The general form a sine b of x minus h plus k, vertical and horizontal translations, the midline, reflections when the leading coefficient is negative, translated and reflected tangent graphs, and models built from them.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 14 — Trigonometric Graphs, Identities, and Equations
Translate and Reflect Trigonometric Graphs
Objectives
Five outcomes. The same wave, moved and flipped.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 915-921 — the lesson these objectives are drawn from
Warm-up
Lesson 14.1 graphed waves that oscillate about the x-axis and begin their cycle at x equal to zero.
Discussion prompt
A Ferris wheel's axle is 90 feet up and its radius is 85 feet. What would the rider's height graph look like? Can y equal to a sine bx describe it?
Hint: Where is the middle of that oscillation?
Answer:
The height swings between 5 and 175 feet, so it oscillates about 90, not about zero. No function of the form a sine bx can do that.
\[ h = 85\sin b(t-h_0)+90 \]
Adding a constant raises the whole wave, and subtracting one inside slides it sideways. Those two moves are exactly the translations you applied to parabolas in Chapter 4, and they work here unchanged.
Concept
In y equal to a sine b of x minus h plus k, the constant a sets the amplitude, b sets the period, h slides the graph horizontally and k slides it vertically. The horizontal line y equal to k is the midline, the level the wave oscillates about.
midline — The horizontal line y equal to k about which a translated sine or cosine graph oscillates. It sits halfway between the maximum and the minimum, and the amplitude is measured from it.
\[ y = a\sin b(x-h)+k \]
When a is negative the graph is reflected in the midline, so its maximums and minimums trade places.
Figure (svg): Two columns comparing the shape constants with the position constants
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 915-917
Section
Section 1
Concept
Adding k to a sine or cosine function shifts every point up by k. The graph then oscillates about the line y equal to k, called the midline, and the amplitude is measured from that line.
\[ y = a\sin bx+k \;\Longrightarrow\; \text{midline } y = k \]
The maximum becomes k plus the amplitude and the minimum becomes k minus it, so the two together locate both k and a.
Figure (svg): A sine graph raised three units above the axis
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 915-915 — Translations of Sine and Cosine Graphs
Picture it
Example 1: y equal to 2 sine 4x plus 3.
Figure (svg): A sine graph raised three units above the axis
The dashed curve is the untranslated graph. Every key point rose by exactly 3, and the midline moved from the axis to y equal to 3.
Worked example
Example 1.
\[ \text{Graph } y = 2\sin 4x+3. \]
Identify the four constants
Why: Two in front, 4 inside, nothing subtracted, 3 added.
\[ a = 2,\text{ period } \frac{\pi}{2}, h = 0, k = 3 \]
Draw the midline
Why: The horizontal line at height 3.
\[ y = 3 \]
Find the untranslated key points
Why: Quarter the period pi/2.
\[ 0, \frac{\pi}{8}, \frac{\pi}{4}, 3 \pi / 8, \frac{\pi}{2} \]
Raise every height by 3
Why: Sine goes 0, a, 0, -a, 0.
\[ 3, 5, 3, 1, 3 \]
Figure (svg): A sine graph raised three units above the axis
\[ \!\left(\tfrac{\pi}{8},5\right), \; \!\left(\tfrac{3\pi}{8},1\right) \]
Verify: check the maximum and minimum against k and a
Why: The graph runs from 1 up to 5, whose midpoint is 3 — the value of k — and whose half-spread is 2, the amplitude. Reading a graph backwards this way is how you recover the equation from a picture.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 915-915
Fill the middle
Example 1.
Fill in the blanks
y = 2\sin 4x+3 \;\Longrightarrow\; \text3 y = ___
Why: Three. The wave now oscillates about y equal to 3 rather than about the x-axis.
