The difference quotient: average rate of change

Explain it like I am five

Think of driving between two mile markers. Your car has traveled some extra miles over some extra hours. Dividing the extra miles by the extra hours gives your average speed. A difference quotient does the same comparison for a function: new output minus old output, divided by the step in input. Difference means the result of subtraction. Quotient means the result of division. The answer tells you how much the output changes for each 1 step of input, on average. On a graph, it is the steepness, called the slope, of the straight line joining the two points. The step is called h, and it must be nonzero.

For f(x) = + 3x − 4, input 1 to 3 gives an output climb of 14 over a step of 2. The dashed line has slope 14 ÷ 2 = 7.

Reminder

Signed arithmetic. A negative squared is positive when enclosed: (−2 $) squared$ = 4, while 3(−2) = −6.
Distributive property. Multiply every term inside: 3(a + h) = 3a + 3h.
Squaring a sum. (a + h $) squared$ = $a^{2}$ + 2ah + $h^{2}$ because it contains four products, not two.
Subtracting parentheses. −( $a^{2}$ + 3a − 4) = − $a^{2}$ − 3a + 4; each sign changes.
Nonzero factor cancellation. $h(2 a + h) over h$ = 2a + h only for h ≠ 0.
Pulling out a common factor. 2xh + $h^{2}$ + 3h = h(2x + h + 3). Distributing back verifies every term.

Why it works. If a car travels 12 extra miles in 2 extra hours, 12 ÷ 2 = 6 miles per hour makes 2 × 6 = 12. The same reasoning compares any output change with its matching input change. The input moves from x to x + h, so its change is (x + h) − x = h. The outputs at those same inputs are f(x) and f(x + h). Subtract in the same direction before dividing. A zero input step gives zero in the denominator, so that quotient has no value.

RuleDifference quotient: , with h ≠ 0 and both inputs in the domain. Read it: f of x plus h, minus f of x, all over h. The starting input can also be named a: .

The same idea, five ways

Say it

f of x plus h, minus f of x, all over h

Write it

Average output change for each 1 step of input, from x to x + h.

In math

$f(x + h) − f(x) over h$
$f(a + h) − f(a) over h$
$\frac{r i s e}{r u n}$
h ≠ 0

Extra miles divided by extra hours measures average speed.

See it

For f(x) = + 3x − 4, input 1 to 3 gives an output climb of 14 over a step of 2. The dashed line has slope 14 ÷ 2 = 7.

The same idea, other ways

As average speed

Your odometer reads 20 miles at hour 1 and 32 miles at hour 3. The extra distance is 32 − 20 = 12 miles. The extra time is 3 − 1 = 2 hours. Average speed is $12 over 2$ = 6 miles per hour. The unit comes from dividing miles by hours.

Two extra hours give 12 extra miles: 6 miles per hour.

As steepness on a picture

For f(x) = $x^{2}$ , the points (1, 1) and (3, 9) have a sideways step of 2 and an upward change of 8. Rise means the output change. Run means the input change. Their joining line has slope $\frac{r i s e}{r u n}$ = $8 over 2$ = 4. The curve itself need not be a straight line.

The two points have rise 8 and run 2. The dashed line has slope 8 ÷ 2 = 4.

As cost per extra unit

For C(n) = 15 + 2n dollars for n gigabytes, C(3) = 21 and C(5) = 25. The difference quotient is $25 − 21 over 5 − 3$ = $4 over 2$ = 2 dollars per extra gigabyte. The fixed 15-dollar fee cancels because both costs contain it.

The same two columns supply both the cost change and the input step.

Why the terms without h cancel

For these polynomial examples, expanding f(x + h) gives a copy of f(x) and extra terms that each contain h. Subtracting f(x) removes that copy. For f(x) = $x^{2}$ + 3x − 4, the $x^{2}$ , 3x, and −4 cancel; 2xh + $h^{2}$ + 3h remain. Factor h or divide each term by h. This polynomial self-check is not a rule for expanding root or reciprocal formulas.

f(x + h): + 2xh + + 3x + 3h − 4

subtract f(x): − − 3x + 4

difference: 2xh + + 3h

quotient: 2x + h + 3, h ≠ 0

Subtracting removes each starting-output term.

