The definition of a rational function, the parent hyperbola y equals 1 over x with its two asymptotes, graphing y equals a over x, translating to y equals a over x minus h plus k, finding both asymptotes of the general linear-over-linear form, and using a rational model for average cost.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 8 — Rational Functions
Graph Simple Rational Functions
Objectives
Five outcomes. One new kind of graph, and two lines it never crosses.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 558-564 — the lesson these objectives are drawn from
Warm-up
Lesson 8.1 gave you inverse variation, y equals a over x, and its two-branch graph.
Discussion prompt
For y equals 1 over x, what happens as x gets closer and closer to 0 from the right? And as x gets very large?
Hint: Try x equal to 0.1, then 0.01, then 100.
Answer:
\[ x = 0.1 \to y = 10; \quad x = 0.01 \to y = 100; \quad x = 0.001 \to y = 1000 \]
\[ x = 10 \to y = 0.1; \quad x = 100 \to y = 0.01; \quad x = 1000 \to y = 0.001 \]
Near zero the outputs run away without bound, and far out they shrink toward zero without reaching it. Those two behaviours are what asymptotes describe, and this lesson gives them names and rules.
Concept
A rational function is one polynomial divided by another. The simplest ones graph as hyperbolas: two separate branches that approach a vertical line and a horizontal line without ever meeting either.
rational function — A function of the form p of x over q of x, where p and q are polynomials and q is not the zero polynomial. The inverse variation function a over x is the simplest case.
\[ f(x) = \frac{p(x)}{q(x)}, \quad q(x) \neq 0 \]
The vertical asymptote comes from the denominator: division by zero is undefined, so the graph has nothing there. The horizontal one describes what happens far out along the axis.
Figure (svg): The parent hyperbola with its vertical and horizontal asymptotes marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 558-558
Section
Section 1
Concept
The graph of f of x equals 1 over x is a hyperbola with two symmetric branches. Its vertical asymptote is the vertical axis and its horizontal asymptote is the horizontal axis, and both the domain and the range are every real number except zero.
\[ f(x) = \frac{1}{x}: \; \text{asymptotes } x = 0 \text{ and } y = 0 \]
Every function of the form a over x, with a not zero, has the same asymptotes, the same domain and the same range. Only the distance of the branches from the axes changes.
Figure (svg): The parent hyperbola with its vertical and horizontal asymptotes marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 558-558 — Parent Function for Simple Rational Functions
Picture it
The parent function, with the lines it approaches.
Figure (svg): The parent hyperbola with its vertical and horizontal asymptotes marked
The branches sit in the first and third quadrants. Each approaches one axis horizontally and the other vertically, without ever touching.
Worked example
Reading the parent function's domain and range.
\[ \text{Explain why } f(x) = \tfrac{1}{x} \text{ excludes } 0 \text{ from both the domain and the range.} \]
Test the domain at zero
Why: One divided by zero is undefined, so no output exists there.
\[ 0\text{ not in the domain} \]
Test the range at zero
Why: Setting 1 over x equal to 0 gives 1 equal to 0, which is false.
\[ 0\text{ not in the range} \]
Name the two lines
Why: The excluded input is a vertical asymptote; the excluded output a horizontal one.
\[ x = 0\text{ and } y = 0 \]
Describe the behaviour
Why: Near the excluded input the outputs run away; far out they shrink toward the excluded output.
Figure (svg): The parent hyperbola with its vertical and horizontal asymptotes marked
\[ \text{domain: } x \neq 0; \quad \text{range: } y \neq 0 \]
Verify: check numerically from both sides
Why: At x equal to 0.001 the output is 1000, and at negative 0.001 it is negative 1000 — the two branches run off in opposite directions. At x equal to 1000 the output is 0.001, close to zero but not zero. No input ever produces exactly zero, which is what excluding it from the range means.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 558-558
Sorting
Both parts must be polynomials.
Sort into buckets
Sort each expression.
The third is a polynomial with an invisible denominator of 1, which counts — just as every integer is also a rational number.