Worked example
Guided Practice 1.
\[ \text{Graph } y = \cos x+4. \]
Identify the constants
Why: One in front, 1 inside, nothing subtracted, 4 added.
\[ a = 1,\text{ period } 2 \pi, h = 0, k = 4 \]
Draw the midline
Why: The horizontal line at height 4.
\[ y = 4 \]
Find the untranslated key points
Why: Cosine goes a, 0, -a, 0, a.
\[ 1, 0, -1, 0, 1 \]
Raise every height by 4
Why: Add 4 throughout.
\[ 5, 4, 3, 4, 5 \]
Figure (svg): The solution to Worked example a cosine raised four units shown as a ladder of expressions, one row per algebraic move
\[ (0,5), \; (\pi,3), \; (2\pi,5) \]
Verify: notice the graph never dips below the axis
Why: Its minimum is 3, comfortably above zero, because the shift of 4 exceeds the amplitude of 1. Whenever k is larger than the amplitude the whole graph stays above the x-axis, which is exactly what a physical quantity like height or pressure usually needs.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 917-917
Trap
\[ y = 2\sin 4x+3 \]
Read the amplitude off the highest point
Why: The graph reaches 5, so 5 is taken as the amplitude.
\[ \text{amplitude } 5 \quad \text{(wrong)} \]
Amplitude is measured from the MIDLINE, not from the axis. The graph rises only 2 above y equal to 3.
\[ \text{amplitude} = \tfrac{1}{2}(M-m) = \tfrac{1}{2}(5-1) = 2 \]
Take half the spread between maximum and minimum
Why: That works whatever the midline is.
\[ k = \tfrac{1}{2}(M+m) = 3 \]
The average of the maximum and minimum gives k, and half their difference gives the amplitude. Two readings, two constants.
Fill the middle
Example 1.
Fill in the blanks
\text5 = k+|a| = 3+2 = ___
Why: Five. The minimum is 3 minus 2, or 1, so the graph runs from 1 to 5 about its midline of 3.
Comparison
Fill the blanks. The same wave, raised.
Comparison matrix
| Question | y = 2 sin 4x | y = 2 sin 4x + 3 |
|---|---|---|
| Midline | y = 0 | y = 3 |
| Maximum | 2 | 5 |
| Amplitude | 2 | 2 |
| Period | pi/2 | pi/2 |
Only the midline and the extreme values changed. Amplitude and period are properties of the shape, and adding a constant does not touch the shape at all.
Prediction
Commit before reasoning.
Predict first
A wave runs between a minimum of 5 and a maximum of 175. What are its midline and amplitude?
Correct: Midline 90, amplitude 85 — the average and half the difference.
\[ k = \tfrac{M+m}{2}, \quad |a| = \tfrac{M-m}{2} \]
Why: The average of 5 and 175 is 90, which is where the wave is centred, and half of 175 minus 5 is 85, which is how far it swings either way. Those two formulas recover k and a from any picture or table of a periodic quantity, which is exactly the step needed to write a model from data. They are also the numbers you will meet again in Lesson 14.4.
Section
Section 2
Concept
Writing the function as a sine b of x minus h shifts the graph h units horizontally, to the right when h is positive. Every key x-value increases by h while the heights stay the same.
\[ y = a\sin b(x-h) \;\Longrightarrow\; \text{shift right } h \]
The b must be factored out in front of the bracket for h to be read directly, which is why the form is written that way.
Figure (svg): A cosine graph slid three pi units to the right
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 916-916 — Graph a horizontal translation
Picture it
Example 2: y equal to 5 cosine 2 of x minus 3 pi.
Figure (svg): A cosine graph slid three pi units to the right
The five key x-values each gained 3 pi, so a cycle that would have started at 0 now starts at 3 pi.
Worked example
Example 2.
\[ \text{Graph } y = 5\cos 2(x-3\pi). \]
Identify the four constants
Why: Five in front, 2 inside, 3 pi subtracted, nothing added.
\[ a = 5,\text{ period } \pi, h = 3 \pi, k = 0 \]
Draw the midline
Why: Since k is zero.