.1Numbers first: what the quotient is for

Compare two readings on the same trip. The difference quotient measures the average rate of change, meaning the average output change for each 1 step of input. This becomes average speed when the output is distance and the input is time. On a graph it measures the slope of the line through the two points. Section 1.3 develops this same rate-of-change idea. Calculus later considers steps closer and closer to zero to describe change at one instant.

Difference means subtraction; quotient means division.
Use the same two inputs for the top change and the bottom change.
Across these two readings, a positive rate means the output at the larger input is higher. A negative rate means it is lower. Zero means the two outputs agree. The average does not assert what happens at every input between the two readings.

For f(x) = + 3x − 4, input 1 to 3 gives an output climb of 14 over a step of 2. The dashed line has slope 14 ÷ 2 = 7.

The same idea, five ways

Say it

f of three minus f of one, all over three minus one, equals seven.

Write it

The two output readings change by 14 while their inputs change by 2, giving 7 output units per input unit.

In math

f(x) = $x^{2}$ + 3x − 4
$f(3) − f(1) over 3 − 1$ = $14 − 0 over 2$ = 7
2 × 7 = 14

Extra miles divided by the matching extra hours gives average speed.

See it

For f(x) = + 3x − 4, input 1 to 3 gives an output climb of 14 over a step of 2. The dashed line has slope 14 ÷ 2 = 7.

Worked exampleSee the quotient with numbers first

For f(x) = $x^{2}$ + 3x − 4, compare input 1 with input 3. The input step is h = 2. Find $f(3) − f(1) over 2$ , the average output change for each 1 step of input.

For f(x) = + 3x − 4, input 1 to 3 gives an output climb of 14 over a step of 2. The dashed line has slope 14 ÷ 2 = 7.

The input moves by 2 while the output moves by 14.

What it asks. Find the two output readings, subtract the starting reading from the ending reading, and divide by the matching input change.

Plan. Evaluate the ending and starting inputs. Subtract new minus old for the outputs and for the inputs. Divide output change by input change and check by multiplication.

Keep the input and output from each reading together.
Subtract ending minus starting in both the top and bottom.
Multiply the rate by the input step to check that it recovers the output change.

f(3) = (3 $) squared$ + 3(3) − 4 = 9 + 9 − 4 = 14.The ending input is 3; replace every x with 3.
f(1) = (1 $) squared$ + 3(1) − 4 = 1 + 3 − 4 = 0.The starting input is 1; replace every x with 1.
The output change is 14 − 0 = 14. The input change is 3 − 1 = 2.Subtract the starting reading from the ending reading in both places.
$f(3) − f(1) over 2$ = $14 − 0 over 2$ = $14 over 2$ = 7.Divide the output change by the nonzero input change.

Answer

The average rate of change is 7 output units per input unit.

Check The calculated output change is 14. Multiplying the quotient 7 by the input change 2 recovers 14, so the division matches the two readings. On the dashed line, two steps right give a rise of 2 × 7 = 14, matching the change on the curve.

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

✗ Not this: Use f(3) + f(2) = 14 + 6 = 20 as the output change.

A change compares new with old by subtraction. Addition totals both readings instead of measuring how much they differ.

✓ Instead: Use f(3) − f(2) = 14 − 6 = 8. The matching input step is 3 − 2 = 1, so the average rate is 8 ÷ 1 = 8.

Tips and tricks

Say new minus old in both places.
As an everyday comparison: Compare two readings on the same trip. The difference quotient measures the average rate of change, meaning the average output change for each 1 step of input. This becomes average speed when the output is distance and the input is time. On a graph it measures the slope of the line through the two points. Section 1.3 develops this same rate-of-change idea. Calculus later considers steps closer and closer to zero to describe change at one instant.
With the worked values: The calculated output change is 14. Multiplying the quotient 7 by the input change 2 recovers 14, so the division matches the two readings. On the dashed line, two steps right give a rise of 2 × 7 = 14, matching the change on the curve.

.2Difference quotient

Imagine checking a car's mileage at the start and end of a trip. You need the difference between the readings, not their sum. Then divide by the time spent traveling. A difference quotient does this for any function: it compares the output at input a + h with the output at input a, then divides by the input change h. Here h is a number measuring a change, not a function name. It may be negative, but it must not be zero.

