Algebra 2 · F.BF.B.3

Transforming the reciprocal graph

Stretch or slide the graph of one over x, and both of the lines it hugs move with it. Where do they go, and how do you draw the curve around them without plotting a long table?

You need

Graphing by transforming a parent graphAlgebra 2 The square root and reciprocal functionsAlgebra 2 Inverse, joint and combined variationAlgebra 2

Where you are 5 in Rational functions

Functions13

Rational functions5

Asymptotes3 Graphing rational functions Domain and zeros of a rational function

Transforming the reciprocal graph

Rational functions in context

When you reach for it

You already know the graph of \displaystyle y = \frac{1}{x} from the reciprocal function: two curves, one in the first quadrant and one in the third, each hugging both axes without touching them. A great many rules are that same graph moved:

\begin{aligned} g(x) &= \frac{3}{x + 2} - 1, \\ C(n) &= \frac{240}{n} + 5, \\ f(x) &= \frac{4x - 1}{x - 3}. \end{aligned}

Reach for this move whenever a rule is, or can be rewritten as, a number over x - h plus a number. From the rule alone you get both asymptotes, the domain, the range, the shape and a handful of exact points, without a table of dozens of values. It is the parent graph method applied to the one parent whose key features are two lines rather than a single point.

The same move gives inverse variation its graph as a function. The rule \displaystyle y = \frac{a}{x} is the reciprocal stretched by a, and when a fixed amount is added (a set-up time, a cost everyone pays) the graph is that curve slid up or across.

The procedure

One form for every move

From graphing by transformations: a vertical stretch by |a| (with a reflection in the x-axis when a < 0), then a shift right by h and up by k, turns the graph of y = f(x) into the graph of y = a f(x - h) + k. With \displaystyle f(x) = \frac{1}{x}, and since \displaystyle a \cdot \frac{1}{x - h} = \frac{a}{x - h}, every such move of the reciprocal has the form

y = \frac{a}{x - h} + k, \qquad a \neq 0.

Number	What it does to the curve	What it does to the asymptotes
a	pulls the branches away from the center by \lvert a \rvert; if a < 0, also reflects	nothing
h	slides right by h (left if h < 0)	x = 0 becomes x = h
k	slides up by k (down if k < 0)	y = 0 becomes y = k

Read h with care. Since x + 2 = x - (-2), the rule \displaystyle \frac{3}{x + 2} - 1 has h = -2 and moves the graph left 2. Its numbers are a = 3, h = -2, k = -1.

In the language of inverse variation the constant is called k, as in \displaystyle y = \frac{k}{x}. Here k already means the shift up, so the constant of variation is called a: inverse variation \displaystyle y = \frac{a}{x} is the case h = 0, k = 0. As a function it assigns one output to every input except 0, and its graph is the reciprocal stretched by a.

The asymptotes depend on h and k alone

The input x = h makes the denominator 0, so it is excluded. For x close to h, the number x - h is small and \displaystyle \frac{a}{x - h} is large in size: with a = 3, any x within 0.001 of h gives \displaystyle \left|\frac{3}{x - h}\right| > 3000. So the curve runs up or down beside the line x = h, which is the vertical asymptote.

For x far from h, in either direction, \displaystyle \frac{a}{x - h} is close to 0, so y is close to k. But a fraction is 0 only when its numerator is, and a \neq 0, so \displaystyle \frac{a}{x - h} is never 0 and y never equals k. The line y = k is the horizontal asymptote, and this graph never meets it.

Multiplying by a keeps large numbers large and small numbers small, and it does not change which input is excluded. So a reshapes the curve but leaves both lines where they are. The asymptotes cross at (h, k), which this lesson calls the center of the graph. Draw the two dashed lines first; the curve is built around them.

Measured from the center, it is inverse variation again

Subtract k from both sides of \displaystyle y = \frac{a}{x - h} + k and multiply by x - h:

(x - h)(y - k) = a.

So for every point on the graph, the distance across from the center times the distance up from the center is the same number a. That is inverse variation, in coordinates whose origin is the center. Two things follow.

