Graph transformations

Translate the whole curve, including its asymptotes

A translation moves every point the same distance and direction. The curve keeps its shape and size. Move its asymptotes too: they belong to the graph, even though they are not part of the curve.

For the rules below, $a>0$ and $(p,q)$ is a point on $y=f(x)$.

Move right or left

• Right by a: replace x by $x-a$, giving $y=f(x-a)$. The point moves by $(p,q)\mapsto(p+a,q)$.
• Left by a: replace x by $x+a$, giving $y=f(x+a)$. The point moves by $(p,q)\mapsto(p-a,q)$.

Why does a minus sign move the graph right? To get the old output $f(p)$ from $f(x-a)$, solve $x-a=p$. The new input is $x=p+a$.

Move up or down

• Up by a: replace y by $y-a$. Then $y-a=f(x)$ gives $y=f(x)+a$, and $(p,q)\mapsto(p,q+a)$.
• Down by a: replace y by $y+a$. Then $y+a=f(x)$ gives $y=f(x)-a$, and $(p,q)\mapsto(p,q-a)$.

Move the asymptotes with the curve

In Example 1, replace x by $x-1$ in the vertical asymptote: $x=0$ becomes $x-1=0$, so the new asymptote is $x=1$. The horizontal asymptote $y=0$ stays fixed.

In Example 2, replace y by $y-1$: the asymptote $y=0$ becomes $y-1=0$, or $y=1$. The vertical asymptote $x=0$ stays fixed. Draw both branches on separate axes, as each question requests.

Check before the examples: a point $(2,3)$ lies on $y=f(x)$. On $y=f(x-1)+2$, it moves to $(3,5)$: one unit right and two up.

Activities and source notes

The opening calculator activities compare three parabolas in each direction. Use their minimum points to check that every point moves by the same amount. The worked tables include the Teacher’s completed minima and labels. Self-Practice 1 asks for two separate straight-line sketches; its worked answers are independently derived because neither source copy prints them here.

The printed Explore transformations! QR resource links to the source exploration. Its current destination content has not been reviewed here.

The later units cover scaling, reflection, combinations, $|f(x)|$, $f(|x|)$ and $1/f(x)$. The source references are Steward, J. (2015), Calculus: Early Transcendentals, 8th ed., Cengage Learning, and Burghes, Cassell, Cross, Deft, Middle and Thallon (1994), Pure Mathematics, Heinemann Educational. The original bibliography spelling is retained in the native transcription.

Scale coordinates, using the reciprocal of the replacement multiplier

Start by deciding which coordinate changes. A multiplier inside f changes horizontal distances; a multiplier outside f changes vertical distances. Use a point to check the factor instead of guessing from the equation.

For the replacement rules below, $a>0$ and $(p,q)$ lies on $y=f(x)$.

Scale horizontal distances

Replace x by $ax$ to get $y=f(ax)$. To reproduce the old output at input p, solve $ax=p$. The new input is $p/a$, so the point map is $(p,q)\mapsto(p/a,q)$.

The scale factor is $1/a$, parallel to the x-axis. Thus $f(2x)$ halves horizontal distances from the y-axis; it does not double them. The y-axis stays fixed.

Scale vertical distances

Replace y by $ay$ to get $ay=f(x)$, hence $y=f(x)/a$. The point map is $(p,q)\mapsto(p,q/a)$: scale by $1/a$ parallel to the y-axis, keeping the x-axis fixed.

Be careful about the equation you are given: $2y=f(x)$ halves y-coordinates, while $y=2f(x)$ doubles them.

Check asymptotes and intercepts

• Example 3 halves x-coordinates. Its vertical asymptote moves from $x=-1$ to $x=-1/2$; $y=1$ stays fixed.
• Example 4 doubles y-coordinates. Its horizontal asymptote moves from $y=1$ to $y=2$; $x=-1$ stays fixed.

