The difference quotient: average rate of change from x to x + h

Explain it like I am five

The difference quotient has two ordinary actions hidden in its name. A difference is the result of subtracting: how much the output changed. A quotient is the result of dividing: share that change over the input change. Picture starting at a road marker and walking a chosen distance. Call your starting input x and your signed step h. A positive h goes right on the number line; a negative h goes left. Your ending input is x + h. The difference quotient measures average climb per input unit over that trip. It is the same average rate you already know, written so one answer can cover many starts and steps.

The whole expression x + h replaces the input of g.

Reminder

Distributive property. A number outside parentheses multiplies each term: −2(x + h) = −2x − 2h.
Subtracting a whole expression. The minus means multiply by −1: −(6 − 2x) = −6 + 2x.
Like terms. Like terms have the same letters with the same powers: −2x + 2x = 0. Terms $x^{2}$ and x cannot be combined.
Canceling a nonzero factor. $−2 h over h$ = −2 · $\frac{h}{h}$ = −2 only for h ≠ 0.

Why it works. Put starting input x and ending input x + h into the endpoint formula. The bottom is (x + h) − x = h, and the top is f(x + h) − f(x). That is why the formula has this form. For an expanded polynomial, f(x + h) contains a copy of f(x); subtracting removes that copy, leaving terms with h. For g(x) = 6 − 2x, that is 6 − 2x − 2h minus 6 − 2x. Other kinds of functions may require combining fractions first, rather than showing cancellation term by term.

RuleDifference quotient = , with h ≠ 0 and both outputs defined. It is the average rate on [x, x + h] if h > 0, and on the sorted interval [x + h, x] if h < 0.

The same idea, five ways

Say it

Say 'difference in outputs divided by the input step'. Read f(x + h) as 'f of x plus h'.

Write it

Start at input x, change the input by a nonzero signed amount h, and compare the two outputs per input unit.

In math

$f(x + h) − f(x) over h$ , h ≠ 0
h = (x + h) − x
[x, x + h] if h > 0; [x + h, x] if h < 0

Average climb per mile between a chosen road marker and another marker a signed step away.

See it

The starting input 2 plus step 3 gives ending input 5.

The same idea, other ways

With numbers first

For $x^{2}$ starting at 3, a step 1 gives $16 − 9 over 1$ = 7. A step 0.5 gives $12.25 − 9 over 0.5$ = 6.5. Steps 0.1 and 0.01 give 6.1 and 6.01. The table shows every rate is 6 + h. The fully worked first rung computes every column.

For starting at 3, the rate in every column is 6 + h.

A line before its algebra

For g(x) = 6 − 2x, start at 1 and step 1: $2 − 4 over 1$ = −2. Start at 3 and step 2: $−4 − 0 over 2$ = −2. Start at −2 and step 0.5: $9 − 10 over 0.5$ = −2. All give −2; the worked subtraction shows why.

For a positive input step h, the line drops 2h; the rate is −2 on every pair of distinct inputs.

What it is for

If d(t) is miles driven after t hours, $d(t + h) − d(t) over h$ gives the average speed across those times when h > 0. For positive s and h on a growing square tile, 2sh + $h^{2}$ is the extra area and 2s + h is extra area per inch of growth. At s = 10 and h = 1, that rate is 21. A reusable answer also covers many intervals: 2x + h at x = 3, h = 0.5 gives 6.5.

A growing square adds two sh strips and one corner.

The same function machine

In f(x + h), f names the machine and x + h is its one input. It is not f multiplied by x + h. For a square machine, put in 3 + 1 = 4 and get 16, rather than adding 1 to the old output 9.

The parentheses contain one input for the named function.

A smaller observation window

The square table's rates 7, 6.5, 6.1, and 6.01 move toward 6 as the positive step shrinks. For smoothly changing travel, shrinking the time window makes an average speed approach the speedometer reading at the starting time. Here the immediate task is building reusable interval rates; no zero step is ever divided by.

