Increasing, decreasing and constant lines

Explain it like I am five

Imagine three jars. You add the same number of coins to the first jar each day. You remove the same number from the second. You leave the third alone. As days pass, the first amount grows, the second shrinks, and the third stays the same. Linear functions can behave in those three ways. An increasing linear function has outputs that grow when inputs grow. A decreasing linear function has outputs that shrink. A constant function keeps the same output. Read their graphs from left to right, because moving right means the input is increasing.

f(x) = 3x − 2: increasing

g(x) = −x + 5: decreasing

h(x) = 7 = 0x + 7: constant

The slope sign determines each kind of linear behavior.

Reminder

Negative coefficient. In − $1 over 2$ x + 5, the minus sign belongs to the coefficient − $1 over 2$ .
Function versus output. Different inputs may share an output: h(0) = h(2) = 7 still gives one output per input.
Evaluation. For f(x) = 3x − 2, f(1) = 3(1) − 2 = 1.
Fraction times a whole number. − $1 over 2$ × 2 = − $2 over 2$ = −1, so g(2) = 4.
Signed subtraction. 1 − (−2) = 3, so an output can increase from a negative start.
Multiplication by zero. 0 × x = 0, so h(x) = 0x + 7 always gives 7.
Domain and counts. Completed-day counts use whole numbers. R(9) = −40 is outside a model for nonnegative remaining texts.

Why it works. For f(x) = mx + b, adding one to the input adds m to the output. A positive m raises the output. A negative m lowers it. A zero m changes nothing. The starting value b shifts the line's height but does not change what one extra input step does. That is why the slope's sign, rather than the starting amount's sign, determines direction.

RuleRule: In f(x) = mx + b, m > 0 gives an increasing linear function, m < 0 gives a decreasing linear function, and m = 0 gives a constant function with a horizontal line.

The same idea, five ways

Say it

Say: positive slope rises, negative slope falls, and zero slope stays level.

Write it

The sign of slope determines whether linear outputs increase, decrease, or stay constant.

In math

m > 0: increasing
m < 0: decreasing
m = 0: constant
$x sub 2$ > $x sub 1$ : compare f( $x sub 2$ ) and f( $x sub 1$ )

A jar fills, empties, or stays unchanged as another day passes.

See it

Read left to right

Positive slope: rising line

Negative slope: falling line

Zero slope: horizontal line

Moving right increases the input, so read each line that way.

The same idea, other ways

As jars

Adding coins gives positive change, removing coins gives negative change, and leaving the jar alone gives zero change.

m > 0: output grows

m < 0: output shrinks

m = 0: unchanged output

The sign of the change determines what happens as input advances.

As graphs

Move right to increase input. An upward line gives larger outputs; a downward line gives smaller outputs; a horizontal line gives the same output.

The solid positive-slope line rises while the dashed negative-slope line falls.

With one extra input

In f(x) = 3x − 2, one extra input adds three. In g(x) = − $1 over 2$ x + 5, it removes half. In h(x) = 7, it adds zero.

f: m = 3, increasing

g: m = −, decreasing

h: m = 0, constant

The input coefficient identifies the behavior.

Slope	The line
m > 0	Increasing: rises from left to right.
m < 0	Decreasing: falls from left to right.
m = 0	Constant: horizontal.

.1Increasing linear function

An increasing linear function is like a jar receiving the same deposit each day. As you move to a later day, you have more in the jar. On the graph, larger inputs are farther right and their outputs are higher. The slope is positive because each extra input step adds a positive amount. The jar could begin with a negative balance, such as a debt, and still increase. Increasing describes the direction of the change, not whether the current amount is above or below zero.

Rule: An increasing linear function has m > 0, because extra input adds positive output change.
Rule: Its graph rises from left to right, because moving right increases the input and the matching output becomes higher.
Rule: T(d) = 60d models sixty texts per completed day, because d equal daily groups contain 60d texts.

Read the line from left to right as the input increases.