Worked example
Applying the definition, including lesson exercise 2.
\[ \text{Which are rational? } \tfrac{1}{x}, \; \tfrac{-3x+5}{2x+1}, \; \tfrac{300m+24000}{m}, \; \sqrt{x}. \]
Check the first
Why: One and x are both polynomials, and x is not the zero polynomial.
Check the second
Why: Both numerator and denominator are first-degree polynomials.
Check the third
Why: The numerator is a polynomial in m and so is the denominator.
Check the fourth
Why: A square root is not a polynomial, so it fails the definition.
Figure (svg): The solution to Worked example is it a rational function shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{1}{x}, \; \tfrac{-3x+5}{2x+1}, \; \tfrac{300m+24000}{m} \text{ are rational} \]
Verify: check the definition both ways
Why: A polynomial alone counts too: x squared plus 1 is that polynomial over the constant polynomial 1. So every polynomial is a rational function, just as every integer is a rational number — and the analogy in the name is not an accident.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 558-561
Trap
\[ \text{a graph never crosses an asymptote} \]
Apply the rule to every rational function
Why: The parent hyperbola never does, so the rule is generalised.
\[ \text{always true} \quad \text{(wrong)} \]
A VERTICAL asymptote is never crossed, since the function is undefined there. A horizontal one describes only far-out behaviour, and can be crossed near the middle.
\[ \text{vertical: never crossed; horizontal: possibly crossed} \]
Separate the two kinds
Why: One marks a gap in the domain; the other describes a limiting height.
\[ \text{for } y = \tfrac{a}{x-h}+k \text{ neither is crossed, but that is special} \]
For the simple hyperbolas of this lesson neither asymptote is crossed. The distinction still matters, because more complicated rational functions in Lesson 8.3 do cross their horizontal asymptotes.
Fill the middle
The parent function.
Fill in the blanks
f(x) = \frac0___ \text___ x = ___
Why: The denominator is zero at x equal to 0, and division by zero is undefined. That single excluded input is exactly where the vertical asymptote sits.
Matching
The parent hyperbola.
Match the pairs
Why: The two asymptotes name exactly the two excluded values. That correspondence holds for every hyperbola in this lesson, which is why finding the asymptotes and stating the domain and range are the same job.
Prediction
Commit before reasoning.
Predict first
For y equals 1 over x, what do the outputs do as x approaches 0 from the left?
Correct: They run to negative infinity.
\[ x \to 0^-: \; y \to -\infty; \qquad x \to 0^+: \; y \to +\infty \]
Why: From the left x is negative and small, so 1 over x is negative and enormous: at negative 0.001 the output is negative 1000. From the right the outputs run to positive infinity instead. The two branches therefore leave in opposite directions, which is exactly what makes a hyperbola two separate pieces rather than one connected curve.
Section
Section 2
Concept
Every function of the form a over x, with a nonzero, has the same asymptotes, domain and range as the parent. A larger absolute value of a pushes the branches farther from the axes, and a negative a moves them to the other pair of quadrants.
\[ y = \frac{6}{x}: \; \text{same asymptotes, farther out} \]
This is a vertical stretch, the same transformation as multiplying any function by a constant. Stretching cannot move a line the graph never touches.
Figure (svg): The parent hyperbola compared with the same hyperbola stretched by a factor of six
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 558-558 — Graph a rational function of the form y = a/x
Picture it
Example 1: the parent dashed, and 6 over x solid.
Figure (svg): The parent hyperbola compared with the same hyperbola stretched by a factor of six
The stretched graph passes through the points where x is 2 and 3 at heights 3 and 2, while the parent passes at 0.5 and one third.
Worked example
Example 1.
\[ \text{Graph } y = \frac{6}{x} \text{ and compare it with } y = \frac{1}{x}. \]
Draw the asymptotes
Why: They are unchanged from the parent.
\[ x = 0\text{ and } y = 0 \]
Plot points on each side
Why: Choose inputs whose reciprocals are easy.
\[ (-3, -2), (-2, -3), (2, 3), (3, 2) \]
Draw the branches
Why: Each passes through its points and bends toward both asymptotes.
Compare with the parent
Why: Every height is 6 times the parent's, so the branches sit farther out.
