Last updated .
4.2 — Shift and Stretch
Human review completed .
Transform reciprocal anchor points and asymptotes, then write and sketch the resulting function.
1. In class: A Solidify Understanding Task
Start with asymptotes and anchor points
Review 4.1: the reciprocal graph. An anchor point is a convenient known point that helps you sketch a curve. It is not an intercept unless one coordinate is zero.
Main 1 answers: horizontal asymptote \(y=0\), vertical asymptote \(x=0\); points \((1,1),(-1,-1),(\frac12,2),(-\frac12,-2),(2,\frac12),(-2,-\frac12)\).
Move the curve by moving its inputs and outputs
Write a transformed reciprocal function as
Start with \(y=\frac1x\). Multiply outputs by b, then translate right h and up k. A negative h means left; a negative k means down. If b is negative, reflect across the x-axis before translating. The vertical scale factor is |b|: greater than 1 stretches, between 0 and 1 compresses.
The denominator becomes zero at x=h, so the vertical asymptote is \(x=h\). At either far end, \(\frac b{x-h}\to0\), so the horizontal asymptote is \(y=k\). Inputs cannot equal h, and outputs cannot equal k. Conversely, every other output is possible: \(x=h+\frac b{y-k}\).
| Main 12 feature | Answer |
|---|---|
| Vertical asymptote | \(x=h\) |
| Horizontal asymptote | \(y=k\) |
| Vertical stretch / reflection | \(|b|\); reflect if b<0. |
| Anchor points | \((h+1,k+b),\ (h-1,k-b)\); also \((h+\frac12,k+2b),\ (h-\frac12,k-2b)\) and \((h+2,k+\frac b2),\ (h-2,k-\frac b2)\). |
| Domain | All real numbers except \(h\) |
| Range | All real numbers except \(k\) |
If b=0, the formula equals the constant k wherever x≠h; it no longer has the two reciprocal branches described here. The packet’s transformed reciprocal family uses b≠0.
Write a rule from a graph or description
- Read the vertical asymptote x=h and horizontal asymptote y=k.
- Use a supplied point (x,y) to solve \(b=(y-k)(x-h)\).
- Check the rule at the remaining marked points.
Original problem: Write a rule for the supplied graph. Its asymptotes are x=5 and y=2, and its marked points are (4,5) and (6,−1).
Worked solution: substitute (4,5) into \(y=2+\frac b{x-5}\):
Answer: \(y=2-\frac3{x-5}\). At x=6 it gives −1, checking the second point. The negative coefficient puts the left branch above y=2 and the right branch below it.
| Main | Equation | Point / asymptote check |
|---|---|---|
| 2 | \(y=5+\frac1x\) | (1,6), (−1,4); x=0, y=5. |
| 3 | \(y=-\frac1x\) | (−1,1), (1,−1); x=0, y=0. |
| 4 | \(y=\frac1{x+3}\) | \((0,\frac13)\); x=−3, y=0. |
| 5 | \(y=\frac1{x-2}\) | (1,−1), (3,1); x=2, y=0. |
| 6 | \(y=\frac2x\) | (1,2), (−1,−2), (2,1), (−2,−1); see correction. |
| 7 | \(y=-4+\frac1{x+1}\) | (0,−3), (−2,−5); x=−1, y=−4. |
| 8 | \(y=2-\frac3{x-5}\) | (4,5), (6,−1); x=5, y=2. |
| 9 | \(y=-3-\frac1{x+6}\) | (−5,−4), (−7,−2); \(x=-\frac{19}3\) at the x-intercept. |
Main 10 — Match the transformations
| Printed equation | Matching letter | Meaning |
|---|---|---|
| \(y=\frac1{x+b}\) | C | Left b; vertical asymptote x=−b. |
| \(y=b+\frac1x\) | B | Up b; horizontal asymptote y=b. |
| \(y=\frac bx\) | D | Multiply outputs by b. |
| \(y=-\frac1x\) | A | Reflect across x-axis. |
| \(y=\frac1{x-b}\) | E | Right b; vertical asymptote x=b. |
Sketch by hand from the asymptotes
- Draw the vertical and horizontal asymptotes as dashed lines.
- Transform several anchor points, checking them in the rule.
- Draw two separate smooth branches through the points. Approach both asymptotes; do not join through the vertical asymptote.
Original problem: Graph \(y=-2+\frac3{x-4}\) without using technology.
Worked solution: stretch the parent outputs by 3, shift right 4 and down 2. Draw x=4 and y=−2. Use (5,1), (3,−5), (6,−\(\frac12\)) and (2,−\(\frac72\)). The positive multiplier keeps the right branch above y=−2 and the left branch below it.
Answer: the sketch below, with domain x≠4 and range y≠−2.
Original problem: Graph \(y=1-\frac1{x+3}\) without using technology.
Worked solution: reflect across the x-axis, shift left 3 and up 1. Draw x=−3 and y=1. Plot (−2,0), (−4,2), (−1,\(\frac12\)) and (−5,\(\frac32\)). The left branch is above y=1, the right below.
Answer: the sketch below, with domain x≠−3 and range y≠1.
2. Ready — Connect polynomial zeros with denominator restrictions
A product is zero when a factor is zero. A real input is excluded from a rational function’s domain when it makes the denominator zero. For a refresher, see 3.7: roots and factors (draft pending human review). Try Ready before checking.
Check your work — Ready 1–8
| Polynomial problem | Zeros | Rational problem | Real domain |
|---|---|---|---|
| 1 | \(-4,2,7\) | 5 | All real numbers except \(-4,2,7\) |
| 2 | \(3,\frac18,5\) | 6 | All real numbers except \(3,\frac18,5\) |
| 3 | \(-\frac13,-3,3\) | 7 | All real numbers except \(-\frac13,-3,3\) |
| 4 | \(\pm5i\) (no real zeros) | 8 | All real numbers |
In 3, factor \(x^2-9=(x-3)(x+3)\). In 4, \(x^2+25=0\) gives \(x=\pm5i\), but real x always gives \(x^2+25>0\), so 8 excludes no real inputs.
