Midline, maximum, minimum and range

Explain it like I am five

Picture a Ferris wheel on a platform. Raise the platform 7 meters and the top, the bottom and the hub all rise 7 meters, but the wheel is no bigger. The number B added outside does that to a wave: its center line becomes the midline, the flat line y = B, every point of it at height B.

Take y = 4 sin x + 7. B = 7, so the midline is y = 7. The amplitude is |4| = 4. Go 4 up from the center for the maximum, the highest output: 7 + 4 = 11. Go 4 down for the minimum, the lowest output: 7 − 4 = 3. The range, every height the wave reaches, is [3, 11], where square brackets mean 3 and 11 are included.

Check both formulas: the average, $11 + 3 over 2$ = 7, gives the midline, and half the gap, $11 − 3 over 2$ = 4, gives the amplitude. The domain, the inputs you may feed in, is still every real number x. Only the outputs are fenced in.

In plain words

Imagine a Ferris wheel standing on a platform. Raising the platform raises its top, bottom and center by the same amount. It does not make the wheel bigger. The outside addition B does that to a wave. Its center becomes the Midline, a Horizontal line written y = B. The Maximum is the highest output; the Minimum is the lowest. A Peak is a highest point and a Trough is a lowest point. The Range lists every output height the wave reaches. Start at the center and move one amplitude up or down to find its limits. Square brackets mean the limiting heights are included.

The wave reaches 3 above and below y = 1, so its extreme heights are −2 and 4.

Reminder

Average. The average of 10 and 2 is (10 + 2) ÷ 2 = 6.
Absolute value. |−3| = 3; use 3 for both vertical distances.
Interval notation. [−2, 4] means −2 ≤ y ≤ 4, with both endpoints included.
Substitution. sin( $π over 2$ ) = 1, so −3 sin( $π over 2$ ) + 1 = −2.

Why it works. The original sine or cosine stays between −1 and 1. After multiplying, the heights lie between −|A| and |A|, regardless of reflection. Adding B makes the extremes B − |A| and B + |A|. Their average is B because the equal positive and negative distances cancel. Their half-difference is |A| because B cancels. This proves both the midline and amplitude formulas and explains why they measure different features. Sine's coordinate is positive for half a circle turn and negative for the other half; cosine has equal positive and negative halves too. Constant inside speed makes these equal horizontal lengths above and below B.

RuleRule: midline y = B, with B = . Minimum = B − |A|; maximum = B + |A|. Range: [B − |A|, B + |A|].

The same idea, five ways

Say it

The midline is the horizontal line through the wave's center; the range is every height it reaches.

Write it

The wave is centered at height B and reaches one amplitude above and below.

In math

Midline: y = B
B = $y sub max + y sub min over 2$
B − |A| ≤ y ≤ B + |A|
[B − |A|, B + |A|]
{y | B − |A| ≤ y ≤ B + |A|}
Graph words: center line and vertical shadow.

Raising a Ferris wheel's platform raises its center and both extremes together.

See it

The wave reaches 3 above and below y = 1, so its extreme heights are −2 and 4.

The same idea, other ways

As a center and radius

A wheel's center height and size are separate. The midline sets the center, and amplitude sets how far the top and bottom sit from it.

The wave reaches 3 above and below y = 1, so its extreme heights are −2 and 4.

With cancellation

Add the extremes: (B + |A|) + (B − |A|) = 2B. Divide by 2 and the equal distances disappear.

With numbers

A top at 10 and bottom at 2 center on 6. They are both 4 away, so midline y = 6 and amplitude 4.

.1Vertical shift and midline

A Vertical shift changes every output by the same added amount. A horizontal line y = B has the same height for every input. Here y(x) means the wave's height at input x; it does not mean y times x.

Rule: adding B outside moves every height by B.
Rule: the midline equation is y = B.
Rule: in a nonconstant sinusoidal wave, half the horizontal length of each complete cycle is above the midline and half is below it, apart from the crossing points.

Adding 2 raises the middle to y = 2 and the extreme heights to 1 and 3, while the amplitude stays 1.

The same idea, five ways

Say it

Vertical shift and midline

Write it

A Vertical shift changes every output by the same added amount. A horizontal line y = B has the same height for every input. Here y(x) names the wave's output height at input x; y(x + $T over 2$ ) names its output half a period later.