Figure (svg): The parent hyperbola compared with the same hyperbola stretched by a factor of six
\[ \text{quadrants I and III; } x \neq 0, \; y \neq 0 \]
Verify: compare a single input
Why: At x equal to 2 the parent gives 0.5 and this graph gives 3, exactly 6 times as much. Multiplying every output by 6 stretches the graph vertically but cannot move a line it never reached, so the asymptotes, domain and range are all untouched.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 558-558
Comparison
Fill the blanks. Some things move and some do not.
Comparison matrix
| Feature | y = 1/x | y = 6/x |
|---|---|---|
| Vertical asymptote | x = 0 | x = 0, unchanged |
| Horizontal asymptote | y = 0 | y = 0, unchanged |
| Value at x = 2 | 0.5 | 3 |
| Quadrants | I and III | I and III, unchanged |
Only the third row changed. A vertical stretch multiplies heights and leaves every asymptote, the domain and the range exactly where they were.
Worked example
Guided Practice 1 and lesson exercises 5 and 9.
\[ \text{Graph } f(x) = \frac{-4}{x} \text{ and state its domain, range and quadrants.} \]
Draw the same asymptotes
Why: The constant never moves them.
\[ x = 0\text{ and } y = 0 \]
Plot points on each side
Why: Negative constant, so signs are opposite.
\[ (-2, 2), (-1, 4), (1, -4), (2, -2) \]
Read the quadrants
Why: Positive inputs give negative outputs and the reverse.
State domain and range
Why: Zero is still excluded twice.
\[ \text{all reals except } 0,\text{ both} \]
Figure (svg): The solution to Worked example a negative constant shown as a ladder of expressions, one row per algebraic move
\[ \text{quadrants II and IV}; \; x \neq 0, \; y \neq 0 \]
Verify: check one point's signs
Why: At x equal to 2 the output is negative 2, so the product xy is negative 4 — negative, as a negative constant of variation requires. This is Lesson 8.1's sign rule seen on a graph: a negative constant puts the branches where the coordinates disagree in sign.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 559-561
Error analysis
A student graphs a stretched hyperbola.
Annotate
On: \( y = \frac{6}{x}: \; \text{horizontal asymptote } y = 6 \)
Only a term added outside the fraction moves the horizontal asymptote. That is the k of the next idea, and it is in a different place entirely.
Sorting
The sign of the constant decides.
Sort into buckets
Sort each function by where its branches sit.
The size of the constant changes only how far out the branches sit; the sign alone decides which quadrants they occupy.
Fill the middle
Example 1.
Fill in the blanks
y = \frac2___ \text___ x = 3 \text___ y = ___
Why: Six over 3 is 2. Choosing inputs that divide the constant evenly is what makes plotting by hand quick, which is why 2 and 3 were chosen rather than 1.5 and 2.5.
Prediction
Commit before reasoning.
Predict first
Compare y equals 15 over x with y equals 0.1 over x in the first quadrant.
Correct: The one with 15 lies farther from both axes.
\[ xy = 15 \text{ vs } xy = 0.1: \text{ same asymptotes, different distance} \]
Why: At x equal to 1 the first gives 15 and the second gives 0.1, so the first branch starts far higher. A larger absolute value of the constant means larger products, and a larger constant product pushes the curve away from both axes at once. Their asymptotes are identical, so this is entirely a matter of distance rather than position.
Section
Section 3
Concept
To graph a over x minus h, plus k, draw the asymptotes at x equals h and y equals k, plot points on each side of the vertical one, and sketch the two branches. The domain excludes h and the range excludes k.
\[ y = \frac{a}{x-h}+k: \; \text{asymptotes } x = h, \; y = k \]
Unlike the exponential and logarithmic families, this one has both kinds of asymptote at once, so h moves one and k moves the other.
Figure (svg): A hyperbola translated left two units and down one, with its moved asymptotes
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 559-559 — Graphing Translations of Simple Rational Functions
Picture it
Example 2, with both asymptotes drawn.
Figure (svg): A hyperbola translated left two units and down one, with its moved asymptotes
The asymptotes cross at the point where x is negative 2 and y is negative 1, which is the centre the branches are arranged around.