Find the untranslated key x-values
Why: Quarter the period pi.
\[ 0, \frac{\pi}{4}, \frac{\pi}{2}, 3 \pi / 4, \pi \]
Add 3 pi to each
Why: The heights are unchanged.
\[ 3 \pi, 13 \pi / 4, 7 \pi / 2, 15 \pi / 4, 4 \pi \]
Figure (svg): A cosine graph slid three pi units to the right
\[ (3\pi,5), \; \!\left(\tfrac{7\pi}{2},-5\right), \; (4\pi,5) \]
Verify: check the shift is a whole number of periods
Why: The period is pi and the shift is 3 pi, so the graph has moved exactly three full cycles — which means it looks identical to the unshifted graph. Not every shift is invisible, but this one is, and noticing that is a useful reality check on the arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 916-916
Fill the middle
Example 2.
Fill in the blanks
\text3 0 \;\Longrightarrow\; \text___ 0+3\pi = ___\pi
Why: Three pi. Every key x-value gains h, so the whole cycle starts 3 pi further along.
Worked example
Guided Practice 2 and 3.
\[ \text{Graph } y = 3\sin\!\left(x-\tfrac{\pi}{2}\right) \text{ and } f(x) = \sin(x+\pi)-1. \]
First: read h
Why: Pi over 2 is subtracted.
\[ \text{shift right } \frac{\pi}{2} \]
First: shift the key points
Why: Zeros at pi/2, 3pi/2, 5pi/2.
\[ \max(\pi, 3), \min(2 \pi, -3) \]
Second: read h and k
Why: Pi is ADDED, so h is negative pi.
\[ \text{shift left } \pi,\text{ down } 1 \]
Second: shift the key points
Why: Midline y = -1.
\[ \max(-\frac{\pi}{2}, 0), \min(\frac{\pi}{2}, -2) \]
Figure (svg): The solution to Worked example two more translations shown as a ladder of expressions, one row per algebraic move
\[ h = \tfrac{\pi}{2}; \qquad h = -\pi, \; k = -1 \]
Verify: watch the sign in the second one
Why: The form is x minus h, so x plus pi means h equals NEGATIVE pi and the shift is to the LEFT. Reading a plus sign as a rightward shift is the commonest error here, and rewriting x plus pi as x minus negative pi makes it obvious.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 917-917
Error analysis
A student states the horizontal shift of y = cos(2x - 6).
Annotate
On: \( h = 6 \)
Without factoring, the number inside is b times h rather than h itself. Factoring b out of the bracket is not cosmetic; it is what makes h readable.
Sorting
The form is x minus h.
Sort into buckets
Sort each function by the direction of its horizontal shift.
Rewriting x plus c as x minus negative c makes every one of these a single glance rather than a decision, and it is worth doing on paper until it becomes automatic.
Prediction
Commit before reasoning.
Predict first
Why does replacing x with x minus 3 move a graph three units to the RIGHT rather than left?
Correct: Because the function now needs x to be 3 larger to produce the value it used to produce at x.
\[ x-3 = 0 \;\Longleftrightarrow\; x = 3 \]
Why: The original graph had its maximum where the input was 0; the new one has its maximum where x minus 3 is 0, that is at x equal to 3. Every feature therefore appears 3 units later along the axis. This is the same reasoning that put the vertex of y equal to x minus 3 squared at x equal to 3 in Chapter 4, and it works identically for every function family.
Matching
Factor b out first if it is not already.
Match the pairs
Why: The second needed factoring: 2x minus 6 is 2 times x minus 3, so the shift is 3 rather than 6. The third is the only one with a vertical shift as well, moving the midline down to y equal to negative 1.
Section
Section 3
Concept
When a is negative the graph is the reflection, in the midline y equal to k, of the graph with the size of a in front. Maximums become minimums and minimums become maximums.
\[ y = -|a|\sin b(x-h)+k \]
Only the extreme points move. The midline crossings stay exactly where they were, since reflecting a point on the mirror leaves it fixed.