In math

Rule: adding B outside moves every height by B.
Rule: the midline equation is y = B.
y(x + $T over 2$ ) − B = −(y(x) − B)

Lift the whole Ferris wheel on a platform without making it larger.

See it

Adding 2 raises the middle to y = 2 and the extreme heights to 1 and 3, while the amplitude stays 1.

Worked exampleAdd outside to lift every height

Compare y = sin x with y = sin x + 2. Find the new middle and two extreme heights.

Adding 2 raises the middle to y = 2 and the extreme heights to 1 and 3, while the amplitude stays 1.

Add 2 to the original middle 0: new middle = 2.An outside addition changes every output by the same amount.
Add 2 to the original minimum −1 and maximum 1: new extremes are 1 and 3.The whole graph slides upward without changing distances.
The midline is y = 2 and amplitude remains 1.Each extreme is still 1 away from the center.

Answer

Midline: y = 2
Minimum = 1
Maximum = 3

Check = 1, so the outside addition did not change amplitude. Here ω = 1, so T = 2π and half the period is π. At x = , y = sin() + 2 = 1 + 2 = 3. Half a period later, x + = + π = , and y = −1 + 2 = 1. Relative to center 2, 3 − 2 = 1 and 1 − 2 = −1. The displacements have opposite signs, as the half-period identity predicts.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: Adding 2 outside doubles amplitude.

Addition raises outputs, while multiplication changes distances.

✓ Instead: sin x + 2 has midline y = 2 and amplitude 1.

Tips and tricks

Tip: sin x + 2 has midline y = 2 and amplitude 1.

.2Upper and lower endpoints

The Range is the vertical shadow of the wave. Interval notation [a, b] includes every height from a through b. Set notation {y | a ≤ y ≤ b} says the same thing: all y satisfying the condition after the bar.

Rule: range [B − |A|, B + |A|].

The wave reaches 3 above and below y = 1, so its extreme heights are −2 and 4.

The same idea, five ways

Say it

Upper and lower endpoints

Write it

The range is the vertical shadow of the wave. Interval [a, b] includes every height from a through b. Set notation {y | a ≤ y ≤ b} says the same thing: all y satisfying the condition after the bar.

In math

Rule: range [B − |A|, B + |A|].

The wheel reaches every height between its lowest and highest rim points.

See it

The wave reaches 3 above and below y = 1, so its extreme heights are −2 and 4.

Worked exampleA reflected wave above a new center

Find the midline and range of y = −3 sin x + 1. This asks for the center line and every possible output height.

The wave reaches 3 above and below y = 1, so its extreme heights are −2 and 4.

B = 1, so the midline is y = 1.The outside addition raises every height by 1.
|A| = |−3| = 3.The negative multiplier reflects the wave, but the distance is positive.
Minimum = B − |A| = 1 − 3 = −2.Move 3 below the center.
Maximum = B + |A| = 1 + 3 = 4.Move 3 above the center.
The range is [−2, 4], including both endpoints.The sine wave passes continuously through every height between its attained extremes.

Answer

Midline: y = 1
Range: [−2, 4]

Check At x = , y = −3(1) + 1 = −2. At x = , y = −3(−1) + 1 = 4, so both claimed endpoints really occur.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: Use [B − A, B + A] when A is negative.

That reverses the endpoints because A carries reflection direction.

✓ Instead: Use |A| so the lower height comes first.

Tips and tricks

Tip: Use |A| so the lower height comes first.

.3Read height and center together

Subtract the extremes to find size. Add the extremes to find center. The two computations answer different questions.

Rule: amplitude = $maximum − minimum over 2$ .
Rule: B = $maximum + minimum over 2$ .

The top 10 and bottom 2 are both 4 from the center line y = 6.

The same idea, five ways

Say it

Read height and center together

Write it

Subtract the extremes to find size. Add the extremes to find center. The two computations answer different questions.

In math

Rule: amplitude = $maximum − minimum over 2$ .
Rule: B = $maximum + minimum over 2$ .

Find the center and radius of a wheel from its top and bottom heights.

See it

The top 10 and bottom 2 are both 4 from the center line y = 6.

Worked exampleFind the center between two extremes

A wave has maximum 10 and minimum 2. Find its midline and amplitude. You want the middle height and the equal distance on each side.