Worked example
Example 2.
\[ \text{Graph } y = \frac{-4}{x+2}-1 \text{ and state the domain and range.} \]
Read h and k
Why: X plus 2 means h is negative 2, and the constant outside is negative 1.
\[ h = -2, k = -1 \]
Draw the asymptotes
Why: One vertical at h, one horizontal at k.
\[ x = -2\text{ and } y = -1 \]
Plot points on both sides
Why: Two to the left of the vertical asymptote and two to the right.
\[ (-3, 3), (-4, 1), (-1, -5), (0, -3) \]
State the domain and range
Why: Each asymptote names one excluded value.
\[ x\text{ not } -2; y\text{ not } -1 \]
Figure (svg): A hyperbola translated left two units and down one, with its moved asymptotes
\[ x \neq -2, \quad y \neq -1 \]
Verify: check one point on each side
Why: At x equal to negative 3 the fraction is negative 4 over negative 1, which is 4, and subtracting 1 gives 3. At x equal to 0 the fraction is negative 2, and subtracting 1 gives negative 3. Both plotted points check, and both sit on opposite sides of the vertical asymptote as they should.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 559-559
Matching
Solve the denominator for zero; read k outside.
Match the pairs
Why: In every row the vertical asymptote's sign is opposite to the sign written in the denominator, while the horizontal asymptote's sign matches the constant outside. That asymmetry is the whole source of the errors here.
Worked example
Guided Practice 2 and 3.
\[ \text{Graph } y = \frac{8}{x}-5 \text{ and } y = \frac{1}{x-3}+2. \]
First: read h and k
Why: There is no shift inside, so h is 0; the constant outside is negative 5.
\[ x = 0, y = -5 \]
First: state the domain and range
Why: Zero is still excluded, and now negative 5 is too.
\[ x\text{ not } 0; y\text{ not } -5 \]
Second: read h and k
Why: X minus 3 gives h equal to 3, and the constant outside is 2.
\[ x = 3, y = 2 \]
Second: state the domain and range
Why: Three and 2 are the excluded values.
\[ x\text{ not } 3; y\text{ not } 2 \]
Figure (svg): The solution to Worked example two more translations shown as a ladder of expressions, one row per algebraic move
\[ x \neq 0, y \neq -5; \quad x \neq 3, y \neq 2 \]
Verify: test the second above and below
Why: At x equal to 4 the value is 1 plus 2, or 3, above the horizontal asymptote. At x equal to 2 it is negative 1 plus 2, or 1, below it. One branch sits above the line y equal to 2 and one below, which is what the two branches of a hyperbola always do.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 559-559
Trap
\[ y = \frac{-4}{x+2}-1 \]
Take the vertical asymptote as x equals 2
Why: The number 2 is read directly from the denominator.
\[ x = 2 \quad \text{(wrong)} \]
Substituting x equal to 2 gives a perfectly ordinary value, negative 2. There is no gap there.
\[ x + 2 = 0 \;\Longrightarrow\; x = -2 \]
Set the denominator equal to zero and solve
Why: The vertical asymptote sits where the function is undefined, not where the constant is written.
\[ \text{asymptote } x = -2 \]
The same sign flip appeared in Lesson 7.4's logarithmic translations and Lesson 6.5's radicals. Solving the inside for zero avoids it every time.
Fill the middle
Example 2.
Fill in the blanks
x + 2 = 0 \;\Longrightarrow\; x = -2
Why: Setting the denominator equal to zero gives x equal to negative 2. Solving rather than reading is what protects against the sign flip built into the form x minus h.
Comparison
Fill the blanks. Each asymptote names an exclusion.
Comparison matrix
| Asymptote | Excludes | For y = 1/(x - 3) + 2 |
|---|---|---|
| vertical, x = h | one value from the domain | domain excludes 3 |
| horizontal, y = k | one value from the range | range excludes 2 |
| how h is found | set the denominator to zero | x - 3 = 0 |
| how k is found | read the constant added outside | the plus 2 |
Finding the asymptotes and stating the domain and range are the same task written twice, so doing one gives the other for free.