Reminder

Subtracting a negative. 1 − (−2) = 1 + 2 = 3, so the output rose by three.

The same idea, five ways

Say it

Say: as input gets larger, output gets larger.

Write it

A positive slope gives an increasing linear function.

In math

m > 0
$x sub 2$ > $x sub 1$ implies f( $x sub 2$ ) > f( $x sub 1$ )
f(x) = 3x − 2
T(d) = 60d

You add the same positive deposit to a jar each day.

See it

Read the line from left to right as the input increases.

Worked exampleA line starting below zero and a growing text count

Show increasing behavior means find a higher output at a larger input. Compare f(0) and f(1) for f(x) = 3x − 2. Then compare T(2) and T(3) for T(d) = 60d within one billing month.

Read the line from left to right as the input increases.

Larger completed-day inputs give larger sent-text counts.

f(0) = 3 × 0 − 2 = −2 and f(1) = 3 × 1 − 2 = 1.Substitution gives the requested outputs.
Compare 1 > −2 and 1 − (−2) = 3, so f is increasing.The input rose by one and the output rose by its positive slope three.
T(2) = 60 × 2 = 120 texts and T(3) = 60 × 3 = 180 texts.Every completed day contributes sixty texts to the total sent.
Compare 180 − 120 = 60 texts for the additional day.The positive daily change shows increasing behavior.

Answer

f is increasing because m = 3 > 0.
T is increasing because its slope is 60 texts per day.
After three completed days, T(3) = 180 texts.

Check Two days' 120 texts plus another day's 60 texts equals 180, matching direct substitution.

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

✗ Not this: Counterexample: f(x) = 3x − 2 is decreasing because its constant term is negative.

The constant term sets starting height, while the coefficient controls change.

✓ Instead: f is increasing because m = 3 > 0.

Tips and tricks

Tip: Circle the input coefficient, including its sign, before classifying.

.2Decreasing linear function

A decreasing linear function is like spending the same amount from a jar each day. As you move to a later day, less remains. Its slope is negative because an extra input step removes some output. On the graph, moving right takes you to a lower point. This can describe texts remaining in an allowance, but you must stop before the formula claims a negative number of remaining texts. A formula can keep decreasing forever; a real allowance cannot contain fewer than zero unused texts.

Rule: A decreasing linear function has m < 0, because increasing the input adds negative output change.
Rule: Its graph falls from left to right, because larger inputs have smaller outputs.
Rule: R(d) = 500 − 60d counts remaining texts from a 500-text allowance, because sent texts are removed from the original allowance.
Rule: For whole completed days, use d in {0, 1, 2, 3, 4, 5, 6, 7, 8}, because R(8) = 20 is nonnegative but R(9) = −40 is not a remaining count.

Read the line from left to right as the input increases.

Reminder

Fraction multiplication and negative addition. − $1 over 2$ × 2 = −1, and 5 + (−1) = 4.

The same idea, five ways

Say it

Say: as input gets larger, output gets smaller.

Write it

A negative slope gives a decreasing linear function on its allowed inputs.

In math

m < 0
$x sub 2$ > $x sub 1$ implies f( $x sub 2$ ) < f( $x sub 1$ )
g(x) = − $1 over 2$ x + 5
R(d) = 500 − 60d
R(d) ≥ 0
d in {0, 1, 2, 3, 4, 5, 6, 7, 8}

A jar loses the same amount at each withdrawal until nothing remains.

See it

The remaining-text table uses whole days and nonnegative outputs.

Worked exampleHalf a unit per step, then a limited allowance

Show decreasing behavior means find a lower output at a larger input. Compare g(0) and g(2) for g(x) = − $1 over 2$ x + 5. Then find texts remaining after three completed days and check the last allowed completed day in R(d) = 500 − 60d.

Read the line from left to right as the input increases.

