) = -\frac{1}{2}, k(4) = 2. So k is negative on (-\infty, -2), positive on (-2, 0), negative on (0, 3) and positive on (3, \infty).
Crossing the horizontal asymptote. The graph meets y = 0 exactly at its zeros, so it crosses its asymptote once, at the origin.
Joining. The left branch is negative, close under the axis far left, and falls to -\infty along x = -2. The middle branch comes down from \infty just right of -2, passes through the origin, and falls to -\infty along x = 3. The right branch comes down from \infty just right of 3 and flattens toward the axis from above. So on each side of the picture the graph settles onto the axis, but from opposite sides: from below on the left, from above on the right.
Day 1: no horizontal asymptote
Sketch the graph of \displaystyle f(x) = \frac{x^2 - 4}{x - 1}.
SolutionThe numerator is a difference of squares, (x - 2)(x + 2), and neither factor matches x - 1, so nothing cancels.
Vertical asymptote. x = 1, an odd power, with C = 1^2 - 4 = -3. As x \to 1^-, x - 1 is a small negative number and \displaystyle \frac{-3}{\text{small negative}} is large and positive, so f(x) \to \infty; as x \to 1^+, f(x) \to -\infty.
Far out. The numerator has degree 2 and the denominator degree 1, so there is no horizontal asymptote. Since the degree is higher by exactly one, divide: x^2 - 4 = (x - 1)(x + 1) - 3, so
Far out the fraction \displaystyle \frac{3}{x - 1} shrinks toward 0, so the graph follows the line y = x + 1, which goes down on the left and up on the right. Far right the fraction is a small positive number being subtracted, so the graph runs just below the line; far left it is a small negative number being subtracted, so the graph runs just above it.
Intercepts and signs. Zeros x = -2 and x = 2, both crossings; y-intercept \displaystyle f(0) = \frac{-4}{-1} = 4. Test values \displaystyle f(-3) = -\frac{5}{4}, f(0) = 4, \displaystyle f(1.5) = -\frac{7}{2}, \displaystyle f(3) = \frac{5}{2} give the signs -, +, -, + on (-\infty, -2), (-2, 1), (1, 2), (2, \infty).
Joining. The left branch comes up from the bottom left along the line, crosses the axis at -2, passes through (0, 4) and climbs to \infty along x = 1. The right branch comes up from -\infty along x = 1, crosses the axis at 2 and rises to the right just under the line.
Day 2: a factor that cancels
Sketch the graph of \displaystyle g(x) = \frac{x^2 - 2x - 24}{x^2 - 9x + 18}.
SolutionTwo numbers with product -24 and sum -2 are -6 and 4; two with product 18 and sum -9 are -6 and -3. So
The domain of g still excludes both 3 and 6, because the original denominator is zero at both.
Hole. The factor x - 6 canceled completely, so x = 6 is a hole, at the height the canceled formula gives there: \displaystyle \frac{6 + 4}{6 - 3} = \frac{10}{3}. The hole is at \displaystyle \left(6, \frac{10}{3}\right).
Asymptotes. The lowest-terms denominator is zero only at x = 3, so that is the one vertical asymptote. There C = 3 + 4 = 7 > 0 with an odd power, so g(x) \to -\infty as x \to 3^- and g(x) \to \infty as x \to 3^+. The degrees are equal (in the original and in the canceled form alike) with leading coefficients 1 and 1, so the horizontal asymptote is y = 1.
Intercepts. The zero is x = -4; the canceled factor gives no zero. The y-intercept is \displaystyle g(0) = \frac{-24}{18} = -\frac{4}{3}.
Signs. The breaks are -4 and 3 only. Test values \displaystyle g(-5) = \frac{1}{8}, \displaystyle g(0) = -\frac{4}{3}, g(4) = 8 give +, -, +. The hole at 6 is not a break: the canceled factor \displaystyle \frac{x - 6}{x - 6} is 1 on both sides of 6.
Crossing. Solving \displaystyle \frac{x + 4}{x - 3} = 1 gives x + 4 = x - 3, which no x satisfies, so the graph never meets y = 1.