Prediction
Commit before reasoning.
Predict first
How do y equals negative 4 over x and y equals negative 4 over x plus 2, minus 1, differ?
Correct: They are the same shape, one slid left 2 and down 1.
\[ y = \tfrac{a}{x-h}+k: \; h \text{ shifts sideways, } k \text{ shifts up or down} \]
Why: Adding to the input shifts horizontally and adding outside shifts vertically, and neither changes the curve's shape — every point simply moves by the same amount. The constant negative 4 is what controls the shape and the quadrants, and it is identical in both. This is the same transformation grammar as Lessons 4.1, 6.5 and 7.4, applied to a new parent.
Section
Section 4
Concept
Every function of the form a linear expression over a linear expression graphs as a hyperbola. The vertical asymptote is where the denominator is zero, and the horizontal asymptote is the ratio of the two leading coefficients.
\[ y = \frac{ax+b}{cx+d}: \; x = -\frac{d}{c}, \quad y = \frac{a}{c} \]
The horizontal rule makes sense from far away: when x is enormous, the constants b and d barely matter and the fraction behaves like ax over cx.
Figure (svg): A rational function in the general linear-over-linear form with both asymptotes
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 559-560 — Other rational functions
Picture it
Example 3: two x plus 1, all over x minus 3.
Figure (svg): A rational function in the general linear-over-linear form with both asymptotes
The vertical asymptote came from solving the denominator, and the horizontal one from dividing 2 by 1. The four plotted points sit two on each side.
Worked example
Example 3.
\[ \text{Graph } y = \frac{2x+1}{x-3} \text{ and state the domain and range.} \]
Find the vertical asymptote
Why: Solve x minus 3 equal to zero.
\[ x = 3 \]
Find the horizontal asymptote
Why: Divide the leading coefficients, 2 over 1.
\[ y = 2 \]
Plot points on both sides
Why: Two to the left of 3 and two to the right.
\[ (2, -5), (0, -\frac{1}{3}), (4, 9), (6, \frac{13}{3}) \]
State the domain and range
Why: Each asymptote names one exclusion.
\[ x\text{ not } 3; y\text{ not } 2 \]
Figure (svg): A rational function in the general linear-over-linear form with both asymptotes
\[ x \neq 3, \quad y \neq 2 \]
Verify: test the horizontal rule far out
Why: At x equal to 1000 the value is 2001 over 997, which is about 2.007 — close to 2. At x equal to negative 1000 it is negative 1999 over negative 1003, about 1.993. Both approach 2 from opposite sides, confirming that the ratio of leading coefficients really does describe the far-out behaviour.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 560-560
Fill the middle
Example 3.
Fill in the blanks
y = \frac1___: \; \text___ y = \frac______} = 2
Why: The denominator's leading coefficient is 1, since x minus 3 is 1 times x minus 3. Dividing 2 by 1 gives the horizontal asymptote y equal to 2.
Worked example
Guided Practice 4 to 6.
\[ \text{Find the asymptotes of } \frac{x-1}{x+3}, \; \frac{2x+1}{4x-2}, \; \frac{-3x+2}{2x-1}. \]
First: solve the denominator and divide the leaders
Why: X plus 3 vanishes at negative 3; 1 over 1 is 1.
\[ x = -3, y = 1 \]
Second: solve the denominator
Why: Four x minus 2 vanishes at one half.
\[ x = \frac{1}{2} \]
Second: divide the leaders
Why: Two over 4 is one half.
\[ y = \frac{1}{2} \]
Third: both rules again
Why: Two x minus 1 vanishes at one half; negative 3 over 2.
\[ x = \frac{1}{2}, y = -\frac{3}{2} \]
Figure (svg): The solution to Worked example three more asymptote pairs shown as a ladder of expressions, one row per algebraic move
\[ (-3, 1), \; \left(\tfrac{1}{2}, \tfrac{1}{2}\right), \; \left(\tfrac{1}{2}, -\tfrac{3}{2}\right) \]
Verify: notice the second's coincidence
Why: In the second function both asymptotes are at one half, which is a coincidence of the numbers rather than a rule — the vertical came from the denominator and the horizontal from the coefficients, by two unrelated computations. In the third, the same denominator gives the same vertical asymptote but a completely different horizontal one.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 561-561
Error analysis
A student finds the horizontal asymptote of a linear-over-linear function.