Which corners. If a > 0, the across and up distances have the same sign, so the branches sit up and to the right of the center and down and to the left, as for \displaystyle \frac{1}{x}. If a < 0 they have opposite signs, and the branches sit up and to the left and down and to the right.

Which points. One step across from the center, the curve is a steps up (or down, if a < 0). When a is an integer, |a| steps across it is one step up or down. And if (h + d,\ k + e) is on the graph, so is (h - d,\ k - e), because (-d)(-e) = de = a. The midpoint of those two points is (h, k): the graph is unchanged by a half turn about its center, so each point found on one branch gives a point on the other.

Domain, range and intercepts

The domain is every number except h. The range is every number except k: solving \displaystyle y = \frac{a}{x - h} + k for x gives

x = h + \frac{a}{y - k},

which produces an input for every y \neq k and none for y = k.

The y-intercept comes from x = 0, possible only when h \neq 0. The x-intercept comes from solving \displaystyle \frac{a}{x - h} + k = 0, that is \displaystyle x - h = -\frac{a}{k}. That needs k \neq 0: when k = 0 the horizontal asymptote is the x-axis itself and the graph never meets it.

Rules that arrive in another shape

A number in front of x. In \displaystyle \frac{a}{bx - c}, factor b out of the denominator:

\begin{aligned} \frac{a}{bx - c} &= \frac{a}{b\left(x - \frac{c}{b}\right)} \\ &= \frac{\frac{a}{b}}{x - \frac{c}{b}}. \end{aligned}

So the stretch is \displaystyle \frac{a}{b} and the shift is \displaystyle \frac{c}{b}. Behind this sit two facts special to the reciprocal. A horizontal squeeze is a vertical one, since \displaystyle \frac{1}{bx} = \frac{1}{b} \cdot \frac{1}{x}; and a reflection in the y-axis is a reflection in the x-axis, since \displaystyle \frac{1}{-x} = -\frac{1}{x}. Every horizontal move is therefore absorbed into a and h.

One fraction, linear over linear. In \displaystyle \frac{px + q}{x - h}, write the numerator as a multiple of the denominator plus a number, px + q = p(x - h) + (q + ph), and split the fraction (fractions with the same denominator add by their numerators):

\frac{px + q}{x - h} = p + \frac{q + ph}{x - h}.

Now k = p and a = q + ph. If the denominator is rx - s instead, factor out r first.

Worked examples

Example 1

A stretched and shifted reciprocal

Graph \displaystyle g(x) = \frac{3}{x + 2} - 1. Give its asymptotes, domain, range and intercepts.

Solution

Write x + 2 as x - (-2): then a = 3, h = -2, k = -1. The graph of \displaystyle \frac{1}{x} is stretched by 3, moved left 2 and moved down 1.

The asymptotes move with the shifts only: x = 0 goes to x = -2 and y = 0 goes to y = -1. They cross at the center (-2, -1). Since a = 3 > 0, the branches sit up and to the right of the center and down and to the left.

Points from the center, using (x + 2)(y + 1) = 3: one across, three up gives (-1, 2); three across, one up gives (1, 0). Their mirror images through the center are (-3, -4) and (-5, -2). Check two of them in the rule: \displaystyle g(-1) = \frac{3}{1} - 1 = 2 and \displaystyle g(-5) = \frac{3}{-3} - 1 = -2.

Intercepts. \displaystyle g(0) = \frac{3}{2} - 1 = \frac{1}{2}, so the graph crosses the y-axis at \displaystyle \left(0, \frac{1}{2}\right). For the x-axis, \displaystyle \frac{3}{x + 2} - 1 = 0 gives \displaystyle \frac{3}{x + 2} = 1, so x + 2 = 3 and x = 1: the point (1, 0) already found.

The domain is every number except -2, (-\infty, -2) \cup (-2, \infty), and the range is every number except -1, (-\infty, -1) \cup (-1, \infty).

Figure 1The asymptotes of \displaystyle g(x) = \frac{3}{x + 2} - 1 cross at (-2, -1). One step across from that point the curve is 3 steps up, because a = 3.