The input change is (a + h) − a = h.
The output change is f(a + h) − f(a).
Cancel only a common factor of the whole numerator and denominator, with h ≠ 0.
h is the size and direction of the input step. A step of zero gives no move and division by zero. Both inputs must belong to the function’s domain.

For f(x) = + 3x − 4, input 1 to 3 gives an output climb of 14 over a step of 2. The dashed line has slope 14 ÷ 2 = 7.

Reminder

Canceling fractions. Only common nonzero factors cancel: $3 × 8 over 3$ = 8, but $3 + 8 over 3$ does not equal 8.

The same idea, five ways

Say it

f of the whole sum a plus h, minus f of a, all over h; h is not zero.

Write it

Subtract the starting output from the ending output, then divide by the same input step.

In math

$f(a + h) − f(a) over h$
(a + h) − a = h
h ≠ 0

Compare an odometer’s ending and starting readings over the matching travel time.

See it

For f(x) = + 3x − 4, input 1 to 3 gives an output climb of 14 over a step of 2. The dashed line has slope 14 ÷ 2 = 7.

Worked exampleReversing the input step reverses both changes

For f(x) = $x^{2}$ + 3x − 4, compare starting input 3 with ending input 1. The input step is h = −2. Find the average output change per input step.

Starting at input 3 and ending at input 1 gives output change −14 and input change −2. Their ratio is 7, the same slope as in the forward direction.

What it asks. Find the two output readings, subtract the starting reading from the ending reading, and divide by the matching input change.

Plan. Evaluate the ending and starting inputs. Subtract new minus old for the outputs and for the inputs. Divide output change by input change and check by multiplication.

Keep the input and output from each reading together.
Subtract ending minus starting in both the top and bottom.
Multiply the rate by the input step to check that it recovers the output change.

f(3) = 14 and f(1) = 0.Both values were found in the numbers-first example.
The input change is 1 − 3 = −2 and the output change is 0 − 14 = −14.Both subtractions use the same ending-minus-starting direction.
$0 − 14 over 1 − 3$ = $−14 over −2$ = 7.A negative divided by a negative is positive, so reversing both changes keeps their ratio the same.

Answer

The average rate is 7 output units per input unit.

Check Multiplying the input step −2 by the rate 7 gives −14, which is the actual output change.

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

✗ Not this: Cancel h directly in by crossing out only one h in one term.

The numerator is a sum, not one product. Cancellation removes a common factor multiplying the entire numerator.

✓ Instead: First write , then cancel h to get 2a + h + 3, with h ≠ 0.

✗ Not this: After cancellation, the original quotient allows h = 0.

The original denominator is still h. At zero it is undefined, even though the simplified expression can be calculated there.

✓ Instead: Keep the condition h ≠ 0 on the quotient's simplified answer.

Tips and tricks

Use three lines: first f(a + h), next f(a), then the quotient. Keep the subtraction parentheses until every sign has changed.
A before and after comparison: The numerator asks how much the output changed. The denominator asks how much the input changed. Both must describe the same move from a to a + h.
Whole groups cancel: If the numerator becomes h(2a + h + 3), it is h times one whole group. Dividing by h removes that repeated nonzero factor. A term such as 3h cannot be canceled alone while other terms stay in the numerator.

.3One letter keeps many step sizes

A letter h saves you from doing the same calculation for every possible step. For f(x) = $x^{2}$ , the algebra gives 2x + h. At starting input x = 3, it becomes 6 + h. Put any nonzero allowed step into that one expression to get its average rate. A smaller step compares points closer together. You are still dividing by a nonzero step, even when the algebra is simplified.

The picture lists exact step sizes and their exact rates for starting input x = 3. The decimal rates shown are exact, not rounded.
a and x are alternate names for the starting input. h names the step.
h(2x + h) means h × (2x + h) here because h is a number, not the function name.

One expression gives the rate for each of these exact nonzero steps.

The same idea, five ways

Say it

For the square rule, the average rate from x to x plus h is two x plus h, with h not zero.

Write it

One expression gives the square rule’s average rate for many different nonzero input steps.