Joining. The gap from the line is \displaystyle \frac{x + 4}{x - 3} - 1 = \frac{7}{x - 3}, negative for x < 3 and positive for x > 3. So the left branch comes in just below y = 1 on the far left, crosses the axis at -4, passes through \displaystyle \left(0, -\frac{4}{3}\right) and falls along x = 3. The right branch comes down along x = 3 and flattens toward y = 1, and on it, at \displaystyle \left(6, \frac{10}{3}\right), sits an open circle. Without that circle the picture would be the graph of \displaystyle \frac{x + 4}{x - 3}, a different function.
Day 2: doubled factors
Sketch the graph of \displaystyle r(x) = \frac{x^2 - 2x + 1}{x^2 + 4x + 4}.
SolutionBoth are perfect squares:
No factor is common. A square is never negative, so r(x) \ge 0 at every input of its domain: the whole graph lies on or above the axis.
Vertical asymptote. x = -2, where x + 2 appears twice, an even power. Here C = (-2 - 1)^2 = 9 > 0, so r(x) \to \infty as x \to -2^- and as x \to -2^+: up on both sides.
Horizontal asymptote. Equal degrees, leading coefficients 1 and 1, so y = 1.
Intercepts. The only zero is x = 1, where x - 1 appears twice, so the graph touches the axis at (1, 0) and turns back without crossing, exactly as the sign said it must. The y-intercept is \displaystyle r(0) = \frac{1}{4}.
Crossing the horizontal asymptote. Solve (x - 1)^2 = (x + 2)^2, that is x^2 - 2x + 1 = x^2 + 4x + 4, so -6x = 3 and \displaystyle x = -\frac{1}{2}. The graph crosses y = 1 at \displaystyle \left(-\frac{1}{2}, 1\right). The gap from the line tells which side each end is on:
The denominator is positive, so the gap has the sign of -6x - 3: positive for \displaystyle x < -\frac{1}{2} and negative for \displaystyle x > -\frac{1}{2}.
Joining. The left branch, on (-\infty, -2), sits above y = 1 everywhere (\displaystyle r(-10) = \frac{121}{64} \approx 1.89) and climbs to \infty along x = -2. The right branch comes down from \infty just right of -2 (r(-1) = 4), crosses y = 1 at \displaystyle x = -\frac{1}{2}, passes through \displaystyle \left(0, \frac{1}{4}\right), touches the axis at (1, 0), and rises again, staying below y = 1 for good (\displaystyle r(4) = \frac{1}{4}).
Day 2: from a graph to a formula
The figure shows the graph of a rational function. It has vertical asymptotes x = -1 and x = 3 and horizontal asymptote y = -2, it meets the horizontal axis only at (1, 0), where it touches and turns back, and it passes through \displaystyle \left(5, -\frac{8}{3}\right). Find a formula for it.
Run the procedure backwards, one feature at a time.
The graph touches the axis at x = 1 and meets it nowhere else, so the numerator’s only real zero is 1, to an even power. The simplest choice is (x - 1)^2.
At x = -1 the graph goes down on the left and up on the right, opposite directions, so x + 1 appears an odd number of times in the denominator. At x = 3 it goes up on the left and down on the right, again opposite, so x - 3 appears an odd number of times. The simplest choice is (x + 1)(x - 3). So try
with a constant a still to find. There is no hole in the picture, so no other factor is needed.
The marked point fixes a: \displaystyle t(5) = \frac{a(4)^2}{(6)(2)} = \frac{4a}{3}, and this must be \displaystyle -\frac{8}{3}, so a = -2.
Check the remaining feature. Numerator and denominator now both have degree 2, with leading coefficients -2 and 1, so the horizontal asymptote is y = -2, as drawn. Check the signs too: in the middle interval (-1, 3) the denominator is negative and -2(x - 1)^2 is negative or zero, so t is positive or zero, matching the U above the axis; outside that interval the denominator is positive and t is negative, matching the outer branches. So
The answer is the simplest formula with these features, not the only one, because a graph shows only whether each power is odd or even. For instance \displaystyle \frac{-2(x - 1)^4}{(x + 1)^3(x - 3)} has the same asymptotes, the same touch at 1 and the same signs, only flatter near the touch. It misses the marked point (its value at 5 is \displaystyle -\frac{32}{27}), so the marked point is what picks out the simpler formula here.