Example 2

A reflection moves the branches to the other corners

Graph \displaystyle m(x) = -\frac{2}{x - 4} + 5, with its asymptotes and intercepts.

Solution

In the form \displaystyle \frac{a}{x - h} + k the numbers are a = -2, h = 4 and k = 5. So the reciprocal is stretched by 2, reflected in the x-axis, moved right 4 and up 5.

The asymptotes are x = 4 and y = 5, crossing at (4, 5). Since a < 0, the across and up distances from the center have opposite signs: one branch sits up and to the left of the center, the other down and to the right.

Points: one across, two down gives (5, 3); two across, one down gives (6, 4). Through the center these give (3, 7) and (2, 6). Check: \displaystyle m(3) = -\frac{2}{-1} + 5 = 7.

Intercepts. \displaystyle m(0) = -\frac{2}{-4} + 5 = \frac{1}{2} + 5 = \frac{11}{2}. For the x-intercept, \displaystyle -\frac{2}{x - 4} + 5 = 0 gives \displaystyle \frac{2}{x - 4} = 5, so \displaystyle x - 4 = \frac{2}{5} and \displaystyle x = \frac{22}{5}. Check: \displaystyle -\frac{2}{2/5} + 5 = -5 + 5 = 0. So the lower right branch dips below the x-axis only for \displaystyle 4 < x < \frac{22}{5}, where it plunges beside its vertical asymptote.

The domain is every number except 4 and the range every number except 5.

Figure 2With a = -2 the branches of \displaystyle m(x) = -\frac{2}{x - 4} + 5 sit in the other two corners around the crossing point (4, 5).

Example 3

From a description to a rule

The graph of \displaystyle y = \frac{1}{x} is stretched vertically by 6, reflected in the x-axis, then moved left 5 and down 2. Write the new rule, its asymptotes and one point on each branch.

Solution

The stretch with reflection gives a = -6; left 5 means h = -5; down 2 means k = -2. The rule is

y = -\frac{6}{x + 5} - 2.

The asymptotes are x = -5 and y = -2. One step right of the center the curve is 6 steps down, at (-4, -8); its mirror image through the center is (-6, 4). Check: \displaystyle -\frac{6}{1} - 2 = -8 and \displaystyle -\frac{6}{-1} - 2 = 4.

The order in the description matters. Had the graph been moved down 2 first and stretched and reflected afterwards, the shift would have been stretched too: \displaystyle -6\left(\frac{1}{x + 5} - 2\right) = -\frac{6}{x + 5} + 12, with horizontal asymptote y = 12 instead.

Example 4

Reading the rule from a graph

The graph below has asymptotes x = 1 and y = 2 and passes through (3, 6). Find its rule, and check it against the other marked points (5, 4), (-3, 0) and (0, -6).

Figure 3The graph in Example 4. The two asymptotes and one labeled point are enough to find the rule.

Solution

The asymptotes give h = 1 and k = 2, so the rule is \displaystyle y = \frac{a}{x - 1} + 2 for some a. Substitute the point (3, 6):

6 = \frac{a}{3 - 1} + 2, \qquad 4 = \frac{a}{2}, \qquad a = 8.

The rule is \displaystyle y = \frac{8}{x - 1} + 2. The center form checks every point at once, since (x - 1)(y - 2) must equal 8: for (3, 6), 2 \cdot 4 = 8; for (5, 4), 4 \cdot 2 = 8; for (-3, 0), (-4)(-2) = 8; for (0, -6), (-1)(-8) = 8. The sign agrees with the picture too: the branches sit up and to the right and down and to the left of (1, 2), which needs a > 0.

Example 5

A single fraction, rewritten

Graph \displaystyle f(x) = \frac{4x - 1}{x - 3} by first writing it in the form \displaystyle \frac{a}{x - h} + k.