Apply the same replacement to each asymptote equation before solving it. In Self-Practice 2, $2f(x)$ and $f(2x)$ happen to give lines with equal slopes, but their y-intercepts differ. Equal slopes do not make the graphs identical.

Check before the examples: if $(4,3)$ is on $y=f(x)$, the corresponding point is $(2,3)$ on $y=f(2x)$ and $(4,6)$ on $y=2f(x)$.

Scaling activity and source notes

The quadratic activities include every Teacher minimum/intercept table entry. Their phrase “shape preserved” means the graphs remain parabolas: scaling changes width or height and need not preserve distances or size. Unlike a translation, it fixes the relevant coordinate axis. Self-Practice 2’s answers are independently derived, rather than printed completions.

Reflect coordinates and asymptotes

A reflection changes the sign of one coordinate. The named axis stays fixed: reflecting in the x-axis changes y; reflecting in the y-axis changes x.

Reflect in the x-axis

Replace y by $-y$. Then $-y=f(x)$ gives $y=-f(x)$, with point map $(p,q)\mapsto(p,-q)$.

Every point moves to the same distance on the opposite side of the x-axis. A minimum becomes a maximum. This is signed scaling by $-1$ parallel to the y-axis.

Reflect in the y-axis

Replace x by $-x$ to get $y=f(-x)$, with point map $(p,q)\mapsto(-p,q)$.

The height stays the same while the horizontal position reverses. This is signed scaling by $-1$ parallel to the x-axis.

Reflect asymptotes too

• In Example 5, the horizontal asymptote $y=1$ becomes $-y=1$, so $y=-1$. The vertical asymptote $x=-1$ stays fixed because its equation does not contain y.
• In Example 6, $x=-1$ becomes $-x=-1$, so the vertical asymptote is $x=1$. The horizontal asymptote $y=1$ stays fixed because its equation does not contain x.

Each example asks for separate graphs with both branches. In Self-Practice 3, compare $-f(x)$ and $f(-x)$ for $f(x)=x+1$: both lines have gradient $-1$, but their intercepts differ.

Check before the examples: $(2,-3)$ becomes $(2,3)$ under an x-axis reflection and $(-2,-3)$ under a y-axis reflection. The coordinate along the mirror axis does not change.

Reflection activity and source notes

The two quadratic activities include the Teacher’s completed stationary-point and intercept tables. Self-Practice 3’s solutions are independently derived; neither source copy prints answers on this page.

Compose replacements in the stated order

Work with the current equation at each step. Write the intermediate equation before applying the next transformation; applying every change to the original equation can give the wrong result.

Changes to different coordinates

In Example 7, horizontal scaling and vertical translation affect different coordinates. Either order gives $f(2x)+1$:

• Replace x by $2x$, then y by $y-1$.
• Or replace y by $y-1$, then x by $2x$.

These are the two sequences compared in the Teacher notes.

Changes to the same coordinate

Order matters when a later scaling also scales an earlier translation. Example 8 reaches $f(2x+1)$ in either of these ways:

1. Scale horizontally by $1/2$ to get $f(2x)$. Translate left by $1/2$ to get $f(2(x+1/2))=f(2x+1)$.
2. Translate left by 1 to get $f(x+1)$. Scale horizontally by $1/2$ to get $f(2x+1)$.

The translation distances differ. Swapping the order while keeping the same distances would change the final graph.

Example 9 makes the same point vertically: reach $2f(x)+1$ by scaling by 2 then translating up 1, or by translating up $1/2$ then scaling by 2. Keep the intermediate replacement equations so you can see which terms the scaling affects.

Choose a sequence, then verify it

The source recommends horizontal translation, horizontal scaling, vertical scaling, then vertical translation. Its mnemonic $T_xS_xS_yT_y$ is a convenient guide, not a restriction on other valid sequences. Check the resulting equation, especially if you change the order.