In math

$s(x + h) − s(x) over h$ = 2x + h, h ≠ 0
At x = 3: 2x + h = 6 + h
h = $1 over 2$ gives 6.5

One adjustable recipe handles many choices of the ingredient amount.

See it

One expression gives the rate for each of these exact nonzero steps.

Worked exampleCheck a square quotient with a negative step

For s(x) = $x^{2}$ , find $s(a + h) − s(a) over h$ , with h ≠ 0. Compare the squares of the ending and starting inputs.

(a + h − = 2ah +

2ah + = h(2a + h)

= 2a + h

h ≠ 0

Factor before canceling, and keep the original restriction.

What it asks. Find any requested outputs first, then compare the ending and starting outputs and divide by their nonzero input step.

Plan. Write the starting output and the output at x + h or a + h. Expand the whole changed input. Subtract the entire starting output in parentheses. In these line and quadratic examples, terms without h cancel. Factor h or divide every remaining term by h. Keep h ≠ 0 and check with numbers.

Every input variable receives the whole sum in parentheses.
A squared sum includes its middle term. An outside coefficient multiplies all three pieces.
Reverse every sign in the subtracted output, then factor the whole numerator before canceling h.

s(a + h) = (a + h $) squared$ = $a^{2}$ + 2ah + $h^{2}$ .Expand the square using its four products.
s(a) = $a^{2}$ .The starting input is a.
s(a + h) − s(a) = ( $a^{2}$ + 2ah + $h^{2}$ ) − $a^{2}$ = 2ah + $h^{2}$ .The starting square $a^{2}$ is removed from the ending output.
$2 ah + h squared over h$ = $h(2 a + h) over h$ .Both terms in the numerator contain a factor h.
$h(2 a + h) over h$ = 2a + h, with h ≠ 0.Cancel the nonzero factor h multiplying the whole numerator.

Answer

= 2a + h, with h ≠ 0

Check Choose a = 2 and h = −1. Directly, = = 3. The simplified expression gives 4 − 1 = 3. A negative nonzero h is allowed.

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

✗ Not this: Use h = 0 in the starting quotient.

It would be , which gives no unique value.

✓ Instead: Use a nonzero h, even after canceling.

Tips and tricks

Write the condition h ≠ 0 beside the answer.
As an everyday comparison: A letter h saves you from doing the same calculation for every possible step. For f(x) = $x^{2}$ , the algebra gives 2x + h. At starting input x = 3, it becomes 6 + h. Put any nonzero allowed step into that one expression to get its average rate. A smaller step compares points closer together. You are still dividing by a nonzero step, even when the algebra is simplified.
With the worked values: Choose a = 2 and h = −1. Directly, $s(1) − s(2) over −1$ = $1 − 4 over −1$ = 3. The simplified expression gives 4 − 1 = 3. A negative nonzero h is allowed.

Strategy: step by step

Write f(x + h) by replacing every x with the whole (x + h). Also write f(x) separately.
Multiply out. A square has the middle term: (x + h $) squared$ = $x^{2}$ + 2xh + $h^{2}$ . A number in front multiplies every piece.
Subtract f(x) in parentheses, then reverse every sign in that subtracted group.
In these polynomial examples, every term without h must cancel. If one remains, recheck the expansion, copied terms, and subtraction signs.
Factor h from the entire numerator, or divide every numerator term by h. Cancel only the shared nonzero factor.
Write the simplified answer and h ≠ 0. Both starting and ending inputs must still be allowed. Check with chosen numbers.

Strategy

Build a difference quotient in six stages

Is the formula a line mx + b?

YesThe output change is mh, so division by h gives m, the slope.

NoCalculate both outputs and subtract.

↓

Does the formula contain a squared sum?

YesExpand two copies, keeping the middle term, before applying an outside coefficient.

NoUse the operations actually in the given formula.

↓

In a polynomial example, did a term without h survive?

YesRecheck expansion and subtraction. Do not cancel h until every numerator term contains it.

NoFactor h or divide each term by h, then keep h ≠ 0.