Figure (svg): The graphs of negative sine and negative cosine
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 917-917 — Reflections
Picture it
The graphs of negative sine x and negative cosine x.
Figure (svg): The graphs of negative sine and negative cosine
Negative sine dips before it rises; negative cosine starts at its minimum. The dashed curves are the originals.
Worked example
Example 4.
\[ \text{Graph } y = -2\sin\tfrac{2}{3}\!\left(x-\tfrac{\pi}{2}\right). \]
Identify the constants
Why: Size of -2; two pi over two thirds; pi over 2 subtracted.
\[ | a | = 2,\text{ period } 3 \pi, h = \frac{\pi}{2}, k = 0 \]
Find the key points of the POSITIVE version
Why: Quarter the period 3 pi and shift by pi/2.
\[ \text{zeros at } \frac{\pi}{2}, 2 \pi, 7 \pi / 2 \]
Locate its extremes
Why: Max at 5 pi/4, min at 11 pi/4.
\[ (5 \pi / 4, 2)\text{ and } (11 \pi / 4, -2) \]
Reflect in the midline
Why: Swap the two extremes.
\[ (5 \pi / 4, -2)\text{ and } (11 \pi / 4, 2) \]
Figure (svg): The graphs of negative sine and negative cosine
\[ \!\left(\tfrac{5\pi}{4},-2\right), \; \!\left(\tfrac{11\pi}{4},2\right) \]
Verify: check the three midline crossings did not move
Why: They remain at pi over 2, 2 pi and 7 pi over 2, because reflecting a point that already lies on the mirror leaves it in place. Only two of the five key points changed, which is a useful economy when drawing by hand.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 917-917
Fill the middle
Example 4, Step 4.
Fill in the blanks
\!\left(\tfrac-2___,2\right) \;\to\; \!\left(\tfrac______,___\right) \text___ y = 0
Why: Negative 2. The maximum became the minimum, while its x-value stayed exactly where it was.
Worked example
The discussion on page 917.
\[ \text{Give the five key points of } y = -\sin x \text{ and } y = -\cos x. \]
Start with sine's key heights
Why: Zero, 1, 0, -1, 0.
Negate each
Why: Zero, -1, 0, 1, 0.
\[ (\frac{\pi}{2}, -1)\text{ and } (3 \pi / 2, 1) \]
Start with cosine's key heights
Why: One, 0, -1, 0, 1.
Negate each
Why: Negative 1, 0, 1, 0, -1.
\[ (0, -1)\text{ and } (\pi, 1) \]
Figure (svg): The solution to Worked example reflect the parent graphs shown as a ladder of expressions, one row per algebraic move
\[ -\sin: (\tfrac{\pi}{2},-1); \qquad -\cos: (0,-1) \]
Verify: notice what negative cosine looks like
Why: Starting at the minimum and rising is exactly the shape needed for a Ferris wheel boarded at the bottom, which is why negative cosine shows up so often in circular-motion models. Recognising the shape you want saves choosing the wrong function.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 917-917
Trap
\[ y = -2\sin 4x+3 \]
Flip the whole graph over the x-axis
Why: The reflection is taken about y equal to zero.
\[ \text{maximum } -1, \; \text{minimum } -5 \quad \text{(wrong)} \]
That reflects the vertical shift too. The graph should still oscillate about y equal to 3, running from 1 to 5.
\[ \text{reflect in the midline } y = 3 \]
Flip only the wave, about y equal to k
Why: The midline itself is the mirror and does not move.
\[ \text{maximum } 5 \text{ at } \tfrac{3\pi}{8}, \; \text{minimum } 1 \text{ at } \tfrac{\pi}{8} \]
The graph occupies exactly the same band of heights as before; only the times of the peaks and troughs have swapped.
Sorting
Look at the sign of a.
Sort into buckets
Sort each function.
A minus sign INSIDE the bracket is a horizontal shift, not a reflection. Only the sign in front of the whole function flips the graph.
Comparison
Fill the blanks. The same wave, flipped.