For starting at 3, the rate in every column is 6 + h.

.1When the start is a number

The letter x only names the starting input. Replacing it by 4 gives an interval between 4 and 4 + h; the method is unchanged.

For h > 0, the interval is [4, 4 + h]. For h < 0, it is [4 + h, 4].
The input difference in starting-to-ending order is always h.

Choosing the starting input 4 leaves the same average line rate 5.

Worked exampleDifference quotient starting at x = 3

Let f(x) = $x^{2}$ − 4x + 1. (a) Find and simplify the difference quotient $f(3 + h) − f(3) over h$ , with h ≠ 0. (b) Use your result with h = −1 to find the average rate of change of f on the matching interval, and name that interval.

Graph of f(x) = − 4x + 1 = (x − 2 − 3, marking the start point (3, −2) and the point (2, −3) reached with h = −1. The secant joining them has slope 1.

Find the fixed output first: f(3) = $3 squared$ − 4(3) + 1 = 9 − 12 + 1 = −2.The start is the number 3, so f(3) is a single number. Computing it once keeps the subtraction simple.
Plug in: f(3 + h) = (3 + h $) squared$ − 4(3 + h) + 1 = (9 + 6h + $h^{2}$ ) − 12 − 4h + 1 = $h^{2}$ + 2h − 2.Every x is replaced by the whole input (3 + h). The square (3 + h $) squared$ is multiplied out to 9 + 6h + $h^{2}$ , and −4 is distributed over both terms of (3 + h).
Subtract: f(3 + h) − f(3) = ( $h^{2}$ + 2h − 2) − (−2) = $h^{2}$ + 2h − 2 + 2.Writing −(f(3)) in parentheses lets the minus sign reach the whole value. Subtracting −2 means adding 2.
Simplify: $h^{2}$ + 2h − 2 + 2 = $h^{2}$ + 2h.The terms that do not depend on h, −2 and +2, cancel. This always happens for a fully expanded polynomial.
Cancel h: $h squared + 2 h over h$ = $h(h + 2) over h$ = h + 2, for h ≠ 0.Factor h shows that h multiplies the entire numerator. Since h ≠ 0, we may divide it out. The restriction h ≠ 0 stays.
Part (b): put h = −1 into h + 2 to get −1 + 2 = 1. The interval runs from 3 + (−1) = 2 to 3. Since h < 0, write it in sorted order as [2, 3].When h is negative, x + h lies to the left of x. The average rate therefore belongs to the interval [x + h, x].

Answer

(a) = h + 2, h ≠ 0. (b) With h = −1, the average rate of change is 1 on the interval [2, 3].

Check Compute directly: f(2) = 4 − 8 + 1 = −3 and f(3) = −2. The average rate on [2, 3] is = = 1, which matches h + 2 at h = −1. A second check with h = 1: f(4) = 16 − 16 + 1 = 1, so = 3, and h + 2 = 3. ✓

Work to write

f(3) = 9 − 12 + 1 = −2
f(3 + h) = (3 + h $) squared$ − 4(3 + h) + 1 = $h^{2}$ + 2h − 2
f(3 + h) − f(3) = ( $h^{2}$ + 2h − 2) − (−2) = $h^{2}$ + 2h
$h squared + 2 h over h$ = $h(h + 2) over h$ = h + 2, h ≠ 0
h = −1: h + 2 = 1; interval [2, 3] since h < 0

(a) $f(3 + h) − f(3) over h$ = h + 2, h ≠ 0. (b) With h = −1, the average rate of change is 1 on the interval [2, 3].

.2When the start is 3 for a square

Use the same replacement in a square formula. Multiplying out creates the output change 6h + $h^{2}$ . Factoring h turns this into the reusable rate 6 + h.

The rate is 6 + h, with h ≠ 0.
This is 2x + h with the starting input set to x = 3.
Section 1.1 used a and a + h; here the general start is named x.