3. Set — Describe and sketch reciprocal transformations
Study Main 11, then draw your own graphs from the asymptotes and anchor points. The key supplies features and points to check your sketches.
Check your work — Set 9–13
| Problem | Rule | Asymptotes | Domain / range | Sample points |
|---|---|---|---|---|
| 9 | \(\frac4x\) | \(x=0,\ y=0\) | \(x\ne0,\ y\ne0\) | (1,4), (−1,−4), (2,2), (−2,−2) |
| 10 | \(\frac3x+2\) | \(x=0,\ y=2\) | \(x\ne0,\ y\ne2\) | (1,5), (−1,−1), (3,3), (−3,1) |
| 11 | \(-\frac5{x-3}\) | \(x=3,\ y=0\) | \(x\ne3,\ y\ne0\) | (4,−5), (2,5), (8,−1), (−2,1) |
| 12 | \(\frac1{x+5}-4\) | \(x=-5,\ y=-4\) | \(x\ne-5,\ y\ne-4\) | (−4,−3), (−6,−5), (−3,−\(\frac72\)), (−7,−\(\frac92\)) |
Plots not shown — Set 9–12: draw the dashed lines and two branches through these points. Positive coefficients place the right branch above the horizontal asymptote; the negative coefficient in 11 reverses this.
13 sample answer: \(f(x)=\frac1{x+15}-6\). More generally, \(f(x)=\frac a{x+15}-6\) works for any \(a\ne0\); the asymptotes do not determine a unique scale.
4. Go — Use a known root to find every factor
A given root r gives the factor x−r. Divide it out, factor the quadratic quotient, and include the original root. For a refresher, see 3.8: known-factor division (draft pending human review). Try Go before checking.
Check your work — Go 14–18
| Problem | Quotient after removing given factor | Remaining roots | Complete factored form |
|---|---|---|---|
| 14 | \(x^2-2x-15\) | \(-3,5\) | \((x+1)(x+3)(x-5)\) |
| 15 | \(x^2-2x-63\) | \(-7,9\) | \((x-1)(x+7)(x-9)\) |
| 16 | \(6x^2-18x-60\) | \(-2,5\) | \(6x(x+2)(x-5)\) |
| 17 | \(x^2-6x+9\) | \(3,3\) | \((x-8)(x-3)^2\) |
In 17, 3 is a double root. 18 sample statement: if \(f(x)=a(x-r_1)(x-r_2)(x-r_3)\), then the constant term is \(f(0)=-a r_1r_2r_3\). The roots are counted with multiplicity. The products give −15, 63, 0 and −72 for 14–17. A root of zero therefore forces a zero constant term.
Additional learning resources
| Resource | What it helps with and scope |
|---|---|
| The Organic Chemistry Tutor — Reciprocal Functions: Basic Introduction | Graph reciprocal functions using vertical and horizontal asymptotes. |
Explore asymptotes more fully
These optional videos focus on asymptotes. They cover more than this individual lesson, including general rational functions and ideas developed elsewhere in Module 4. Start with the first video for the basic idea; use the others to explore vertical asymptotes, holes, and end behavior in more depth.
| Resource | What it helps with and scope |
|---|---|
| Khan Academy — Graphing rational functions 1 | Start here for an explanation of vertical and horizontal asymptotes using a graph and nearby input values. Broader than this lesson: uses a shifted rational function rather than just 1/x and also finds intercepts. |
| Khan Academy — Graphs of rational functions: vertical asymptotes | More practice connecting a rational expression to its vertical asymptotes. Broader than this lesson: examples involve factored expressions beyond the basic reciprocal graph. |
| Khan Academy — Graphing rational functions according to asymptotes | Explains why an excluded input may produce a hole rather than a vertical asymptote, and identifies horizontal asymptotes. Broader than this lesson: combines factoring, cancellation, and several graph features; useful as a preview or later review. |
| Khan Academy — Analyzing horizontal asymptotes of rational functions | A deeper explanation of horizontal asymptotes through end behavior and leading terms. Broader than this lesson: compares multiple rational functions and degree cases rather than focusing on one graph. |
Teacher/parent review of the source packet
Main 6, p. 10: the point (−2,1) is inconsistent with the other points and specified transformed reciprocal family. Replace it with (−2,−1); the rule is \(y=\frac2x\).
For Main 12, the reciprocal-family description assumes b≠0; |b| is the scale magnitude and a negative b also reflects. Set 13 deliberately admits any nonzero coefficient.
Coverage
| Packet component | Support in this lesson |
|---|---|
| Main 1 · p. 8 | Both asymptotes and all six anchor outputs. |
| Main 2–9 · pp. 9–10 | All equations, point checks, worked rule derivation; Main 6 sign corrected. |
| Main 10–12 · p. 11 | Five matches, general features and two worked hand-sketch examples with graphs. |
| Ready 1–8 · p. 12 | All polynomial zeros, real denominator restrictions and linked review. |
| Set 9–13 · pp. 12–13 | All four feature sets, sample points; plots not shown. General family and sample rule for 13. |
| Go 14–18 · p. 14 | All remaining roots, factors and constant-term explanation. |
Data and brief prompts adapted from Mathematics Vision Project, Secondary Mathematics III, Module 4 student packet, Section 4.2, printed pp. 8–14, under CC BY 4.0. Explanations, solutions, and diagrams were prepared for this supplement.
Last updated .