Write f(x + h) by replacing every x with the whole (x + h). Also write f(x) separately.
Multiply out. A square has the middle term: (x + h $) squared$ = $x^{2}$ + 2xh + $h^{2}$ . A number in front multiplies every piece.
Subtract f(x) in parentheses, then reverse every sign in that subtracted group.
In these polynomial examples, every term without h must cancel. If one remains, recheck the expansion, copied terms, and subtraction signs.
Factor h from the entire numerator, or divide every numerator term by h. Cancel only the shared nonzero factor.
Write the simplified answer and h ≠ 0. Both starting and ending inputs must still be allowed. Check with chosen numbers.

Worked exampleEvaluate, subtract, then divide the quadratic outputs

Let f(x) = $x^{2}$ + 3x − 4. Find f(2), f(a), f(a + h), and $f(a + h) − f(a) over h$ for h ≠ 0. The first three ask for outputs at stated inputs. The last compares two outputs and divides by their input change.

The square expansion contributes , two ah terms, and before the subtraction.

For f(x) = + 3x − 4, input 1 to 3 gives an output climb of 14 over a step of 2. The dashed line has slope 14 ÷ 2 = 7.

The square expansion contributes , two ah terms, and before the subtraction.

What it asks. Find any requested outputs first, then compare the ending and starting outputs and divide by their nonzero input step.

Every input variable receives the whole sum in parentheses.
A squared sum includes its middle term. An outside coefficient multiplies all three pieces.
Reverse every sign in the subtracted output, then factor the whole numerator before canceling h.

f(2) = (2 $) squared$ + 3(2) − 4 = 4 + 6 − 4 = 6.Replace every x with 2, then do the power and multiplication before adding.
f(a) = $a^{2}$ + 3a − 4.The input is the letter a, so every x becomes a.
f(a + h) = (a + h $) squared$ + 3(a + h) − 4.Each x receives the whole expression a + h.
(a + h $) squared$ = (a + h)(a + h) = $a^{2}$ + ah + ha + $h^{2}$ = $a^{2}$ + 2ah + $h^{2}$ .The four products include two copies of ah.
3(a + h) = 3a + 3h, so f(a + h) = $a^{2}$ + 2ah + $h^{2}$ + 3a + 3h − 4.The 3 multiplies both terms of the input sum.
f(a + h) − f(a) = ( $a^{2}$ + 2ah + $h^{2}$ + 3a + 3h − 4) − ( $a^{2}$ + 3a − 4).Subtract the entire starting output, not only its first term.
= $a^{2}$ + 2ah + $h^{2}$ + 3a + 3h − 4 − $a^{2}$ − 3a + 4.The subtraction changes all three signs in the second parentheses.
= 2ah + $h^{2}$ + 3h. $a^{2}$ − $a^{2}$ , 3a − 3a and −4 + 4 each equal zero.
$f(a + h) − f(a) over h$ = $2 ah + h squared + 3 h over h$ = $h(2 a + h + 3) over h$ .Each numerator term has a factor h: 2ah = h(2a), $h^{2}$ = h(h), and 3h = h(3).
$h(2 a + h + 3) over h$ = 2a + h + 3, with h ≠ 0.The same nonzero factor h multiplies the whole numerator and denominator, so it cancels.

Answer

f(2) = 6
f(a) = $a^{2}$ + 3a − 4
f(a + h) = $a^{2}$ + 2ah + $h^{2}$ + 3a + 3h − 4
$f(a + h) − f(a) over h$ = 2a + h + 3, with h ≠ 0
With starting input named x, the same quotient is 2x + h + 3, with h ≠ 0.

Check Choose a = 2 and h = 1. Direct evaluation gives f(3) = 14 and f(2) = 6, so the quotient is = 8. The simplified result gives 2 × 2 + 1 + 3 = 8. Also, multiplying h(2a + h + 3) back out gives exactly 2ah + + 3h.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1DQ rung 1: a constant

For k(x) = 7, simplify $k(x + h) − k(x) over h$ , with h ≠ 0. This asks for the average output change per input step.

The line changes by 0 for one step of input.

What it asks. Evaluate any given numerical inputs first. Then compare the constant outputs and divide their difference by the nonzero input step.

Plan. For each requested evaluation, apply its own recipe. A constant recipe returns the same fixed value at a and at a + h. Subtract those equal outputs to obtain zero, then divide zero by nonzero h. Keep h ≠ 0 because the original fraction still excludes a zero step.