Annotate
On: \( y = \frac{2x+1}{x-3}: \; \text{horizontal asymptote } y = \frac{1}{-3} \)
Substituting a huge value settles it in one line: at x equal to 1000 the function gives about 2.007, nowhere near negative one third.
Sorting
Two different computations.
Sort into buckets
Sort each task.
The two asymptotes answer different questions — where there is a gap, and where the graph is heading — so it is unsurprising that they need different computations.
Two truths and a lie
All three are about linear-over-linear functions.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Setting the numerator equal to zero finds an x-INTERCEPT, not an asymptote — for the example it gives x equal to negative one half, a point the graph passes through. The two are easy to confuse because both involve setting a part of the fraction to zero, but one locates a gap and the other locates a crossing.
Prediction
Commit before reasoning.
Predict first
Why does 2x plus 1, over x minus 3, approach 2 for large x?
Correct: Because for huge x the constants 1 and -3 are negligible beside 2x and x.
\[ \frac{2x+1}{x-3} \approx \frac{2x}{x} = 2 \text{ for } |x| \text{ large} \]
Why: At x equal to a million, the numerator is 2,000,001 and the denominator 999,997 — each within a hair of 2x and x. So the fraction is within a hair of 2x over x, which is 2. The constants matter enormously near the origin and not at all far away, which is why they set the intercepts but not the asymptote.
Section
Section 5
Concept
Spreading a fixed cost over a growing number of items gives a rational function. Its horizontal asymptote is the per-item cost, the floor the average approaches but can never reach while the fixed cost still has to be paid.
\[ c = \frac{300m + 24{,}000}{m} \]
Only the first-quadrant branch means anything here, because neither the number of models nor the average cost can be negative. Reading a model's useful domain is part of using it.
Figure (svg): Average cost per printed model falling toward a floor as more models are printed
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 560-560 — Multi-step problem
Picture it
Example 4: a 24,000 dollar printer and 300 dollars of material per model.
Figure (svg): Average cost per printed model falling toward a floor as more models are printed
At 60 models the average is 700 dollars. The curve keeps falling toward 300 and never gets there, because the printer always costs something per model.
Worked example
Example 4.
\[ \text{A printer costs } 24{,}000 \text{ and each model } 300. \text{ Find } m \text{ when the average is } 700. \]
Write the total cost
Why: Material for m models plus the one-off printer.
\[ 300 m + 24, 000 \]
Divide by the number printed
Why: Average cost is total cost over quantity.
\[ c = \frac{300 m + 24, 000}{m} \]
Set the average to 700 and solve
Why: Seven hundred m equals 300m plus 24,000.
\[ 400 m = 24, 000 \]
Read the asymptote
Why: Far out, the average approaches the material cost alone.
\[ m = 60; c\text{ approaches } 300 \]
Figure (svg): Average cost per printed model falling toward a floor as more models are printed
\[ m = 60; \quad c \to 300 \]
Verify: substitute 60 back
Why: Three hundred times 60 is 18,000, plus 24,000 is 42,000, and dividing by 60 gives exactly 700. And the asymptote makes sense in context: the printer's 24,000 dollars spread over m models contributes 24,000 over m per model, which shrinks toward nothing but never reaches it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 560-560
Fill the middle
Example 4.
Fill in the blanks
700m = 300m + 24400000 \;\Longrightarrow\; ___m = 24___000
Why: Subtracting 300m from both sides leaves 400m, and dividing gives 60 models. The 400 is the amount each model contributes toward paying off the printer at that average price.