Solution

Write the numerator as a multiple of x - 3 plus a number: 4(x - 3) = 4x - 12, and 4x - 1 = (4x - 12) + 11. So

\begin{aligned} f(x) &= \frac{4(x - 3) + 11}{x - 3} \\ &= 4 + \frac{11}{x - 3} = \frac{11}{x - 3} + 4. \end{aligned}

The split is allowed because fractions with the same denominator add by their numerators, and \displaystyle \frac{4(x - 3)}{x - 3} = 4 for every x \neq 3. Check at x = 4: \displaystyle \frac{15}{1} = 15 and \displaystyle \frac{11}{1} + 4 = 15.

So a = 11, h = 3, k = 4: asymptotes x = 3 and y = 4, branches up and to the right and down and to the left of (3, 4). One across, eleven up gives (4, 15); eleven across, one up gives (14, 5); their mirror images are (2, -7) and (-8, 3).

For the intercepts the original form is easier. \displaystyle f(0) = \frac{-1}{-3} = \frac{1}{3}. A fraction is 0 exactly when its numerator is 0 and its denominator is not, so 4x - 1 = 0 gives \displaystyle x = \frac{1}{4}. The horizontal asymptote y = 4 is the ratio of the two x-coefficients, which is what the rule for horizontal asymptotes says in general.

Example 6

A number in front of x, with a minus sign

Find the asymptotes, the moves and the intercepts of \displaystyle y = \frac{4}{10 - 2x} - 3.

Solution

Factor the denominator so that x stands alone: 10 - 2x = -2(x - 5). Then

\frac{4}{10 - 2x} = \frac{4}{-2(x - 5)} = -\frac{2}{x - 5},

y = -\frac{2}{x - 5} - 3.

So a = -2, h = 5, k = -3: the reciprocal stretched by 2, reflected in the x-axis, moved right 5 and down 3. The asymptotes are x = 5 (where 10 - 2x = 0) and y = -3, and the branches sit up and to the left and down and to the right of (5, -3). Check at x = 6: \displaystyle \frac{4}{10 - 12} - 3 = -2 - 3 = -5, and \displaystyle -\frac{2}{1} - 3 = -5.

Read straight from the original, the -2x suggests a horizontal squeeze by 2 and a reflection in the y-axis. For the reciprocal those are the same as a vertical squeeze by 2 and a reflection in the x-axis, which is why the factored form needs only a and h.

Intercepts. At x = 0, \displaystyle y = \frac{4}{10} - 3 = -\frac{13}{5}. For y = 0, \displaystyle -\frac{2}{x - 5} = 3 gives \displaystyle x - 5 = -\frac{2}{3}, so \displaystyle x = \frac{13}{3}. Check: \displaystyle 10 - \frac{26}{3} = \frac{4}{3}, and \displaystyle \frac{4}{4/3} - 3 = 3 - 3 = 0.

Example 7

Inverse variation with a fixed part added

A job needs 18 person-hours of work, so n people working together take \displaystyle \frac{18}{n} hours. Before starting, everyone sits through a 1-hour safety briefing, so the whole job takes \displaystyle T(n) = \frac{18}{n} + 1 hours. Find T(6), the fewest people who can finish in at most 2.5 hours, and what the horizontal asymptote says.

Solution

The working time \displaystyle \frac{18}{n} varies inversely with n: people times hours is always 18. As a function its graph is the reciprocal stretched by 18, and only the branch with n > 0 means anything here. Adding the briefing moves it up 1, so T has asymptotes n = 0 and T = 1.

T itself does not vary inversely with n: T(1) = 19 and 1 \cdot 19 = 19, but T(2) = 10 and 2 \cdot 10 = 20. What stays fixed is n(T - 1) = 18, the center form with center (0, 1).

\displaystyle T(6) = \frac{18}{6} + 1 = 4 hours. For at most 2.5 hours, \displaystyle \frac{18}{n} + 1 \le 2.5, so \displaystyle \frac{18}{n} \le 1.5. Since n > 0, multiplying both sides by n keeps the direction: 18 \le 1.5n, so n \ge 12. Twelve people, and T(12) = 1.5 + 1 = 2.5 confirms it.