For positive lengths x and h, the four areas explain the square expansion; the algebra also works for negative inputs.

Worked exampleDifference quotient of a quadratic starting at x = 2

Let f(x) = 2 $x^{2}$ + 4x − 2. (a) Find and simplify the difference quotient $f(2 + h) − f(2) over h$ , with h ≠ 0. (b) Use your result with h = $1 over 2$ to find the average rate of change of f on the matching interval, and name that interval.

The parabola f(x) = 2 + 4x − 2 = 2(x + 1 − 4, with the points (2, 14) and (2.5, 20.5). The secant line through them has slope 13, the average rate of change on [2, 2.5].

Find f(2) first: f(2) = 2(2 $) squared$ + 4(2) − 2 = 8 + 8 − 2 = 14.The fixed output f(2) is needed for the subtraction, and the start is a plain number.
Plug in 2 + h: f(2 + h) = 2(2 + h $) squared$ + 4(2 + h) − 2. Expand (2 + h $) squared$ = 4 + 4h + $h^{2}$ , so f(2 + h) = 8 + 8h + 2 $h^{2}$ + 8 + 4h − 2 = 2 $h^{2}$ + 12h + 14.Every x is replaced by the whole input (2 + h). The square must be multiplied out before it is doubled.
Subtract: f(2 + h) − f(2) = (2 $h^{2}$ + 12h + 14) − (14) = 2 $h^{2}$ + 12h.The parentheses let the minus reach the whole of f(2). The constant terms, which do not depend on h, cancel.
Factor and cancel h: $2 h squared + 12 h over h$ = $h(2 h + 12) over h$ = 2h + 12, for h ≠ 0.h multiplies the entire numerator, so the nonzero factor h divides out. The restriction h ≠ 0 stays.
Part (b): with h = $1 over 2$ , the rate is 2( $1 over 2$ ) + 12 = 13. Because h > 0, the interval is [2, 2 + $1 over 2$ ] = [2, $5 over 2$ ].The difference quotient is the average rate on [x, x + h] when h > 0.

Answer

(a) = 2h + 12, h ≠ 0. (b) The average rate of change is 13 on the interval [2, ].

Check Direct computation gives f() = 2() + 4() − 2 = + 10 − 2 = . Then = = = 13, which matches 2h + 12 at h = .

Work to write

f(2) = 14
f(2 + h) = 2(4 + 4h + $h^{2}$ ) + 4(2 + h) − 2 = 2 $h^{2}$ + 12h + 14
f(2 + h) − f(2) = (2 $h^{2}$ + 12h + 14) − (14) = 2 $h^{2}$ + 12h
$h(2 h + 12) over h$ = 2h + 12, h ≠ 0
h = $1 over 2$ : average rate = 13 on [2, $5 over 2$ ]

(a) $f(2 + h) − f(2) over h$ = 2h + 12, h ≠ 0. (b) The average rate of change is 13 on the interval [2, $5 over 2$ ].

.3A negative h goes left

h is a signed input change, not necessarily a positive length. The ending input is still start plus h. Keep the output subtraction in that same travel order; sorting the interval reverses both differences and preserves the ratio.

h < 0 means the ending input is smaller than the starting input.
For x = 4 and h = −3, the endpoints are 4 and 1, so the sorted interval is [1, 4].

A signed step h = −3 goes left from 4 to 1; the sorted interval is [1, 4].

Worked exampleDifference quotient of a downward parabola, then a step backward with h = −3

Let f(x) = − $x^{2}$ + 6x − 5. (a) Find and simplify the difference quotient $f(x + h) − f(x) over h$ , with h ≠ 0. (b) Use your result with x = 5 and h = −3 to find the average rate of change of f on the matching interval, and name that interval.