Examples 10–11 show each replacement. Completing the square gives $5(x-2)^2-4$; the transformed cubic has exact intercepts $(0,-10)$ and $(2+\sqrt[3]{12},0)$. In Example 11, “stretch” by a factor of $1/2$ means halve vertical distances.

Undo a sequence

Reverse the order and invert each transformation. If you first scale vertically by 2 and then move up 1, undo the move first, then divide the height by 2.

Example 12 undoes C, B, A to obtain $y=\ln(-x)$. It also shows the Teacher’s alternative point map $(p,q)\mapsto(p/2+3,-q)$; the real-domain check is an independently authored supplement.

Check before the examples: after scaling vertically by 2 and moving up 1, an original height of 3 becomes 7. Undoing those steps gives $(7-1)/2=3$, not $7/2-1$.

Self-Practice 4 includes all three two-step answers printed in both copies. Horizontal translation by 2 before scaling can correspond to translation by $2/3$ after scaling. The verification equations are authored explanations of the printed answers.

Keep output modulus and input modulus distinct

First locate the modulus bars. In $|f(x)|$, they act on the output; in $f(|x|)$, they act on the input. These rules retain different parts of the original graph.

Output modulus: keep the upper part

For $y=|f(x)|$:

1. Keep every portion with $y\ge0$.
2. Reflect each portion with $y<0$ in the x-axis.

The outputs become non-negative, the zeros stay in place, and the original domain is preserved. The quadratic activity folds the lower arc of $x^2-1$ upwards. Example 13 keeps the cubic’s zeros $-1,2,3$ and reflects its negative arcs.

In Example 14, the rational graph keeps its x-intercept $a/2$ and y-intercept 1. Its horizontal asymptote changes from $-2$ to 2, while $x=-a$ remains excluded. Self-Practice 5 gives $|x+4|$, with vertex $(-4,0)$; its sketch is independently derived.

Input modulus: keep the right-hand part

For $y=f(|x|)$:

1. Keep only the original portion with $x\ge0$.
2. Reflect that retained portion in the y-axis to create the left-hand part. Discard the original left-hand part.

For $x\ge0$, $|x|=x$, so use $f(x)$. For $x<0$, $|x|=-x$, so use $f(-x)$, wherever defined. The result has y-axis symmetry, but outputs can still be negative. Its domain consists of inputs x for which $|x|$ belongs to the original domain.

The logarithm activity keeps $(0,\ln2)$ and mirrors the right-hand part. The resulting $\ln(|x|+2)$ is defined for all real x and has no vertical asymptote, unlike $\ln(x+2)$.

In Example 15, discard the original root $-1$. Keep roots 2 and 3, and mirror them to $-2,-3$. The negative troughs stay below the x-axis; the sketch includes all four printed approximate extremum labels. Self-Practice 6 gives $|x|+4$, with vertex $(0,4)$ and no x-intercept. Its independently derived sketch differs from Self-Practice 5.

Read a schematic graph carefully

Example 16 supplies a sketch, not a formula. Mirror its positive-input features: poles at $\pm1$, roots at $\pm3$, horizontal asymptote $y=1$ and central point $(0,3)$. These features do not determine a unique formula.

The centre is a corner local minimum, although it is not a differentiable stationary point. The Teacher’s phrase “is not a turning point” uses a narrower meaning; it should not be read as denying the minimum visible in the sketch. The original wording remains in the source review and native text.

Check before the examples: a point $(2,-3)$ becomes $(2,3)$ under output modulus. Under input modulus it stays at $(2,-3)$ and is copied to $(-2,-3)$. Input modulus does not make the output positive.

Source correction: the logarithm activity’s later “Hence to sketch” line omits the modulus bars. Use the displayed piecewise rule above; the original typo remains in the native text.

Reciprocate outputs without extending the domain

For $g(x)=1/f(x)$, change each non-zero output y to $1/y$. Keep its input x unchanged. Before sketching, mark every input excluded from f and every zero of f: none of these inputs belongs to g.