The asymptote T = 1 says that however many people join, the job never takes 1 hour or less: the briefing alone takes that long. The inputs are whole numbers of people, so the true graph is a row of separate dots lying on the curve, and the vertical asymptote at n = 0 is outside the story altogether.

Boundaries

a must not be 0. With a = 0 the rule is y = k, a horizontal line, with no excluded input and no asymptotes.
Linear over linear is a moved reciprocal only when it is not constant. In \displaystyle \frac{px + q}{x - h}, if q + ph = 0 the numerator is p(x - h) and the rule is the constant p with the single input h removed: a line with a hole, no asymptote. Finding such excluded inputs is the subject of the domain of a rational function.
Only these moves. Vertical stretches, reflections and shifts in both directions keep the form. A squared denominator, as in \displaystyle \frac{1}{(x - h)^2}, is a different parent whose two branches sit on the same side of the horizontal asymptote; a denominator with two different zeros gives two vertical asymptotes and is not a moved reciprocal at all.
Never crossing is special to this family. The graph of \displaystyle \frac{a}{x - h} + k never meets y = k, because \displaystyle \frac{a}{x - h} \neq 0. Other rational functions can cross their horizontal asymptote, which only describes the far left and far right.
Inverse variation lives in the center’s coordinates. \displaystyle y = \frac{a}{x - h} + k with h or k not 0 does not make xy constant; it makes (x - h)(y - k) constant.
In a story the inputs may be restricted. A count of people or items is a positive whole number, so only part of one branch is used, as dots.

Where this goes next

Most rational functions are not moved reciprocals, but the habit learned here carries over: find the excluded inputs and the far-off level first, draw them as dashed lines, then place the curve. Asymptotes treats vertical, horizontal and slant asymptotes in general, and horizontal asymptotes gives the rule by degrees and leading coefficients that Example 5 previewed. Graphing rational functions adds holes and several vertical asymptotes. Rational functions in context takes the fixed-part-plus-shared-part models of Example 7 further: average cost, concentration, work. And solving \displaystyle \frac{a}{x - h} + k = c for x, as in the range argument, is the first case of solving a rational equation.

Exercises

Asymptotes, domain and range

For each rule give both asymptotes, the domain, the range, and which corners around the center hold the branches.

\displaystyle y = \frac{5}{x - 7} + 2
\displaystyle y = -\frac{1}{x + 6} - 3
\displaystyle y = \frac{8}{x} - 4
\displaystyle y = \frac{7}{x + 9}

Graph by moving the parent

Name the moves, give the asymptotes, find four points from the center form (two on each branch), and find the intercepts.

\displaystyle y = \frac{2}{x - 3} + 1
\displaystyle y = -\frac{4}{x + 1} - 2
\displaystyle y = \frac{6}{x + 4} + 3

Intercepts

Find every intercept, or explain why there is none.

\displaystyle y = \frac{9}{x - 2} - 3
\displaystyle y = -\frac{5}{x + 5} + 1
\displaystyle y = \frac{1}{x - 8}

Write the rule

The graph of \displaystyle y = \frac{1}{x} is stretched vertically by 5, reflected in the x-axis, then moved right 2 and up 6. Write the rule.
A moved reciprocal has asymptotes x = -3 and y = 4 and passes through (-1, 7). Write its rule.
A moved reciprocal has asymptotes x = 2 and y = -1 and passes through (0, 3). Write its rule.
y varies inversely with x, and (3, -7) is on the graph. Write the rule and name the quadrants its branches lie in. Then write the rule after the graph is moved right 3 and up 1, with its asymptotes.

Rewrite in the moved form

Write each rule as \displaystyle \frac{a}{x - h} + k and give its asymptotes.

\displaystyle y = \frac{3x + 10}{x + 3}
\displaystyle y = \frac{2x - 9}{x - 4}
\displaystyle y = \frac{5}{2x + 8}
\displaystyle y = \frac{x + 7}{2 - x}

In context

A group hires a minibus for the day for 240 dollars, shared equally, and each person also pays 5 dollars for lunch. Write the cost per person C(n) in dollars for n people. Find C(12), the fewest people for a cost of at most 15 dollars each, and explain why no group size brings the cost down to 5 dollars.
A 90 km drive at a steady v km per hour includes a half-hour stop. Write the total time T(v) in hours, find T(45), and find the speed that gives a total of 2 hours.