Worked example
Guided Practice 7.
\[ \text{Repeat with a printer costing } 21{,}000 \text{ instead.} \]
Rewrite the model
Why: Only the fixed cost changes.
\[ c = \frac{300 m + 21, 000}{m} \]
Find the asymptote
Why: The material cost per model is unchanged.
\[ \text{still } c = 300 \]
Solve for an average of 700
Why: Four hundred m equals 21,000.
\[ m = 52.5 \]
Interpret in context
Why: Models come in whole numbers.
\[ \text{about } 53\text{ models} \]
Figure (svg): The solution to Worked example a cheaper printer shown as a ladder of expressions, one row per algebraic move
\[ c = \frac{300m+21{,}000}{m}; \quad m \approx 53 \]
Verify: compare the two printers
Why: The cheaper printer reaches a 700 dollar average after about 53 models instead of 60, and its whole curve sits lower — but both share the same horizontal asymptote at 300, because the material cost is identical. A smaller fixed cost lowers the curve without moving its floor.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 561-561
Trap
\[ c = \frac{300m+24{,}000}{m} \]
Solve for the number of models making the average 300
Why: The asymptote is treated as an attainable value.
\[ 300m = 300m + 24{,}000 \;\Longrightarrow\; 0 = 24{,}000 \quad \text{(impossible)} \]
There is no solution. The equation is telling you that 300 is a limit rather than a value the function takes.
\[ c = 300 + \frac{24{,}000}{m} \]
Rewrite to see the gap explicitly
Why: The average is always the material cost PLUS a positive share of the printer.
\[ \frac{24{,}000}{m} > 0 \text{ for every } m > 0 \]
Rewriting the model this way makes the asymptote obvious and explains it at once: the second term shrinks toward zero but stays positive, so the average stays above 300 forever.
Ranking
Smallest number of models first, largest average cost.
Put in order
Why: Each doubling of the count halves the printer's share per model: 2400, then 1200, 800, 400, 200, added each time to the fixed 300 of material. The averages fall fast at first and then crawl, which is exactly the shape of the curve approaching its asymptote.
Sorting
Read the formula against the situation.
Sort into buckets
Sort each piece of c equals 300m plus 24,000, all over m.
Every symbol in the model corresponds to something in the situation, which is what makes both asymptotes interpretable rather than merely computable.
Prediction
Commit before reasoning.
Predict first
The model has a vertical asymptote at m equal to 0. What does that say?
Correct: That the average cost of zero models is undefined, since you cannot divide 24,000 among nobody.
\[ m \to 0^+: \; c \to +\infty \]
Why: An average needs something to average over. Buying the printer and making nothing leaves 24,000 dollars spread across zero models, which has no meaning — and the model says so by being undefined there. The graph runs away upward as m approaches zero, which matches the intuition that making very few models is ruinously expensive per model.
Comparison
Fill the blanks. Both graph as hyperbolas.
Comparison matrix
| Question | y = a/(x - h) + k | y = (ax + b)/(cx + d) |
|---|---|---|
| Vertical asymptote | x = h | solve cx + d = 0 |
| Horizontal asymptote | y = k | y = a over c |
| Domain | all reals except h | all reals except the zero of the denominator |
| Range | all reals except k | all reals except a over c |
Both rows of asymptotes name exactly the excluded values, so once the asymptotes are found the domain and the range require no extra work.
Pattern
One routine, whichever form you are handed.
In a real model, keep only the branch that makes sense in context — usually the one in the first quadrant.
OpenStax Algebra and Trigonometry 2e, §5.6 Rational Functions §5.6
Check
Translations. Solve the denominator.
Check your understanding
What are the asymptotes of y = -4/(x + 2) - 1?
Answer: A
Why: Setting x + 2 to zero gives -2, and the constant outside is -1.
Check
Linear over linear. Two different rules.
Check your understanding
What are the asymptotes of y = (2x + 1)/(x - 3)?
Answer: A
Why: The denominator vanishes at 3, and the leading coefficients give 2 over 1.
Check
The model. What does the floor mean?
Check your understanding
For c = (300m + 24,000)/m, what does the horizontal asymptote c = 300 represent?
Answer: A
Why: Rewriting gives c = 300 + 24,000/m, so the average is always above 300.
Real world
A drug's concentration in the bloodstream t hours after a dose can be modelled by C equals 5t over t squared plus 4, in milligrams per litre.