Comparison matrix
| Question | y = 2 sin 4x + 3 | y = -2 sin 4x + 3 |
|---|---|---|
| Midline | y = 3 | y = 3 |
| First quarter-point | a maximum of 5 | a minimum of 1 |
| Amplitude | 2 | 2 |
| Midline crossings | 0, pi/4, pi/2 | 0, pi/4, pi/2 |
Three of the four rows are unchanged. Reflecting swaps only the peaks and troughs, which is why the graph occupies exactly the same band of heights.
Prediction
Commit before reasoning.
Predict first
You need a wave that begins at its lowest point and rises. Which is the natural choice?
Correct: y = -cos x, since cosine starts at its maximum and the minus flips that to a minimum.
\[ -\cos 0 = -1 \text{, the minimum} \]
Why: Cosine begins each cycle at the top of its range, so negating it begins at the bottom. That is exactly the shape needed for a Ferris wheel boarded at ground level, or for a mass hanging on a spring released from below its rest position. Choosing between sine, negative sine, cosine and negative cosine is simply choosing which of the four starting heights you want, and it saves adding a horizontal shift to fix the phase later.
Section
Section 4
Concept
For y equal to a tangent b of x minus h plus k, the period is pi over b, the midline is y equal to k, and the asymptotes shift by h. A negative a reflects the graph in the midline.
\[ y = a\tan b(x-h)+k \]
With no horizontal shift the asymptotes stay put, since only h can move them. A vertical shift never moves them at all.
Figure (svg): The graphs of negative sine and negative cosine
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 918-918 — Combine a translation and a reflection
Picture it
The reflection idea, applied above the axis.
Figure (svg): The graphs of negative sine and negative cosine
The mirror is the midline, wherever it happens to sit. For a tangent graph that means the halfway points trade places while the intercept stays on the midline.
Worked example
Example 5.
\[ \text{Graph } y = -3\tan x+5. \]
Identify the constants
Why: Size of -3; b is 1; nothing subtracted; 5 added.
\[ \text{period } \pi, h = 0, k = 5 \]
Draw the midline and asymptotes
Why: No horizontal shift, so the asymptotes stay.
\[ y = 5; x =\text{ plus and minus } \frac{\pi}{2} \]
Find the unreflected key points
Why: Intercept on the midline; halfway points at plus and minus a.
\[ (0, 5), (-\frac{\pi}{4}, 2), (\frac{\pi}{4}, 8) \]
Reflect in the midline
Why: Swap the two halfway heights.
\[ (-\frac{\pi}{4}, 8), (\frac{\pi}{4}, 2) \]
Figure (svg): The solution to Worked example a translated and reflected tangent shown as a ladder of expressions, one row per algebraic move
\[ (0,5), \; \!\left(-\tfrac{\pi}{4},8\right), \; \!\left(\tfrac{\pi}{4},2\right) \]
Verify: check the branch now falls rather than rises
Why: An ordinary tangent branch climbs from left to right, but the minus in front makes this one descend, from 8 down through 5 to 2. That is the visible effect of a reflection on a graph with no maximum or minimum to swap.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 918-918
Fill the middle
Example 5, Step 3.
Fill in the blanks
\!\left(\tfrac8___, |___|+5\right) = \!\left(\tfrac______,___\right) \text___
Why: Eight. After reflecting in the midline y equal to 5 it becomes 2, since 5 minus 3 is 2.
Worked example
Example 6.
\[ \text{Standing } 120 \text{ ft from a } 260 \text{ ft building, model a friend's distance } d \text{ below the top against the angle } \theta. \]
Write the tangent relationship
Why: Opposite over adjacent.
\[ \tan \theta = \frac{260 - d}{120} \]
Clear the fraction
Why: Multiply both sides by 120.
\[ 120 \tan \theta = 260 - d \]
Solve for d
Why: Rearrange.
\[ d = -120 \tan \theta + 260 \]
Read the constants
Why: Negative in front, and 260 added.