A constant output ignores the input.
Equal ending and starting outputs have difference zero.
Zero divided by a nonzero number is zero; division by zero has no value.

k(x + h) = 7 and k(x) = 7.A constant rule returns the same 7 regardless of input, so both the starting and ending outputs are 7.
k(x + h) − k(x) = 7 − 7 = 0.The new and old outputs are equal, and subtracting equal numbers gives zero.
$0 over h$ = 0, h ≠ 0.Zero divided by a nonzero input step is zero; the quotient still requires h ≠ 0.

Answer

0, with h ≠ 0.

Check At x = 1 and h = 1, k(1) = 7 and k(2) = 7. Directly the quotient is = 0. The expression 0 also gives 0.

Rung 2DQ rung 2: a rising line

For L(x) = 6x + 11, simplify $L(x + h) − L(x) over h$ , with h ≠ 0. This asks for the average output change per input step.

The line changes by 6 for one step of input.

What it asks. Find any requested outputs first, then compare the ending and starting outputs and divide by their nonzero input step.

Plan. Complete any requested evaluations. In the linear recipe, substitute the whole sum a + h or x + h, distribute its multiplier, and subtract the entire starting output. The terms unchanged by h cancel. Divide the remaining multiple of h by nonzero h and check with numbers.

A multiplier outside the changed input reaches both its terms.
The minus before the starting output changes every sign in that output.
Cancel only after the entire numerator is a multiple of nonzero h.

L(x + h) = 6(x + h) + 11 = 6x + 6h + 11.Replace x with the whole x + h; distributing 6 gives 6x + 6h.
L(x + h) − L(x) = (6x + 6h + 11) − (6x + 11) = 6h.Subtracting the whole old output reverses both signs. The 6x terms and 11 terms cancel, leaving 6h.
$6 h over h$ = 6, h ≠ 0.The same nonzero h multiplies the whole numerator and denominator, so dividing removes that factor.

Answer

6, with h ≠ 0.

Check At x = 1 and h = 1, L(1) = 17 and L(2) = 23. Directly the quotient is = 6. The expression 6 also gives 6.

Rung 3DQ rung 3: a falling line

For g(x) = 9 − 5x, simplify $g(x + h) − g(x) over h$ , with h ≠ 0. This asks for the average output change per input step.

The line changes by −5 for one step of input.

What it asks. Find any requested outputs first, then compare the ending and starting outputs and divide by their nonzero input step.

A multiplier outside the changed input reaches both its terms.
The minus before the starting output changes every sign in that output.
Cancel only after the entire numerator is a multiple of nonzero h.

g(x + h) = 9 − 5(x + h) = 9 − 5x − 5h.Replace x with x + h. The multiplier −5 reaches both x and h, giving −5x and −5h.
g(x + h) − g(x) = (9 − 5x − 5h) − (9 − 5x) = 9 − 5x − 5h − 9 + 5x = −5h.Subtracting 9 − 5x gives −9 + 5x. The 9 terms and x terms then cancel, leaving −5h.
$−5 h over h$ = −5, h ≠ 0.The numerator is −5 times nonzero h. Division by h returns −5.

Answer

−5, with h ≠ 0.

Check At x = 1 and h = 1, g(1) = 4 and g(2) = −1. Directly the quotient is = −5. The expression −5 also gives −5.

Rung 4DQ rung 4: a square

For s(x) = $x^{2}$ , simplify $s(x + h) − s(x) over h$ , with h ≠ 0. This asks for the average output change per input step.

The dashed line joins the two outputs used by the numerical check.

What it asks. Find any requested outputs first, then compare the ending and starting outputs and divide by their nonzero input step.

Every input variable receives the whole sum in parentheses.
A squared sum includes its middle term. An outside coefficient multiplies all three pieces.
Reverse every sign in the subtracted output, then factor the whole numerator before canceling h.

s(x + h) = (x + h $) squared$ = $x^{2}$ + 2xh + $h^{2}$ .Multiply two whole copies of x + h. The cross products xh and hx add to 2xh.
s(x + h) − s(x) = $x^{2}$ + 2xh + $h^{2}$ − $x^{2}$ = 2xh + $h^{2}$ .The starting output $x^{2}$ is subtracted from the ending output, so $x^{2}$ − $x^{2}$ cancels to zero.
2xh + $h^{2}$ = h(2x + h).Both 2xh and $h^{2}$ contain h: they are h × 2x and h × h.
$h(2 x + h) over h$ = 2x + h, h ≠ 0.Cancel the common factor h multiplying the entire numerator, retaining the nonzero restriction.