Graph of f(x) = − + 6x − 5 = −(x − 3 + 4, with vertex (3, 4). The points (2, 3) and (5, 0) mark the step from x = 5 back to x + h = 2. The secant through them has slope −1.

f(x + h) = −(x + h $) squared$ + 6(x + h) − 5 = −( $x^{2}$ + 2xh + $h^{2}$ ) + 6x + 6h − 5 = − $x^{2}$ − 2xh − $h^{2}$ + 6x + 6h − 5Plug in: every x is replaced by the whole input (x + h). The square is expanded before the leading minus is applied to all three of its terms.
f(x + h) − (f(x)) = (− $x^{2}$ − 2xh − $h^{2}$ + 6x + 6h − 5) − (− $x^{2}$ + 6x − 5) = − $x^{2}$ − 2xh − $h^{2}$ + 6x + 6h − 5 + $x^{2}$ − 6x + 5Subtract: f(x) goes in parentheses so the minus reaches every term. − $x^{2}$ becomes + $x^{2}$ , +6x becomes −6x, and −5 becomes +5.
Numerator = −2xh − $h^{2}$ + 6hSimplify: − $x^{2}$ + $x^{2}$ = 0, 6x − 6x = 0 and −5 + 5 = 0. The terms independent of h cancel, as they must for a fully expanded polynomial.
$−2 xh − h squared + 6 h over h$ = $h(−2 x − h + 6) over h$ = −2x − h + 6, with h ≠ 0Cancel h: h is a factor of the entire numerator and h ≠ 0, so it divides out. The restriction h ≠ 0 stays.
With x = 5 and h = −3: −2(5) − (−3) + 6 = −10 + 3 + 6 = −1Substitute the given values into the simplified quotient. Subtracting a negative h adds 3.
x + h = 5 + (−3) = 2, so the interval is [2, 5]Since h < 0, the step goes left from x = 5 to x + h = 2. The interval is written sorted, from the smaller endpoint to the larger.

Answer

(a) = −2x − h + 6, for h ≠ 0. (b) The average rate of change of f on [2, 5] is −1.

Check Compute directly: f(5) = −25 + 30 − 5 = 0 and f(2) = −4 + 12 − 5 = 3. Then = = −1, which matches. The original form gives the same value: = = −1.

Work to write

f(x + h) = − $x^{2}$ − 2xh − $h^{2}$ + 6x + 6h − 5
f(x + h) − f(x) = −2xh − $h^{2}$ + 6h
$f(x + h) − f(x) over h$ = −2x − h + 6, h ≠ 0
x = 5, h = −3: −10 + 3 + 6 = −1
Interval [2, 5]; average rate of change = −1

(a) $f(x + h) − f(x) over h$ = −2x − h + 6, for h ≠ 0. (b) The average rate of change of f on [2, 5] is −1.

.4An area-growth use

A difference quotient compares the growth of a square's area with the growth of its side. The picture splits the added area into two strips and a corner when the lengths and growth are positive.

A square of side s has area $s^{2}$ .
Growing the side by h adds 2sh + $h^{2}$ square units.
For s > 0 and h > 0, area gain per side-length gain is 2s + h.

A growing square adds two sh strips and one corner.

Worked exampleWhat the difference quotient measures on a tile

This asks how much area is gained per inch of side growth. A square tile's side grows from 10 inches to 11 inches. Find the average area gain per inch.

A growing square adds two sh strips and one corner.

Starting area: $10 squared$ = 100 square inches. Ending area: $11 squared$ = 121 square inches.A square's area is its side times itself.
Area change: 121 − 100 = 21 square inches. Side change: 11 − 10 = 1 inch.Subtract matching endpoints in the same order.
$21 over 1$ = 21 square inches per inch.The difference quotient divides the area difference by the side-length difference.

Answer

21 square inches per inch of side growth.

Check The two strips add 10 · 1 + 1 · 10 = 20 square inches, and the corner adds = 1, giving 21. The general result 2s + h gives 2(10) + 1 = 21.