Boundaries
- A sketch shows shape, not exact turning points. The method finds every break, intercept, crossing and sign, which fixes the shape. It does not find the exact top of the small hump in Example 1, which sits a little above \displaystyle \frac{4}{9}; locating such a high or low point exactly needs the derivative, from calculus.
- Cancel completely, and only then apply the rules. The asymptote rule and the multiplicity rules are statements about the lowest-terms formula. A zero of the original denominator may be a hole (Example 4).
- A hole is never a zero, even on the axis. For \displaystyle \frac{(x - 2)^2}{x^2 - 2x} the canceled formula is 0 at x = 2, so the hole sits at (2, 0) on the axis, but the function has no value at 2 and so no zero there.
- Only real zeros make breaks. A denominator such as x^2 + 4 is never zero for a real input, so it gives no vertical asymptote and no change of sign.
- No horizontal asymptote when the top has the higher degree. If the degree is higher by exactly one, the graph follows a slant line (Example 3); higher by more, it follows no line at all, and far out it behaves like the power \displaystyle \frac{a}{b} x^{\,n - m}.
- A graph shows only whether each power is odd or even. Example 6 found the simplest formula; higher powers of the same parity, with other factors and constants adjusted, can give graphs with the same asymptotes, intercepts and signs.
- The method is for rational functions. For other functions a zero denominator, or a sign change, can behave differently: \displaystyle \frac{|x|}{x} jumps from -1 to 1 at x = 0 with no asymptote.
Where this goes next
The sign table built here is the whole of Polynomial and rational inequalities with sign charts: to solve \displaystyle \frac{P(x)}{Q(x)} > 0 you find the same breaks and read off the intervals where the sign is +. The slanted line of Example 3 is studied in Slant asymptotes. In Rational functions in context the same sketch, cut down to the inputs that make sense, shows an average cost falling toward its asymptote or a mixture’s concentration leveling off. The polynomial version of this lesson, without asymptotes, is Graphing polynomial functions. In calculus the arrows near each break become infinite one-sided limits and the far-out behavior becomes limits at infinity, and the derivative supplies the exact turning points this sketch leaves open.
Exercises
Exercises 1 to 7 go with Day 1 and 8 to 16 with Day 2. In Exercises 1 to 6 and 8 to 13 and 16, find every hole, vertical asymptote, horizontal asymptote (or slant line), and intercept, give the sign on each interval, and sketch the graph.
Day 1: sketching
- \displaystyle \frac{2x + 6}{x - 1}
- \displaystyle \frac{x + 2}{x^2 - 9}
- \displaystyle \frac{x^2 - x - 2}{x^2 + 2x - 3}, including where it crosses its horizontal asymptote
- \displaystyle \frac{-4}{x^2 - 2x - 3}
- \displaystyle \frac{8}{x^2 + 4}
- \displaystyle \frac{x^2 + 3x}{x - 2}
- Find the vertical and horizontal asymptotes of \displaystyle \frac{x^2 + 5x}{x^2 - 4}, and the point where the graph crosses the horizontal one.
Day 2: holes
- \displaystyle \frac{x^2 - 4x}{x^2 - x - 12}
- \displaystyle \frac{x^2 + x - 6}{x^2 - 4}
- \displaystyle \frac{2x^2 - 2x}{x^2 - 1}
Day 2: doubled factors
- \displaystyle \frac{x^2}{x^2 - 6x + 9}
- \displaystyle \frac{x + 1}{(x - 2)^2}
- \displaystyle \frac{-x^2}{x^2 + 2x - 8}
Day 2: building and mixing
- Write a rational function whose graph has vertical asymptotes x = -2 and x = 5, horizontal asymptote y = 3, and crosses the horizontal axis at x = 1 and x = 4 and nowhere else.
- Write a rational function whose graph has vertical asymptote x = 0, horizontal asymptote y = 0, a hole at x = 1, no zeros, and passes through (2, 3). Give the coordinates of the hole.