Work in this order

1. Mark excluded inputs and zeros before moving any points.
2. Reciprocate the heights of labelled points. Keep the sign of each height.
3. Check the limits on each side of an excluded input and at both infinite ends.
4. Sketch each continuous branch, using the direction and turning-point rules below.

For example, $(2,4)$ becomes $(2,1/4)$ and $(3,-2)$ becomes $(3,-1/2)$. An original zero has no reciprocal point.

Reciprocal rules, one feature at a time

Point $(x,y)$ with $y\ne0$

$(x,1/y)$.

y-intercept $(0,a)$ with $a\ne0$

$(0,1/a)$.

Zero at $x=b$

Exclude b; a vertical asymptote occurs when nearby non-zero f values tend to 0.

Vertical divergence at excluded $x=c$

Reciprocal tends to $0^+$ for $f(x)\to+\infty$ and $0^-$ for $f(x)\to-\infty$, but c remains excluded; this is a missing limit point, not automatically an x-intercept.

Non-horizontal oblique asymptote $y=dx+e$

At the relevant infinite end, f grows in magnitude and the reciprocal tends to 0.

Finite non-zero horizontal limit L

Reciprocal horizontal limit $1/L$.

Non-zero local maximum $(g,h)$

Local minimum $(g,1/h)$ on a continuous non-zero neighbourhood.

Non-zero local minimum $(i,j)$

Local maximum $(i,1/j)$ on such a neighbourhood.

Positive/negative output

Same sign.

Increasing/decreasing on a continuous non-zero interval

Decreasing/increasing on that interval.

$f(x)=\pm1$

The two graphs intersect at the same point.

Keep holes distinct from intercepts

The native source table calls zero limits at former vertical asymptotes “x-intercepts.” That printed statement remains in the review. On your sketch, do not fill excluded points: $1/f(x)$ cannot be zero at an input where f has a finite non-zero value, and simplifying a formula does not restore a missing input. Likewise, a zero of an arbitrary discontinuous f does not alone prove a nearby pole; inspect the limit.

The source sidebar’s invariance claim concerns non-degenerate translations/scalings: they preserve the turning-point count, while metric shape and the position of a symmetry axis need not stay fixed. Reciprocation can change the branch structure and the number/type of extrema.

Apply the rules to the examples

The calculator activity compares a line with root −2 and a quadratic with roots −1 and 3. Its four sketches include the Teacher labels.

Example 17 supplies three schematic graphs. Check these features in their reciprocals: the positive bowl has maximum $(0,1/a)$; the rational sketch has a hole at $(0,0)$; the oblique sketch has a hole at $(-1,0)$ and extrema $(-2,-1/4),(1/2,1/2)$. Do not infer an unknown numeric y-intercept from schematic geometry.

In Example 18, follow the labelled points A and B: A becomes the reciprocal maximum $(3,-1)$, B gives the pole $x=1$, the left horizontal asymptote becomes $y=1/2$, and the original excluded origin stays a hole. None of these question sketches supplies a unique formula for f.

Track retained, duplicated and excluded labelled points

First decide whether a transformation moves every point, retains only half of a graph, or excludes an input. Then apply the rule to each labelled point separately. A correct outline is not enough if a requested point has been lost.

Self-Practice 7 applies the reciprocal rule to $x+4$: the independent sketch has pole $x=-4$, horizontal asymptote 0 and y-intercept $1/4$. Example 19 combines the earlier methods in eight separate sketches, preserving all intermediate stages. Use a coordinate map for the current equation, as the Teacher recommends:

Apply each target to an original point (p, q)

$f(x-2)$

$(p,q)\mapsto(p+2,q)$.

$f(-x)+1$

$(p,q)\mapsto(-p,q+1)$.

$f(2x+1)$

$(p,q)\mapsto((p-1)/2,q)$.