Strategy: step by step

1. Plug in: replace every x in the formula by the whole input (x + h); multiply out the needed products.
2. Subtract: write −(f(x)) with parentheses, then let that minus reach every term.
3. Simplify: combine like terms; for a fully expanded polynomial, the terms independent of h cancel.
4. Cancel h: when h multiplies the entire numerator, divide that nonzero factor out; keep every original restriction.

Strategy

The difference quotient in four moves

Is the starting input a fixed number instead of x?

YesUse that number in f(start + h) and f(start); the bottom is still h.

NoKeep x as the unspecified starting input.

↓

Is there a squared input?

YesMultiply (x + h)(x + h) into + 2xh + before applying any outside coefficient.

NoCheck whether another power or a fraction needs work.

↓

Is there a cubed input?

YesUse the cubing-a-sum refresher to get + 3h + 3x + .

NoContinue with the given formula.

↓

Is the input in a fraction bottom?

YesKeep each whole bottom grouped, find a common bottom, and subtract the tops before dividing by h.

NoCombine polynomial like terms directly.

↓

After full polynomial expansion, does a term independent of h remain?

YesRecheck substitution and signs. This check applies to expanded polynomials; other functions can need different simplification.

NoFactor h from the whole numerator if possible.

↓

Is h negative?

YesKeep its sign in the quotient and describe the sorted interval [x + h, x].

NoFor h > 0 use [x, x + h]; h = 0 is excluded.

Plug in the whole input x + h.
Subtract the whole starting output f(x).
Simplify using the right arithmetic for the formula.
Cancel only a whole nonzero common factor, and state the restrictions.

Worked exampleDifference quotient of a line

This asks for the average rate over any nonzero input step h starting at any input x. Find $g(x + h) − g(x) over h$ for g(x) = 6 − 2x. The plan is plug in, subtract, simplify, then cancel h.

For a positive input step h, the line drops 2h; the rate is −2 on every pair of distinct inputs.

Require h ≠ 0. Write g(x + h) = 6 − 2(x + h).x + h is one whole input. The −2 must multiply all of it; writing 6 − 2x + h would multiply only x.
g(x + h) = 6 − 2x − 2h.The distributive property means the multiplier reaches each term: (−2) · x = −2x and (−2) · h = −2h.
Write the output change as (6 − 2x − 2h) − (6 − 2x).Subtract the entire starting output, not one selected term.
−(6 − 2x) = −6 + 2x, so the output change is 6 − 2x − 2h − 6 + 2x.The hidden sign on 6 is +. Multiplying the whole starting expression by −1 changes +6 to −6 and −2x to +2x.
6 − 2x − 2h − 6 + 2x = −2h.The +6 and −6 cancel, and −2x and +2x cancel. The expanded polynomial contains a copy of g(x), removed by subtraction.
$−2 h over h$ = −2 · $\frac{h}{h}$ = −2 · 1 = −2.The nonzero factor h divides by itself to give 1.

Answer

−2, with h ≠ 0.
For each 1 unit right, g drops 2 output units. This rate holds for both positive and negative signed steps.

Check Use x = 2 and h = 3: g(2) = 6 − 4 = 2 and g(5) = 6 − 10 = −4, so = = −2. Also g(0) = 6 and g(1) = 4 show a drop of 2 for a move of 1 right.

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

Rung 1Difference quotient with numbers: f(x) = − 3x + 5 at x = 1, h = 3

Let f(x) = $x^{2}$ − 3x + 5. Use x = 1 and h = 3, so x + h = 4. The table lists f(1) = 3 and f(4) = 9. Find the difference quotient $f(x + h) − f(x) over h$ and state the interval it is the average rate of change on.

Values of f(x) = − 3x + 5 at x = 1 and x + h = 4: f(1) = 3 and f(4) = 9, the two outputs used in the difference quotient.