Figure (svg): The solution to Worked example a glass elevator shown as a ladder of expressions, one row per algebraic move
\[ d = -120\tan\theta+260 \]
Verify: check the value at a level line of sight
Why: At theta equal to 0 the tangent is 0, so d is 260 — the friend is 260 feet below the top, that is at ground level, which is exactly right for a horizontal line of sight from ground level. The reflection is what makes d DECREASE as the angle rises, which matches the elevator climbing toward the top.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 918-918
Error analysis
A student graphs y = -3 tan x + 5 and reports the asymptotes.
Annotate
On: \( x = \pm\tfrac{\pi}{2}+5 \)
Only h can move a vertical asymptote, because only h changes x-values. The textbook flags this explicitly beside Example 5.
Fill the middle
Example 6.
Fill in the blanks
120\tan\theta = 260-d \;\Longrightarrow\; d = -120\tan\theta+260
Why: A minus sign. Moving 120 tangent theta across and d back gives d equal to 260 minus 120 tangent theta, which is the reflected form.
Matching
For a tangent graph.
Match the pairs
Why: Only h and b affect x-values, so only they can move a vertical asymptote. The constant k moves every point up or down along those fixed lines.
Prediction
Commit before reasoning.
Predict first
Why does d = -120 tan theta + 260 rather than +120 tan theta + 260?
Correct: Because d measures distance BELOW the top, and that decreases as the angle of elevation rises.
\[ \theta \uparrow \;\Longrightarrow\; \tan\theta \uparrow \;\Longrightarrow\; d \downarrow \]
Why: As the elevator climbs, the angle you look up at increases while the friend's distance below the roof shrinks toward zero. A quantity that falls as its input rises needs a negative coefficient, and the algebra delivers one automatically when you solve for d. Had the model measured height above the GROUND instead, the sign would have come out positive — which is a reminder to read carefully what a variable is actually measuring.
Section
Section 5
Concept
In a circular-motion model the amplitude is the radius and the midline is the height of the centre. The period is the time for one revolution, and the horizontal shift places the starting point.
\[ h = 85\sin\tfrac{\pi}{20}(t-10)+90 \]
Maximum equals midline plus amplitude and minimum equals midline minus amplitude, so the two extremes can be read off without graphing.
Figure (svg): A Ferris wheel ride graphed as a translated sine wave
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 916-916 — Graph a model for circular motion
Picture it
Example 3: 180 seconds on a wheel with a 40 second period.
Figure (svg): A Ferris wheel ride graphed as a translated sine wave
Four and a half cycles fit into the ride, and the rider swings between 5 and 175 feet about a midline of 90.
Worked example
Example 3, both parts.
\[ \text{For } h = 85\sin\tfrac{\pi}{20}(t-10)+90 \text{ over } 180 \text{ seconds, graph it and find the extreme heights.} \]
Find the amplitude and period
Why: Eighty-five; two pi over pi over 20.
\[ 85\text{ and } 40\text{ seconds} \]
Count the cycles
Why: One eighty over 40.
\[ 4.5\text{ cycles} \]
Find the five key points
Why: Shift by 10 and quarter the period.
\[ (10, 90), (20, 175), (30, 90), (40, 5), (50, 90) \]
Read the extremes
Why: Midline plus and minus amplitude.
\[ 175\text{ and } 5\text{ feet} \]
Figure (svg): A Ferris wheel ride graphed as a translated sine wave
\[ \text{max } 175, \; \text{min } 5 \]
Verify: check the numbers against a real wheel
Why: A midline of 90 with amplitude 85 means the wheel's axle is 90 feet up and its radius is 85, putting the lowest car 5 feet off the ground — sensible for boarding. The shift of 10 seconds says the rider was at axle height and rising 10 seconds into the ride, a quarter period before the top at 20 seconds.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 916-916
Fill the middle
Example 3.
Fill in the blanks
T = \frac40___ = 2\pi \cdot \frac______ = ___
Why: Forty seconds. The pi cancels top and bottom, which is why models of timed cycles almost always have a pi inside.