The top 10 and bottom 2 are both 4 from the center line y = 6.

B = $10 + 2 over 2$ = $12 over 2$ = 6.The average of the two extremes locates the middle height.
Amplitude = $10 − 2 over 2$ = $8 over 2$ = 4.Half their separation is the distance on one side.
Write the midline as y = 6.A line equation says every point on that horizontal line has height 6.
Check 6 − 4 = 2 and 6 + 4 = 10.Adding and subtracting the distance rebuilds the original extremes.

Answer

Midline: y = 6
Amplitude = 4

Check The distances are 10 − 6 = 4 and 6 − 2 = 4, confirming the center.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: The midline is y = 4 for maximum 10 and minimum 2.

4 is half the span, not the midpoint height.

✓ Instead: The midpoint is 6, and the amplitude is 4.

Tips and tricks

Tip: The midpoint is 6, and the amplitude is 4.

Strategy: step by step

Find the center B from the outside addition or the average of the extremes.
Find the positive amplitude |A|.
Subtract the amplitude from B for the minimum.
Add the amplitude to B for the maximum.
Write the midline as an equation and the range as an included interval.

Strategy

Strategy: recover heights and center

Are maximum and minimum given?

YesCompute their average and half-difference.

NoRead A and B from the equation.

From an equation, read B and |A|.
From two extremes, average them for B and halve their difference for |A|.
Use B ± |A| to rebuild the extremes.
Keep the horizontal period separate from these vertical measurements.

Worked exampleA reflected wave above a new center

Find the midline and range of y = −3 sin x + 1. This asks for the center line and every possible output height.

The wave reaches 3 above and below y = 1, so its extreme heights are −2 and 4.

B = 1, so the midline is y = 1.The outside addition raises every height by 1.
|A| = |−3| = 3.The negative multiplier reflects the wave, but the distance is positive.
Minimum = B − |A| = 1 − 3 = −2.Move 3 below the center.
Maximum = B + |A| = 1 + 3 = 4.Move 3 above the center.
The range is [−2, 4], including both endpoints.The sine wave passes continuously through every height between its attained extremes.

Answer

Midline: y = 1
Range: [−2, 4]

Check At x = , y = −3(1) + 1 = −2. At x = , y = −3(−1) + 1 = 4, so both claimed endpoints really occur.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: the midline of −3 sin x + 1 is 1.

1 is the height, but a midline is an entire horizontal line.

✓ Instead: Write y = 1.

✗ Not this: Counterexample: its range is [4, −2].

Intervals put the lower endpoint first. A negative multiplier does not make an amplitude negative.

✓ Instead: Use [1 − 3, 1 + 3] = [−2, 4].

Tips and tricks

Tip: Memory device: center uses the sum, size uses the difference.
Tip: Check that the midpoint of your range equals B.

Trap. The Domain is the input set; the Range is the output set. A transformed sine or cosine still accepts every real x, while its output heights are bounded.

Keep in mind

The midline is a line, so write it as an equation: y = 7, not 7.
The midline is the average of the maximum and the minimum: a wave from −1 to 9 has midline y = $9 + (−1) over 2$ = 4.
A negative A flips the wave, but the range still uses |A|: y = −4 sin x + 7 also has range [3, 11].
The range is the outputs and the domain is the inputs: y = 4 sin x + 7 accepts every real x, yet its outputs stay in [3, 11].

Memory hookCenter plus or minus reach: range = [B − |A|, B + |A|]. The average of top and bottom gives the midline, and half their gap gives the amplitude.

Flash cards: say the answer out loud, then flip

What is the midline?

The horizontal center line y = B, halfway between the maximum and the minimum.

What is the range of a function?

The set of all its output values.

Find the midline and range of y = 9 cos x + 1.

Midline: y = 1
Range: [−8, 10]

A wave peaks at 10 and bottoms out at −2. Find its midline and amplitude.

Midline: y = $10 + (−2) over 2$ = 4
Amplitude: $10 − (−2) over 2$ = 6

What is the range of y = −8 sin x + 6?

[6 − 8, 6 + 8] = [−2, 14]

Is the range of y = 4 sin x + 7 all real numbers?

No. All real numbers is the domain, the inputs. The range is [3, 11].