The remaining-text table uses whole days and nonnegative outputs.

g(0) = 5 and g(2) = − $1 over 2$ × 2 + 5 = −1 + 5 = 4.Two input steps remove two halves, totaling one.
Name g decreasing because 4 < 5 as the input moves from zero to two.Its negative slope produces smaller outputs at larger inputs.
After three days, sent texts are 60 × 3 = 180; remaining texts are R(3) = 500 − 180 = 320.Remaining texts equal the initial allowance minus texts already sent.
R(8) = 500 − 60 × 8 = 500 − 480 = 20 texts.Eight completed days still leave a nonnegative count.
R(9) = 500 − 60 × 9 = 500 − 540 = −40, so exclude day nine from this formula for remaining texts.The calculation shows where the model stops describing an available allowance. A negative remaining count is not possible.

Answer

g is decreasing with slope − $1 over 2$ .
After three completed days, 320 texts remain.
Day eight leaves 20 texts.
Day nine is outside this completed-day remaining-text model.

Check At day three, 180 sent plus 320 remaining equals 500. At day eight, 480 sent plus 20 remaining equals 500. Both account for the whole allowance.

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

✗ Not this: Counterexample: g(x) = −x + 5 increases because its starting value is positive.

The starting value sets height; the negative coefficient sets the change.

✓ Instead: g decreases because m = − < 0.

✗ Not this: Counterexample: R(9) = −40 means negative forty texts remain available.

The algebraic formula has passed the allowance's meaningful range.

✓ Instead: Restrict this completed-day remaining model to days zero through eight. Later days require a different model.

Tips and tricks

Tip: Read a falling line from left to right and require remaining counts to stay at least zero.

.3Constant function

A constant function is like a flat monthly membership price. The day you look at the bill can change, but the monthly price stays the same. The output does not depend on which allowed input you choose. You can write h(x) = 7 as h(x) = 0x + 7 to see its slope: zero. Its graph is horizontal because every point has the same output height. A $50 unlimited-texting plan works this way during one billing month. Fifty dollars is the monthly charge, not a daily charge.

Rule: A constant function has m = 0 and f(x) = b, because the input contributes zero to the output.
Rule: Its graph is horizontal, because every input has the same output height.
Rule: C(d) = 50 is the flat $50 texting charge for one billing month, because changing that month's day does not change its charge.
Rule: A constant function belongs to the mx + b line family but is not degree 1, because its x term has coefficient zero.

Read the line from left to right as the input increases.

Reminder

Multiplication by zero. 0 × 2 + 7 = 7, so changing the input adds nothing.

The same idea, five ways

Say it

Say: output stays the same when input changes.

Write it

A constant function has zero slope and a horizontal graph.

In math

m = 0
f(x) = b
f(x) = 0x + b
h(x) = 7
C(d) = 50

A fixed membership price stays the same throughout the billing month.

See it

Read the line from left to right as the input increases.

Worked exampleSeven everywhere and fifty dollars for the month

Constant means equal outputs at different inputs. Find h(0) and h(2) for h(x) = 7. Then find the monthly texting charge on completed days one and three of the same billing month.

Read the line from left to right as the input increases.

The monthly charge remains fifty dollars at each selected day.

Rewrite h(x) = 7 as h(x) = 0x + 7.The input has no contribution, so its coefficient is zero.
h(0) = 0 × 0 + 7 = 7 and h(2) = 0 × 2 + 7 = 7.Zero multiplication removes either input.
C(d) = 50, so C(1) = 50 dollars and C(3) = 50 dollars.The question measures the fixed charge for one billing month.
Read slope zero dollars per day and classify C as constant.An extra day does not change that month's charge.

Answer

h(0) = h(2) = 7, so h is constant with slope zero.
C(1) = C(3) = $50 for the same billing month, so C is constant with slope zero dollars per day.

Check The charge difference is 50 − 50 = 0 dollars, confirming zero change and a horizontal graph.

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

✗ Not this: Counterexample: h(x) = 7 has slope seven.

Seven is the output, not its change. It does not multiply x.

✓ Instead: h(x) = 0x + 7 has slope zero and initial value seven.