Plug in: f(1 + 3) = (1 + 3 $) squared$ − 3(1 + 3) + 5. Multiply out: (1 + 3 $) squared$ = 1 + 2·1·3 + $3 squared$ = 1 + 6 + 9, and −3(1 + 3) = −3 − 9. So f(1 + 3) = 1 + 6 + 9 − 3 − 9 + 5 = 9.Every x in the formula is replaced by the whole input (x + h) = (1 + 3). Expanding before adding shows which pieces come from h. The total matches the table value f(4) = 9.
Subtract: f(1 + 3) − (f(1)) = (1 + 6 + 9 − 3 − 9 + 5) − (1 − 3 + 5) = 1 + 6 + 9 − 3 − 9 + 5 − 1 + 3 − 5.Writing −(f(x)) with parentheses lets the minus sign reach every term of f(1) = 1 − 3 + 5.
Simplify: the pairs 1 − 1, −3 + 3 and 5 − 5 cancel, leaving 6 + 9 − 9 = 6. So the numerator is 6. As a check, 9 − 3 = 6.The terms that do not involve h (the pieces of f(1)) cancel. Only terms built from h = 3 remain.
Cancel h: $6 over 3$ = 2. Here h = 3 ≠ 0, and both f(1) and f(4) are defined.Dividing by h needs h ≠ 0. Since h > 0, the interval is [x, x + h] = [1, 4].

Answer

= = 2. This is the average rate of change of f on [1, 4].

Check The general difference quotient is = = 2x + h − 3 for h ≠ 0. At x = 1, h = 3 this gives 2 + 3 − 3 = 2, which matches. The slope of the secant line through (1, 3) and (4, 9) is = 2.

Work to write

x + h = 1 + 3 = 4
f(4) = 16 − 12 + 5 = 9 and f(1) = 1 − 3 + 5 = 3
f(4) − (f(1)) = 9 − 3 = 6
$6 over 3$ = 2, with h = 3 ≠ 0
Average rate of change on [1, 4] is 2

$f(4) − f(1) over 3$ = $9 − 3 over 3$ = 2. This is the average rate of change of f on [1, 4].

Rung 2Difference quotient of the constant function f(x) = 4

Let f(x) = 4 for every real number x. Its graph is the horizontal line y = 4, which passes through (−2, 4) and (3, 4). (a) Find and simplify the difference quotient $f(x + h) − f(x) over h$ , with h ≠ 0. (b) Use your result with x = 3 and h = −5, so x + h = −2. Find the average rate of change of f on the matching interval, and name that interval.

The horizontal line y = 4, the graph of the constant function f(x) = 4, through (−2, 4) and (3, 4). The rise between the points is 0, so the average rate of change is 0.

Plug in: f(x + h) = 4.The rule for f has no x in it. Replacing every x by (x + h) changes nothing, so the output is 4 for any input.
Subtract with parentheses: f(x + h) − (f(x)) = 4 − (4).Writing −(f(x)) with parentheses makes the minus sign reach the whole output f(x) = 4.
Simplify the numerator: 4 − 4 = 0, so the difference quotient is $0 over h$ .The two outputs are equal, so they cancel completely. Every term is independent of h, so nothing is left.
Divide: $0 over h$ = 0 for h ≠ 0.Zero divided by any nonzero number is 0. The restriction h ≠ 0 stays, because the quotient is undefined at h = 0.
Part (b): with x = 3 and h = −5, the difference quotient is 0. Since h < 0, sort the endpoints x + h = −2 and x = 3 to get the interval [−2, 3].The rule says that when h < 0, the quotient is the average rate on the sorted interval [x + h, x].

Answer

(a) = 0 for all h ≠ 0. (b) The average rate of change of f on [−2, 3] is 0.

Check Compute directly from the two points: = = = 0. This matches part (b). It also agrees with the graph, because a horizontal line has slope 0 and the output never changes.