$2f(\lvert x\rvert)$

Retain $p\ge0$, duplicate at $x=\pm p$ and double the output.

$f(-\lvert x\rvert)$

Retain $p\le0$ and duplicate at $x=\pm\lvert p\rvert$.

$\lvert f(x-2)\rvert$

$(p,q)\mapsto(p+2,\lvert q\rvert)$.

$f(\lvert x-2\rvert)$

Retain $p\ge0$, duplicate at $x=2\pm p$; symmetry axis $x=2$.

$1/f(x)$

Retain only non-zero outputs, using $(p,1/q)$.

Count images and check exclusions

A point on the retained fold line has coincident images and is counted once.

Discarded original inputs have no corresponding point in an input-modulus construction; other points may have two images.

Keep every requested A–E correspondence rather than listing only surviving turning points.

For part (c), the Teacher notes that the method is the same as Example 8 and need not be retaught in detail; reuse its replacements while retaining all coordinates here.

Keep the precision of the given coordinates

The given B and D coordinates are printed approximations. Transform these consistently without inventing extra precision. In part (b), reflection keeps B’s x-magnitude $0.549$; the source’s final label 0.5 is a typo, retained in native/review evidence. In part (d), the final heights 4.226 confirm the factor 2 omitted from the source final curve label.

Parts (f) and (g) distinguish taking output modulus after a shift from shifting an input-modulus graph. In part (h), original roots A/C/E become poles, while B/D acquire reciprocal heights about $-1.585$ and $0.473$. All source stages, including the original graph repeated in part (h), remain available.

Assess the source mastery objectives twice

The source provides separate self-assessment columns at the end of the lecture and at the end of the tutorial. Check each objective at both stages; reading a solution does not establish mastery. From $y=f(x)$:

• Obtain/describe $y=f(x-a)$.
• Obtain/describe $y=f(x+a)$.
• Obtain/describe $y-a=f(x)$.
• Obtain/describe $y+a=f(x)$.
• Obtain/describe $y=f(ax)$.
• Obtain/describe $ay=f(x)$.
• Obtain/describe $-y=f(x)$.
• Obtain/describe $y=f(-x)$.
• State the variable replacement required for translation, scaling and reflection.
• Describe a suitable sequence of transformations.
• Obtain $y=\lvert f(x)\rvert$.
• Obtain $y=f(\lvert x\rvert)$.
• Obtain $y=1/f(x)$, retaining its domain exclusions.

Use the complete source study map

The printed references below identify the source’s intended practice sequence. The closing Discussion 1–9 and Review 1–9 have now been compared with the existing canonical records; their two lecture-copy appearances retain independent pagination through source aliases, without creating duplicate exercises. The lesson’s Practice, Discussion and Review navigation opens those canonical exercises.

The prior-topic rows retain Graphing Techniques → Discussion 5(i)(ii)(iii),7(i)(ii)(iii)(a), and Functions (Piece-wise) → Discussion 9(ii)(iii). The Graphing Techniques reference is reproduced as printed rather than silently resolving its compressed subpart notation. The source row says “Identity error”; the teaching label uses “Identify and correct” while retaining the original wording in the native transcription.

Closing source appearances and printed qualifications

The closing question asks, graph features, attributions and Review mark allocations match the retained canonical Discussion/Review exercises. Teacher PDF 34 / printed 106 is a blank body page, so later question pages differ between copies.

The short keys answer only the source-listed items; the canonical worked solutions retain their fuller standalone-solution provenance.

Discussion 8 uses $f(3-x/2)$, not the expression suggested by flattened fraction extraction; Discussion 7 asks for an absolute reciprocal.

Review 1’s given B diagram label(2,0) conflicts with its prose(0,2); the canonical coordinate follows the prose and y-axis position.

Review 5’s original graph, absent from the standalone solution PDF, is present in both lecture copies and now directly confirms the reconstruction.

Original typos and reciprocal-domain qualifications remain recorded rather than silently erased.