- \displaystyle \frac{x^3 - x}{x^2 - 3x + 2}
Reasoning
- Explain why one test value in each interval between consecutive zeros and vertical asymptotes is enough to know the sign of a rational function on the whole interval.
- Explain why the graph of a rational function may cross its horizontal asymptote but can never meet one of its vertical asymptotes.
- Brisa says the graph of \displaystyle \frac{x^2 - 9}{x - 3} has a vertical asymptote at x = 3. Explain what the graph really looks like there.
- Give a rational function whose graph has a hole lying on the horizontal axis, and explain why the first coordinate of that hole is not a zero of the function.
Answers
- Vertical asymptote x = 1, horizontal asymptote y = 2, zero (-3, 0), y-intercept (0, -6). Signs: + on (-\infty, -3), - on (-3, 1), + on (1, \infty). Near 1, C = 8 > 0, so the graph falls along x = 1 on the left and comes down from the top on the right. It never crosses y = 2, since 2x + 6 = 2x - 2 has no solution.
- Vertical asymptotes x = -3 and x = 3, horizontal asymptote y = 0, zero (-2, 0), y-intercept \displaystyle \left(0, -\frac{2}{9}\right). The denominator is (x - 3)(x + 3). Signs: - on (-\infty, -3), + on (-3, -2), - on (-2, 3), + on (3, \infty), from the test values -4, -2.5, 0, 4.
- Vertical asymptotes x = -3 and x = 1, horizontal asymptote y = 1, zeros (-1, 0) and (2, 0), y-intercept \displaystyle \left(0, \frac{2}{3}\right). It equals \displaystyle \frac{(x - 2)(x + 1)}{(x + 3)(x - 1)}, nothing cancels. Signs: +, -, +, -, + on (-\infty, -3), (-3, -1), (-1, 1), (1, 2), (2, \infty). Crossing: x^2 - x - 2 = x^2 + 2x - 3 gives \displaystyle x = \frac{1}{3}, at \displaystyle \left(\frac{1}{3}, 1\right).
- Vertical asymptotes x = -1 and x = 3, horizontal asymptote y = 0, no zeros, y-intercept \displaystyle \left(0, \frac{4}{3}\right). The denominator is (x - 3)(x + 1). Signs: - on (-\infty, -1), + on (-1, 3), - on (3, \infty). The middle branch is a U above the axis, rising to \infty at both asymptotes; the outer branches come up toward the axis from below. The numerator is never zero, so the graph never meets its asymptote y = 0.
- No vertical asymptote (the denominator is at least 4), horizontal asymptote y = 0, no zeros, y-intercept (0, 2). Positive everywhere. The denominator is smallest at x = 0, so (0, 2) is the highest point, and the graph is a single hill falling toward the axis on both sides.
- Vertical asymptote x = 2, no horizontal asymptote, zeros (-3, 0) and (0, 0), y-intercept (0, 0). Since x^2 + 3x = (x - 2)(x + 5) + 10, the function is \displaystyle x + 5 + \frac{10}{x - 2} and the graph follows the slant line y = x + 5. Near 2, C = 10 > 0, so the graph falls along x = 2 on the left and comes down from the top on the right. Signs: - on (-\infty, -3), + on (-3, 0), - on (0, 2), + on (2, \infty).
- Vertical asymptotes x = -2 and x = 2, horizontal asymptote y = 1. Setting x^2 + 5x = x^2 - 4 gives 5x = -4, so the crossing is at \displaystyle \left(-\frac{4}{5}, 1\right).
- Hole at \displaystyle \left(4, \frac{4}{7}\right), vertical asymptote x = -3, horizontal asymptote y = 1, zero and y-intercept both (0, 0). It is \displaystyle \frac{x(x - 4)}{(x - 4)(x + 3)} = \frac{x}{x + 3} for x \ne 4. Signs: + on (-\infty, -3), - on (-3, 0), + on (0, \infty), with the open circle on the right branch.