Discussion prompt
Find the concentration at 1, 2 and 8 hours, describe what happens over a long time, and say what feature of the graph that behaviour corresponds to.
Hint: Compare the degrees of the numerator and the denominator.
Answer:
\[ C(1) = \tfrac{5}{5} = 1; \quad C(2) = \tfrac{10}{8} = 1.25; \quad C(8) = \tfrac{40}{68} \approx 0.59 \]
The concentration rises to a peak of 1.25 at two hours and then falls back, approaching zero as time goes on.
\[ C = \frac{5t}{t^2+4} \approx \frac{5t}{t^2} = \frac{5}{t} \text{ for large } t \]
For large t the denominator's t squared outruns the numerator's 5t, so the whole fraction shrinks toward zero — the horizontal asymptote is C equal to 0, which is exactly the statement that the drug eventually clears. Notice that this graph DOES cross its horizontal asymptote's level behaviour in a way the simple hyperbolas never did: it rises from zero, peaks, and returns. Lesson 8.3 handles rational functions whose numerator and denominator have different degrees, and this is one of them.
Commit first
Answer, then rate your confidence honestly.
Predict first
Do the graphs of y equals 1 over x and y equals 6 over x have different asymptotes?
Correct: No — both have x equal to 0 and y equal to 0.
\[ \frac{6}{x} = 6 \cdot \frac{1}{x}: \text{ a stretch, not a shift} \]
Why: Multiplying a function by a constant is a vertical stretch: it multiplies every output but cannot move a line the graph never reaches. Both are undefined only at x equal to 0, and both have outputs shrinking toward 0 far out. What changes is distance, not position: at x equal to 2 one gives 0.5 and the other 3. Only a term ADDED outside the fraction moves the horizontal asymptote, and that is the k of the translated form.
Explain it
They have graphed lines and parabolas and have never seen a graph in two pieces.
Discussion prompt
In four sentences or fewer, explain why the graph of 1 over x has a gap in the middle.
Hint: Ask what the function does at x equal to 0.
Answer:
At x equal to 0 the function asks you to divide 1 by 0, which has no answer, so there is simply no point on the graph above that spot.
Just to the right of zero the outputs are enormous and positive; just to the left they are enormous and negative. So the curve shoots up on one side and down on the other, and the two pieces never join. The vertical line at x equal to 0 that they hug is called an asymptote.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the sign, always set the denominator equal to zero and solve rather than reading a number off. For the horizontal asymptote, divide the coefficients of x, never the constants. For plotting, pick two inputs on each side within a couple of units of the vertical asymptote. For interpreting, rewrite the model as a constant plus a shrinking fraction and the meaning of the floor becomes plain.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a hyperbola page. Top left: sketch the parent 1 over x with both asymptotes labelled and the domain and range written beside it, plus one sentence explaining why zero is excluded twice. Top right: sketch 6 over x and negative 4 over x on the same axes as the parent, and write what changed and what did not. Middle: graph Example 2 in full — asymptotes first as dashed lines, then two points on each side, then the branches — and state the domain and range. Bottom left: take three linear-over-linear functions of your own and find both asymptotes of each, showing the two different computations. Bottom right: write the average-cost model, rewrite it as a constant plus a fraction, and use that rewriting to explain in one sentence why the average never reaches 300.
If any vertical asymptote you wrote has the same sign as the number in the denominator, recheck it: solving x plus 2 equal to zero gives negative 2, not 2.
Recap
Five things, all organised around two lines.
| If you see | Then |
|---|---|
| y = a/x | Asymptotes x = 0 and y = 0, whatever a is |
| y = a/(x - h) + k | Asymptotes x = h and y = k |
| y = (ax + b)/(cx + d) | Solve the denominator; divide the leading coefficients |
| A vertical asymptote at h | The domain excludes h |
| A horizontal asymptote at k | The range excludes k |
| An average-cost model | The horizontal asymptote is the per-item cost |
Lesson 8.3 handles rational functions whose numerator and denominator can have any degree, where comparing those degrees decides what the horizontal asymptote is — or whether there is one at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.2 Graph Simple Rational Functions §8.2, pp. 558-564 — everything on these slides traces back here
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