Worked example
The same model, read in reverse.
\[ \text{A wheel of radius } 85 \text{ ft has its axle } 90 \text{ ft up and turns once every } 40 \text{ s. Write a model.} \]
The radius gives the amplitude
Why: Eighty-five feet.
\[ a = 85 \]
The axle height gives the midline
Why: Ninety feet.
\[ k = 90 \]
The revolution time gives b
Why: Two pi over 40.
\[ b = \frac{\pi}{20} \]
Choose the starting point
Why: Sine starts at the midline rising.
Figure (svg): The solution to Worked example read a model back from a description shown as a ladder of expressions, one row per algebraic move
\[ a = 85, \; k = 90, \; b = \tfrac{\pi}{20} \]
Verify: check b against the period
Why: Two pi divided by pi over 20 is 40, the stated revolution time. Every one of the four constants came from a physical measurement, which is what makes these models so easy to write once the correspondence is learned.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 916-916
Trap
\[ \text{a wheel } 170 \text{ ft across, axle } 90 \text{ ft up} \]
Take the amplitude as the width
Why: The number 170 is the wheel's size.
\[ h = 170\sin bt+90 \quad \text{(wrong)} \]
That would swing the rider from negative 80 to 260 feet. The amplitude is the RADIUS, half the diameter.
\[ a = \tfrac{170}{2} = 85 \]
Halve the diameter
Why: Amplitude is measured from the midline outward.
\[ h = 85\sin bt+90, \; \text{range } 5 \text{ to } 175 \]
A negative height is the giveaway. Checking that a model's minimum is physically possible catches this at once.
Fill the middle
Example 3b.
Fill in the blanks
\text175 = k+a = 90+85 = ___
Why: One hundred seventy-five feet. The minimum is 90 minus 85, which is 5 feet — just above the ground, as a boarding platform should be.
Matching
In a circular-motion model.
Match the pairs
Why: Every constant in the equation corresponds to something you could measure with a tape and a stopwatch. That correspondence is what makes the model trustworthy rather than merely fitted.
Ranking
Smallest first.
Put in order
Why: The values are 5, 90, 90, 175 and 170 feet. The second and third tie at 90 because t equal to 30 is a midline crossing, which is exactly axle height — and the diameter, though not a height, is larger than any of them.
Comparison
Fill the blanks. Each does one job.
Comparison matrix
| Constant | What it controls | Read it from |
|---|---|---|
| a | the amplitude, and a flip if negative | half the spread between max and min |
| b | the period | how far apart the cycles are |
| h | the horizontal shift | where a cycle begins |
| k | the midline | the average of max and min |
Reading a graph is the reverse of drawing one: the extremes give a and k, the spacing gives b, and the starting point gives h.
Pattern
Constants, midline, key points.
A plus sign inside the bracket means a shift to the LEFT, since the form subtracts h.
OpenStax Algebra and Trigonometry 2e, §8.1 Graphs of the Sine and Cosine Functions §8.1
Check
The constant added at the end names the midline.
Check your understanding
What are the amplitude and midline of y = 2 sin 4x + 3?
Answer: A
Why: The coefficient in front is 2 and the constant added is 3.
Check
Factor b out of the bracket first.
Check your understanding
How is the graph of y = cos(2x - 6) translated from y = cos 2x?
Answer: A
Why: 2x - 6 = 2(x - 3), so h is 3 and the shift is 3 units right.
Check
Only h moves a vertical asymptote.
Check your understanding
Where are the asymptotes of y = -3 tan x + 5?
Answer: A
Why: There is no horizontal shift, and a vertical shift cannot move a vertical line.
Real world
In a certain city the number of daylight hours peaks at 15.3 on 21 June, the 172nd day of the year, and bottoms out at 9.1 hours six months later.
Discussion prompt
Write a model for daylight hours as a function of the day number, and find the daylight on 1 March, day 60.
Hint: Find the midline and amplitude from the two extremes.