Answer

2x + h, with h ≠ 0.

Check At x = 1 and h = 1, s(1) = 1 and s(2) = 4. Directly the quotient is = 3. The expression 2x + h also gives 3.

Rung 5DQ rung 5: multiply the entire squared sum

For P(x) = 4 $x^{2}$ , simplify $P(x + h) − P(x) over h$ , with h ≠ 0. This asks for the average output change per input step.

The dashed line joins the two outputs used by the numerical check.

What it asks. Find any requested outputs first, then compare the ending and starting outputs and divide by their nonzero input step.

Every input variable receives the whole sum in parentheses.
A squared sum includes its middle term. An outside coefficient multiplies all three pieces.
Reverse every sign in the subtracted output, then factor the whole numerator before canceling h.

P(x + h) = 4(x + h $) squared$ .Replace the x in the square by the whole sum x + h while leaving its outside multiplier 4 outside.
First square: (x + h $) squared$ = $x^{2}$ + 2xh + $h^{2}$ .The square is computed first. Its four products are $x^{2}$ , xh, hx, and $h^{2}$ .
Then multiply every term by 4: P(x + h) = 4 $x^{2}$ + 8xh + 4 $h^{2}$ .Distribution applies the outside 4 to each term: 4 × 2xh = 8xh and 4 × $h^{2}$ = 4 $h^{2}$ .
P(x + h) − P(x) = 4 $x^{2}$ + 8xh + 4 $h^{2}$ − 4 $x^{2}$ = 8xh + 4 $h^{2}$ .The old output is 4 $x^{2}$ , so subtracting it removes the ending expression’s 4 $x^{2}$ term.
8xh + 4 $h^{2}$ = h(8x + 4h).8xh = h × 8x and 4 $h^{2}$ = h × 4h. Reverse distribution to place the shared h outside and the group 8x + 4h inside.
$h(8 x + 4 h) over h$ = 8x + 4h, h ≠ 0.The whole numerator is h times one group, so nonzero h cancels from that product.

Answer

8x + 4h, with h ≠ 0.

Check At x = 1 and h = 1, P(1) = 4 and P(2) = 16. Directly the quotient is = 12. The expression 8x + 4h also gives 12.

Rung 6DQ rung 6: quadratic, linear, and constant terms

For Q(x) = 3 $x^{2}$ − 2x + 6, simplify $Q(x + h) − Q(x) over h$ , with h ≠ 0. This asks for the average output change per input step.

The dashed line joins the two outputs used by the numerical check.

What it asks. Find any requested outputs first, then compare the ending and starting outputs and divide by their nonzero input step.

Every input variable receives the whole sum in parentheses.
A squared sum includes its middle term. An outside coefficient multiplies all three pieces.
Reverse every sign in the subtracted output, then factor the whole numerator before canceling h.

Q(x + h) = 3(x + h $) squared$ − 2(x + h) + 6.Both occurrences of x receive the complete sum x + h.
Q(x + h) = 3 $x^{2}$ + 6xh + 3 $h^{2}$ − 2x − 2h + 6.Expand (x + h $) squared$ first, multiply all three terms by 3, and distribute −2 to both terms of x + h.
Q(x + h) − Q(x) = (3 $x^{2}$ + 6xh + 3 $h^{2}$ − 2x − 2h + 6) − (3 $x^{2}$ − 2x + 6).The entire old output has three terms. Parentheses keep all three together for subtraction.
= 3 $x^{2}$ + 6xh + 3 $h^{2}$ − 2x − 2h + 6 − 3 $x^{2}$ + 2x − 6 = 6xh + 3 $h^{2}$ − 2h.Negating the old output gives −3 $x^{2}$ + 2x − 6; these cancel the matching $x^{2}$ , x, and constant terms.
6xh + 3 $h^{2}$ − 2h = h(6x + 3h − 2).Each of the three remaining terms contains h; taking it out leaves 6x, 3h, and −2.
$h(6 x + 3 h − 2) over h$ = 6x + 3h − 2, h ≠ 0.Divide out the same nonzero h from the whole numerator and denominator.