Reasoning

Explain why changing a in \displaystyle y = \frac{a}{x - h} + k never moves either asymptote, while changing h or k does.
Show that \displaystyle \frac{1}{-x} = -\frac{1}{x} for every x \neq 0, and explain why reflecting the graph of \displaystyle y = \frac{1}{x} in the y-axis gives the same graph as reflecting it in the x-axis.
Prove that if (h + d,\ k + e) lies on the graph of \displaystyle y = \frac{a}{x - h} + k, then so does (h - d,\ k - e), and explain what this says about the shape of the graph.
Calder thinks \displaystyle g(x) = \frac{3}{x + 2} - 1 might miss some outputs other than -1. Solve \displaystyle y = \frac{3}{x + 2} - 1 for x to show that every number except -1 is an output, each exactly once.
Dovan says \displaystyle T(n) = \frac{18}{n} + 1 from Example 7 is inverse variation, because T goes down as n goes up. Explain why it is not, and say which quantity does vary inversely with n.

Answers

Asymptotes x = 7 and y = 2; domain every number except 7, range every number except 2. Here a = 5 > 0, so the branches sit up and to the right and down and to the left of (7, 2).
Asymptotes x = -6 and y = -3; domain all x \neq -6, range all y \neq -3. Here a = -1 < 0, so the branches sit up and to the left and down and to the right of (-6, -3).
Asymptotes x = 0 and y = -4; domain all x \neq 0, range all y \neq -4. Here a = 8 > 0, so up and to the right and down and to the left of (0, -4).
Asymptotes x = -9 and y = 0; domain all x \neq -9, range all y \neq 0. Here a = 7 > 0, so up and to the right and down and to the left of (-9, 0).
Stretch by 2, right 3, up 1; asymptotes x = 3 and y = 1. Points (4, 3), (5, 2), (2, -1), (1, 0). The x-intercept is (1, 0), since \displaystyle \frac{2}{x - 3} = -1 gives x = 1; the y-intercept is \displaystyle \left(0, \frac{1}{3}\right), since \displaystyle \frac{2}{-3} + 1 = \frac{1}{3}.
Stretch by 4, reflect in the x-axis, left 1, down 2; asymptotes x = -1 and y = -2. Points (0, -6), (1, -4), (-2, 2), (-3, 0), with branches up and to the left and down and to the right of (-1, -2). The intercepts are (0, -6) and (-3, 0): \displaystyle -\frac{4}{x + 1} = 2 gives x + 1 = -2.
Stretch by 6, left 4, up 3; asymptotes x = -4 and y = 3. Points (-3, 9), (-2, 6), (-5, -3), (-6, 0). The x-intercept is (-6, 0), since \displaystyle \frac{6}{x + 4} = -3 gives x + 4 = -2; the y-intercept is \displaystyle \left(0, \frac{9}{2}\right), since \displaystyle \frac{6}{4} + 3 = \frac{9}{2}.
x-intercept (5, 0): \displaystyle \frac{9}{x - 2} = 3 gives x - 2 = 3. y-intercept \displaystyle \left(0, -\frac{15}{2}\right): \displaystyle \frac{9}{-2} - 3 = -\frac{15}{2}.
Both intercepts are the origin (0, 0): \displaystyle -\frac{5}{0 + 5} + 1 = 0, and \displaystyle \frac{5}{x + 5} = 1 has only the solution x = 0.
No x-intercept, since \displaystyle \frac{1}{x - 8} is never 0 (its numerator is 1), so the graph never meets its asymptote y = 0. y-intercept \displaystyle \left(0, -\frac{1}{8}\right).
\displaystyle y = -\frac{5}{x - 2} + 6.
\displaystyle y = \frac{6}{x + 3} + 4: from \displaystyle 7 = \frac{a}{-1 + 3} + 4, \displaystyle \frac{a}{2} = 3 and a = 6. Check with the center form: (-1 + 3)(7 - 4) = 6.