Answer:
\[ k = \tfrac{15.3+9.1}{2} = 12.2, \quad a = \tfrac{15.3-9.1}{2} = 3.1 \]
\[ b = \tfrac{2\pi}{365} \;\Longrightarrow\; L(d) = 3.1\sin\!\left(\tfrac{2\pi}{365}(d-81)\right)+12.2 \]
\[ L(60) = 3.1\sin\!\left(\tfrac{2\pi}{365}(-21)\right)+12.2 \approx 3.1(-0.359)+12.2 \approx 11.1 \]
The model predicts about 11.1 hours of daylight on 1 March.
The shift of 81 is where the real work is. A sine graph peaks a quarter period after its shift, and a quarter of 365 is about 91, so peaking on day 172 requires h equal to 172 minus 91, which is 81. That day is the spring equinox, when daylight passes through its average of 12.2 hours on the way up — exactly what a sine graph does at its own starting point. The astronomy and the algebra agree, which is the best sign that a model has been set up honestly rather than merely fitted.
Commit first
Answer, then rate your confidence honestly.
Predict first
The graph of y = sin(x + pi/2) is shifted which way from y = sin x?
Correct: Left pi/2 — the form is x minus h, so x plus pi/2 means h is negative pi/2.
\[ \sin\!\left(x+\tfrac{\pi}{2}\right) = \cos x \]
Why: Rewriting x plus pi over 2 as x minus negative pi over 2 makes h explicitly negative, and a negative h shifts left. The check is easy: the shifted graph has its maximum where x plus pi over 2 equals pi over 2, that is at x equal to 0, which is to the LEFT of the original maximum at pi over 2. Incidentally this particular shift turns the sine graph into the cosine graph exactly, which is why sine of x plus pi over 2 equals cosine of x.
Explain it
They can graph y equals x squared plus 3 and have just met sine waves.
Discussion prompt
In four sentences or fewer, explain what the k does in y equals a sine bx plus k.
Hint: Compare it with what plus 3 did to a parabola.
Answer:
It lifts the whole graph up by k, exactly as adding 3 lifted the parabola. The shape is untouched; every point just moves up the same distance.
So instead of waving about the x-axis, the graph waves about the line y equals k. That line is called the midline, and the amplitude is measured from it rather than from zero.
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 amplitude, take half the difference between maximum and minimum. For direction, rewrite x plus c as x minus negative c and let the sign of h decide. For factoring, always write the bracket as b times x minus h before reading anything. For reflecting, remember the mirror is y equal to k, so the midline crossings never move.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a translations page. Top left: write the general form large, label all four constants, and beside each write in a few words what it does and how to read it off a graph. Top right: graph one function with a vertical shift, drawing the midline as a dashed line and marking all five key points with their translated coordinates. Middle: graph one function with a horizontal shift, showing the untranslated key x-values and the shifted ones side by side, and include one function written in unfactored form that you factor first. Bottom left: take one graph and reflect it, marking which key points moved and which did not, and write one sentence on why the midline crossings stayed put. Bottom right: write a circular-motion model of your own from an invented radius, centre height and revolution time, and state its maximum and minimum without graphing.
If any reflected graph on your page changed the heights of its midline crossings, the mirror was in the wrong place: it is y equals k, not the x-axis.
Recap
Five things, and waves that can sit anywhere on the plane.
| If you see | Then |
|---|---|
| A constant added at the end | It is k, and the midline is y equal to it |
| A quantity subtracted inside | It is h, and the graph shifts right |
| A quantity added inside | h is negative and the graph shifts left |
| b not factored out | Factor it before reading h |
| A negative coefficient in front | Reflect in the midline, not the axis |
| A circular-motion problem | Amplitude is the radius, midline the centre height |
Lesson 14.3 leaves graphing behind and turns to identities: relationships among the six functions that hold for every angle at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.2 Translate and Reflect Trigonometric Graphs §14.2, pp. 915-921 — everything on these slides traces back here
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