- Hole at \displaystyle \left(2, \frac{5}{4}\right), vertical asymptote x = -2, horizontal asymptote y = 1, zero (-3, 0), y-intercept \displaystyle \left(0, \frac{3}{2}\right). It is \displaystyle \frac{(x + 3)(x - 2)}{(x - 2)(x + 2)} = \frac{x + 3}{x + 2} for x \ne 2. Signs: + on (-\infty, -3), - on (-3, -2), + on (-2, \infty).
- Hole at (1, 1), vertical asymptote x = -1, horizontal asymptote y = 2, zero and y-intercept both (0, 0). It is \displaystyle \frac{2x(x - 1)}{(x - 1)(x + 1)} = \frac{2x}{x + 1} for x \ne 1. Signs: + on (-\infty, -1), - on (-1, 0), + on (0, \infty).
- Vertical asymptote x = 3, up on both sides, horizontal asymptote y = 1, zero and y-intercept (0, 0), where the graph touches. It is \displaystyle \frac{x^2}{(x - 3)^2}, never negative. Crossing: x^2 = x^2 - 6x + 9 gives \displaystyle x = \frac{3}{2}, at \displaystyle \left(\frac{3}{2}, 1\right).
- Vertical asymptote x = 2, up on both sides (C = 3 > 0, even power), horizontal asymptote y = 0, zero (-1, 0), a crossing, y-intercept \displaystyle \left(0, \frac{1}{4}\right). Signs: - on (-\infty, -1), + on (-1, 2) and on (2, \infty).
- Vertical asymptotes x = -4 and x = 2, horizontal asymptote y = -1, zero (0, 0), where the graph touches. It is \displaystyle \frac{-x^2}{(x + 4)(x - 2)}. Signs: - on (-\infty, -4), + on (-4, 0) and on (0, 2), - on (2, \infty). Crossing: -x^2 = -x^2 - 2x + 8 gives x = 4, at (4, -1).
- One answer: \displaystyle \frac{3(x - 1)(x - 4)}{(x + 2)(x - 5)}. The denominator gives the asymptotes, the numerator the crossings (odd powers), and equal degrees with leading coefficients 3 and 1 give y = 3.
- One answer: \displaystyle \frac{6(x - 1)}{x(x - 1)}, with the hole at (1, 6). It equals \displaystyle \frac{a}{x} for x \ne 1; passing through (2, 3) needs \displaystyle \frac{a}{2} = 3, so a = 6, and the hole’s height is \displaystyle \frac{6}{1} = 6.
- Hole at (1, -2), vertical asymptote x = 2, no horizontal asymptote, zeros (-1, 0) and (0, 0), y-intercept (0, 0). It is \displaystyle \frac{x(x - 1)(x + 1)}{(x - 1)(x - 2)} = \frac{x(x + 1)}{x - 2} for x \ne 1, which is \displaystyle x + 3 + \frac{6}{x - 2}, so the graph follows the slant line y = x + 3. Signs: - on (-\infty, -1), + on (-1, 0), - on (0, 2), + on (2, \infty).
- In lowest terms the function is a constant times a product of linear factors x - c (and factors with no real zeros, which never change sign), some on top and some below. A factor x - c is negative for x < c and positive for x > c, so it can change sign only at c. Inside an interval containing no zero and no asymptote, no factor changes sign, so the quotient keeps one sign throughout, and any one value shows it.
- A horizontal asymptote is about the outputs far out, and the graph meets y = L wherever the equation p(x) = L \cdot q(x) has a solution in the domain, which can happen (Example 1). A vertical asymptote x = a sits at a zero of the lowest-terms denominator, so a is not an allowed input and no point of the graph has first coordinate a.
- There is a hole, not an asymptote. Since x^2 - 9 = (x - 3)(x + 3), the function equals x + 3 for x \ne 3, so the graph is the line y = x + 3 with the single point (3, 6) removed, drawn as an open circle. The outputs near 3 settle on 6 rather than running off.
- One answer: \displaystyle \frac{(x - 2)^2}{x^2 - 2x}. It equals \displaystyle \frac{x - 2}{x} for x \ne 2, which is 0 at x = 2, so the hole is at (2, 0). A zero is an input where the function’s value is 0, and the function has no value at x = 2, because its denominator x^2 - 2x is zero there. So the graph never reaches (2, 0), and 2 is not a zero.