\displaystyle y = -\frac{8}{x - 2} - 1: from \displaystyle 3 = \frac{a}{0 - 2} - 1, \displaystyle -\frac{a}{2} = 4 and a = -8. Check: (0 - 2)(3 + 1) = -8.
\displaystyle y = -\frac{21}{x}, since a = xy = 3 \cdot (-7) = -21; the product is negative, so the branches lie in the second and fourth quadrants. Moved: \displaystyle y = -\frac{21}{x - 3} + 1, with asymptotes x = 3 and y = 1.
\displaystyle y = \frac{1}{x + 3} + 3, since 3x + 10 = 3(x + 3) + 1. Asymptotes x = -3 and y = 3.
\displaystyle y = -\frac{1}{x - 4} + 2, since 2x - 9 = 2(x - 4) - 1. Asymptotes x = 4 and y = 2.
\displaystyle y = \frac{5/2}{x + 4}, that is \displaystyle \frac{5}{2} \cdot \frac{1}{x + 4}, since 2x + 8 = 2(x + 4). Asymptotes x = -4 and y = 0.
\displaystyle y = -\frac{9}{x - 2} - 1. First \displaystyle \frac{x + 7}{2 - x} = -\frac{x + 7}{x - 2}, and x + 7 = (x - 2) + 9, so it is \displaystyle -\left(1 + \frac{9}{x - 2}\right). Asymptotes x = 2 and y = -1. Check at x = 3: \displaystyle \frac{10}{-1} = -10 and -9 - 1 = -10.
\displaystyle C(n) = \frac{240}{n} + 5. C(12) = 20 + 5 = 25, so 25 dollars each. \displaystyle \frac{240}{n} \le 10 with n > 0 gives n \ge 24: 24 people. The share \displaystyle \frac{240}{n} is never 0, so C(n) is always more than 5 dollars; C = 5 is the horizontal asymptote.
\displaystyle T(v) = \frac{90}{v} + \frac{1}{2}. \displaystyle T(45) = 2 + \frac{1}{2} = \frac{5}{2} hours. For T = 2, \displaystyle \frac{90}{v} = \frac{3}{2}, so v = 60 km per hour.
The vertical asymptote is the excluded input, where x - h = 0; that depends on h only, and multiplying the fraction by a excludes the same input. The horizontal asymptote is the level k that y nears because \displaystyle \frac{a}{x - h} nears 0; multiplying a number near 0 by a fixed a leaves it near 0, so the level is still k. Changing h or k moves the whole graph, lines included.
\displaystyle (-x) \cdot \left(-\frac{1}{x}\right) = 1, so \displaystyle -\frac{1}{x} is the reciprocal of -x, which is \displaystyle \frac{1}{-x}. Reflecting in the y-axis replaces x by -x, giving \displaystyle y = \frac{1}{-x}; reflecting in the x-axis negates the output, giving \displaystyle y = -\frac{1}{x}. These are the same rule, so the same graph.
On the graph, (x - h)(y - k) = a, so de = a. Then (-d)(-e) = de = a, so (h - d,\ k - e) also satisfies the equation, and its x is not h because d \neq 0. The two points have midpoint (h, k), so turning the graph half a turn about the center maps it onto itself: each branch is the other turned round.
Add 1 to both sides: \displaystyle y + 1 = \frac{3}{x + 2}. Then for y \neq -1, \displaystyle x + 2 = \frac{3}{y + 1} and \displaystyle x = \frac{3}{y + 1} - 2. This gives exactly one input for each y \neq -1; for y = -1 it would need \displaystyle \frac{3}{x + 2} = 0, which never happens.
Inverse variation needs nT to be constant, but 1 \cdot T(1) = 19 and 2 \cdot T(2) = 20. Falling as n rises is not enough; the product must stay fixed. The working time \displaystyle T - 1 = \frac{18}{n} does vary inversely with n, since n(T - 1) = 18.