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Transforming sec, cosec and cot graphs

Transform sec, cosec and cot graphs using stretches, shifts and reflections. Map asymptotes, find ranges and intersections, and preserve holes after simplification.

Before you startBase sec, cosec and cot graphs; function transformations

01 / Move the inputs and outputs separately

A point mapping keeps the order of transformations clear.

For y = a f(b(x − h)) + k, assume a and b are nonzero. If (u,v) lies on the original graph y = f(u), match the input b(x − h) to u, then transform the output.

(u,v) → (u/b + h, av + k)

The same horizontal rule moves an excluded input u = α to the new excluded input x = α/b + h. Vertical changes move the branches but do not change which x-values are allowed.

Use the stages in the model to separate these two jobs. A negative b reflects horizontally; a negative a reflects vertically.

Map the points and the asymptotesExplore

y = 2sec x + 1

At u = 0: v = 1; mapped point (0, 3).

Period 2π. Range y ≤ −1 or y ≥ 3.

Choose a stage yourself. The scale stays fixed between stages of the selected example. Dashed lines are excluded inputs; they have no gold point.

ExampleTransformation stageBase angle u

Watch a secant point and its asymptotes move

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Stretch, reflect and shift vertically

Apply the output rule to both range bands.

Find the range of y = 2sec x + 1Worked example

Base range: sec x ≤ −1 or sec x ≥ 1

There are two separate bands.

Multiply by 2: y −1 ≤ −2 or y −1 ≥ 2

A positive scale preserves each inequality direction.

Range: y ≤ −1 or y ≥ 3

The gap has moved and widened. It is not simply y ≥ 3.

Find the range of y = −3cosec x + 2Worked example

Base edge values −1 and 1 map to 5 and −1

The negative multiplier reverses their vertical order.

Range: y ≤ −1 or y ≥ 5

The lower band remains unbounded below and the upper band above.

For transformed sec or cosec:
range y ≤ k − |a| or y ≥ k + |a|

Cot has every real output, so a nonzero vertical scale and any vertical shift keep its range as all real numbers.

03 / Read horizontal scaling from the whole input

Solve b(x − h) = u rather than shifting by the un-factored constant.

Interpret y = sec(2x − π/3)Worked example

Write 2x − π/3 = 2(x − π/6)

The horizontal shift is right π/6, not right π/3.

Point mapping: x = u/2 + π/6

Horizontal distances shrink by a factor 1/2.

Period = 2π/2 = π

Sec repeats when its complete input increases by 2π.

Sec/cosec period = 2π/|b|
Cot period = π/|b|

In y = cot(−x), b = −1 produces reflection in the y-axis. Since cot is odd, this is also y = −cot x. Sec is even, so sec(−x) = sec x.

04 / Map excluded inputs first

Solve for x where the original denominator vanishes.

Sec: b(x − h) = π/2 + nπ
Cosec/cot: b(x − h) = nπ

Find the asymptotes of y = sec(2(x − π/6)) − 2Worked example

2(x − π/6) = π/2 + nπ

The cosine denominator is 0.

x = 5π/12 + nπ/2

Solve the whole input equation.

Turning points: x = π/6 + nπ/2

Map the base turning inputs u = nπ.

At even n, y = −1; at odd n, y = −3

Apply the vertical shift −2 to the base values ±1.

The vertical translation does not move the asymptotes. Mark excluded x-values before connecting smooth branches.

05 / Transform cotangent zeros carefully

After a vertical shift, the base zeros lie at the new midline, not necessarily y = 0.

Sketch the structure of y = 3cot((x + π/2)/2) + 1Worked example

b = 1/2, h = −π/2, a = 3, k = 1

Use the factored form.

Asymptotes: x = −π/2 + 2nπ

Map u = nπ using x = 2u −π/2.

Period: 2π; range: all real numbers

Cot’s base period π doubles horizontally.

Base zeros map to (π/2 + 2nπ, 1)

The vertical shift moves them up to y = 1.

For x-intercepts, solve cot((x + π/2)/2) = −1/3

They are not the same as the mapped base zeros.

A positive a/b combination here preserves the decreasing branch direction. Reflecting in one axis reverses it; reflecting in both preserves it.

06 / Count intersections using range and period

A horizontal line can meet several branches in a specified interval.

Solve 2cosec(2x) − 1 = 3 for 0 ≤ x ≤ 2πWorked example

cosec(2x) = 2, so sin(2x) = 1/2

The reciprocal is valid because neither side is 0.

0 ≤ 2x ≤ 4π

The inner angle covers two sine periods.

2x = π/6, 5π/6, 13π/6, 17π/6

There are two intersections per period.

x = π/12, 5π/12, 13π/12, 17π/12

All four satisfy the original equation and avoid its asymptotes.

For y = 2sec x + 1, a horizontal line y = c has no intersections if −1 < c < 3. At a boundary height, a branch is touched at a turning point. Outside the gap, count the actual branches in the requested interval; endpoint inclusion can change the total.

07 / Recognise the same graph in another form

Use sine/cosine identities before comparing pictures.

Rewrite y = 2cosec(x − π/2) − 1Worked example

sin(x − π/2) = −cos x

This is a quarter-turn shift of sine.

cosec(x − π/2) = −sec x

Take reciprocals where cos x ≠ 0.

y = −2sec x − 1

The two expressions have the same domain and values.

Range: y ≤ −3 or y ≥ 1

The output reflection and shift agree with the transformed cosecant graph.

Equivalent-looking formulas must agree on domains too. A simplification that cancels an undefined intermediate expression can enlarge the domain if its exclusions are forgotten.

08 / Keep holes when simplifying a reciprocal

A removable gap is different from an unbounded asymptote.

Consider R(x) = 1/(3 − 4sec x). The original expression requires cos x ≠ 0. Its outer denominator never vanishes for a real allowed x, because sec x = 3/4 is impossible.

R(x) = cos x / (3cos x − 4),
with the original restriction cos x ≠ 0.

The simplified quotient can be evaluated at cos x = 0 and would give 0. That does not fill the holes in R. At x = π/2 + nπ, the original secant is undefined, so R is still undefined. Nearby values approach 0; there is no unbounded vertical asymptote there.

Find its range without losing the gapWorked example

Set c = cos x, so c belongs to [−1,0) or (0,1]

Zero is excluded by the original domain.

The expression c/(3c −4) decreases as c increases

For c₁ < c₂, the difference is −4(c₂ −c₁)/((3c₂ −4)(3c₁ −4)) < 0.

c = −1 gives 1/7; c = 1 gives −1; c → 0 gives 0

The point (π/3,2) lies on sec u. Where does it map on y = −sec((x + π/2)/2)/2 + 2?

Hint

Here b = 1/2, h = −π/2, a = −1/2 and k = 2.

Worked solution

x = 2(π/3) −π/2 = π/6; y = −(1/2)(2) +2 = 1. The mapped point is (π/6,1).

10 / Recap

Use one mapping for points, exclusions and periods.

For y = a f(b(x − h)) + k, map (u,v) to (u/b + h, av + k).
Vertical transformations leave asymptote x-values unchanged.
Divide the base period by |b|.
Map both range bands of sec and cosec.
A shifted cot zero need not be an x-intercept.
Keep the original domain after algebraic simplification.

Section 1 of 10 · Move the inputs and outputs separately