Answer

6x + 3h − 2, with h ≠ 0.

Check At x = 1 and h = 1, Q(1) = 7 and Q(2) = 14. Directly the quotient is = 7. The expression 6x + 3h − 2 also gives 7.

Rung 7DQ rung 7: a negative squared coefficient

For R(x) = −2 $x^{2}$ + 9, simplify $R(x + h) − R(x) over h$ , with h ≠ 0. This asks for the average output change per input step.

The dashed line joins the two outputs used by the numerical check.

What it asks. Find any requested outputs first, then compare the ending and starting outputs and divide by their nonzero input step.

Every input variable receives the whole sum in parentheses.
A squared sum includes its middle term. An outside coefficient multiplies all three pieces.
Reverse every sign in the subtracted output, then factor the whole numerator before canceling h.

R(x + h) = −2(x + h $) squared$ + 9.The whole x + h replaces x inside the square; the multiplier −2 remains outside.
R(x + h) = −2 $x^{2}$ − 4xh − 2 $h^{2}$ + 9.Expand the square and multiply every term by −2: 2xh becomes −4xh and $h^{2}$ becomes −2 $h^{2}$ .
R(x + h) − R(x) = (−2 $x^{2}$ − 4xh − 2 $h^{2}$ + 9) − (−2 $x^{2}$ + 9).Subtract the entire starting output −2 $x^{2}$ + 9 as one group.
= −2 $x^{2}$ − 4xh − 2 $h^{2}$ + 9 + 2 $x^{2}$ − 9 = −4xh − 2 $h^{2}$ .Subtracting −2 $x^{2}$ + 9 adds 2 $x^{2}$ and subtracts 9. Both matching old-output terms cancel.
−4xh − 2 $h^{2}$ = h(−4x − 2h).Both remaining terms contain h, and dividing each by h leaves −4x and −2h.
$h(−4 x − 2 h) over h$ = −4x − 2h, h ≠ 0.The shared nonzero h cancels from the whole product, preserving the quotient value.

Answer

−4x − 2h, with h ≠ 0.

Check At x = 1 and h = 1, R(1) = 7 and R(2) = 1. Directly the quotient is = −6. The expression −4x − 2h also gives −6.

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

✗ Not this: f(a + h) = f(a) + h.

The change happens to the input before the function recipe runs. The output generally changes by a different amount. Here f(3) = 14 while f(1) + 2 = 2.

✓ Instead: Replace every x by a + h: f(a + h) = (a + h + 3(a + h) − 4.

✗ Not this: −( + 3a − 4) = − + 3a − 4.

Subtracting a whole group multiplies every term by −1, not only the first term.

✓ Instead: −( + 3a − 4) = − − 3a + 4.

✗ Not this: The simplified quotient means h can equal zero.

Canceling preserves the value where the original quotient is defined. It does not create a value for division by zero.

✓ Instead: Write 2a + h + 3 with h ≠ 0.

✗ Not this: For g(x) = 9 − 5x, write −(9 − 5x) as −9 − 5x.

The second term has a minus already. Subtracting that whole term gives plus 5x; failing to reverse it leaves an incorrect −10x term without h.

✓ Instead: −(9 − 5x) = −9 + 5x. The output difference is −5h, giving quotient −5 for h ≠ 0.

✗ Not this: 4(x + h = 4 + 2xh + , or (4x + 4h.

The first version multiplies only one term by 4. The second moves 4 inside the square, multiplying the squared result by 16. At x = h = 1 they give 7 and 64, while 4(2 = 16.

✓ Instead: 4(x + h = 4 + 8xh + 4.

Tips and tricks

New output minus old output, over the step h. Say this before writing the symbols.
For the polynomial ladder, circle any remaining term without h after subtraction. Check all expansion and subtraction signs if one survives.
a and x both mean the starting input. h is the step.
On the exam, write the simplified expression and h ≠ 0. Rebuild the algebra rather than memorize a quadratic’s final answer.

Trap. Subtracting only the first term of the starting output. Keep −(f(x)) as one group, then change every sign inside.