✗ Not this: Counterexample: A flat $50 monthly plan costs 50d after d days.

That would describe a daily charge rather than a monthly one.

✓ Instead: The current billing month's charge is C(d) = 50 dollars.

Tips and tricks

Tip: Insert a zero input coefficient when the input letter is absent.

Strategy: step by step

1. Write the formula as mx + b and keep the sign attached to m.
2. Read the input coefficient; if the term is absent, m = 0.
3. Use positive, negative, or zero slope to classify the function.
4. Compare outputs at two increasing allowed inputs.
5. Keep counting inputs whole and remaining counts nonnegative.

Strategy

Strategy: Classify a line by slope

Is m positive?

YesThe function is increasing.

NoCheck whether m is negative.

↓

Is m negative?

YesThe function is decreasing.

Nom is zero, so the function is constant.

↓

Does the output count a remaining allowance?

YesStop the decreasing model before its count becomes negative.

NoApply the story's other limits.

1. Identify the input coefficient.
2. Use its sign to name increasing, decreasing, or constant.
3. Check with outputs at increasing allowed inputs.
4. Apply limits from the real situation.

Worked exampleThree formulas, then three text-message models

Classify means say what happens to output as input grows. First classify f(x) = 3x − 2, g(x) = − $1 over 2$ x + 5, and h(x) = 7. Then model sixty texts sent per day, texts remaining from a 500-text allowance at that rate, and a flat $50 monthly texting charge.

Read the line from left to right as the input increases.

The remaining-text table uses whole days and nonnegative outputs.

For f, m = 3 > 0; f(0) = −2 and f(1) = 1, so the output increases.The positive coefficient adds three per step even though the start is negative.
For g, m = − $1 over 2$ < 0; g(0) = 5 and g(2) = −1 + 5 = 4, so the output decreases.Two input steps remove two halves, totaling one.
For h, write h(x) = 0x + 7; h(0) = h(1) = 7, so the output stays constant.An absent input term has coefficient zero.
Let d count completed days within one billing month. Write sent texts as T(d) = 60d, with positive slope sixty texts per day.Every completed day adds sixty to a starting sent count of zero.
Write remaining texts as R(d) = 500 − 60d. At day three, R(3) = 500 − 180 = 320 texts, with negative slope −60 texts per day.Sent texts are removed from the original allowance.
R(8) = 500 − 480 = 20, but R(9) = 500 − 540 = −40. Keep the completed-day remaining model only through day eight.A remaining count cannot be negative.
Write C(d) = 50 dollars for the unlimited plan's charge within one billing month, with slope zero.The monthly charge does not vary with which day is observed.

Answer

f(x) = 3x − 2: increasing.
g(x) = − $1 over 2$ x + 5: decreasing.
h(x) = 7: constant.
Sent texts: T(d) = 60d, increasing, using whole completed days within the month.
Remaining texts: R(d) = 500 − 60d, decreasing, using d in {0, 1, 2, 3, 4, 5, 6, 7, 8}.
Current monthly charge: C(d) = 50 dollars, constant.

Check At day three, 180 sent plus 320 remaining equals 500. At day eight, 480 sent plus 20 remaining also equals 500. The monthly charge remains fifty dollars on either day.

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

✗ Not this: Counterexample: A function is increasing whenever its outputs are positive.

Increasing compares outputs at larger inputs. Positive outputs can still become smaller.

✓ Instead: g(x) = −x + 5 decreases because its slope is negative, even where its outputs are positive.

✗ Not this: Counterexample: A horizontal line's slope equals its height.

Height is output; slope is output change per input unit.

✓ Instead: h(x) = 7 has initial value seven and slope zero.

Tips and tricks

Tip: Read the slope sign, then read the graph left to right.
Tip: For this text-count model, use whole completed days and exclude negative remaining counts.

Trap. Tip: Classify by the input coefficient rather than the starting value. In 3x − 2, the positive three gives increasing behavior.