Work to write

f(x + h) = 4
f(x + h) − (f(x)) = 4 − (4) = 0
$f(x + h) − f(x) over h$ = $0 over h$ = 0, h ≠ 0
h = −5 < 0, so the interval is [−2, 3]
Average rate of change on [−2, 3] = 0

(a) $f(x + h) − f(x) over h$ = 0 for all h ≠ 0. (b) The average rate of change of f on [−2, 3] is 0.

Rung 3Rung 3: multiplication changes the whole input

This asks for a rate formula for every start and every nonzero step. Find the difference quotient of f(x) = 4x.

This line rises 4 for each unit right.

f(x + h) = 4(x + h) = 4x + 4h.The whole new input replaces x, and 4 multiplies each piece.
Output change: (4x + 4h) − (4x) = 4x + 4h − 4x = 4h.Subtract the entire original output; +4x and −4x cancel.
$4 h over h$ = 4 · $\frac{h}{h}$ = 4, with h ≠ 0.A common nonzero multiplier h divides out.

Answer

4, with h ≠ 0.

Check At x = 2 and h = 3, f(5) = 20 and f(2) = 8, so = = 4.

Rung 4Rung 4: a negative slope with a constant term

This asks for the average rate over any nonzero input step h starting at any input x. Find $g(x + h) − g(x) over h$ for g(x) = 6 − 2x. The plan is plug in, subtract, simplify, then cancel h.

For a positive input step h, the line drops 2h; the rate is −2 on every pair of distinct inputs.

Require h ≠ 0. Write g(x + h) = 6 − 2(x + h).x + h is one whole input. The −2 must multiply all of it; writing 6 − 2x + h would multiply only x.
g(x + h) = 6 − 2x − 2h.The distributive property means the multiplier reaches each term: (−2) · x = −2x and (−2) · h = −2h.
Write the output change as (6 − 2x − 2h) − (6 − 2x).Subtract the entire starting output, not one selected term.
−(6 − 2x) = −6 + 2x, so the output change is 6 − 2x − 2h − 6 + 2x.The hidden sign on 6 is +. Multiplying the whole starting expression by −1 changes +6 to −6 and −2x to +2x.
6 − 2x − 2h − 6 + 2x = −2h.The +6 and −6 cancel, and −2x and +2x cancel. The expanded polynomial contains a copy of g(x), removed by subtraction.
$−2 h over h$ = −2 · $\frac{h}{h}$ = −2 · 1 = −2.The nonzero factor h divides by itself to give 1.

Answer

−2, with h ≠ 0.
For each 1 unit right, g drops 2 output units. This rate holds for both positive and negative signed steps.

Check Use x = 2 and h = 3: g(2) = 6 − 4 = 2 and g(5) = 6 − 10 = −4, so = = −2. Also g(0) = 6 and g(1) = 4 show a drop of 2 for a move of 1 right.

Rung 5Difference quotient of a downward parabola starting at x = 1

Let f(x) = − $x^{2}$ − 2x + 5. Its graph is the downward-opening parabola y = −(x + 1 $) squared$ + 6, with vertex (−1, 6). It passes through (1, 2) and (3, −10). (a) Find and simplify the difference quotient $f(1 + h) − f(1) over h$ , with h ≠ 0. (b) Use your result with h = 2 to find the average rate of change of f on the matching interval, and name that interval.

The parabola y = −(x + 1 + 6 with vertex (−1, 6). The marked points (1, 2) and (3, −10) are the endpoints of the interval [1, 3], where the average rate of change is −6.

Find the starting output: f(1) = −(1 $) squared$ − 2(1) + 5 = −1 − 2 + 5 = 2.The difference quotient needs f(1), so the fixed start x = 1 is evaluated first. The result matches the point (1, 2) on the graph.
Plug in 1 + h: f(1 + h) = −(1 + h $) squared$ − 2(1 + h) + 5 = −(1 + 2h + $h^{2}$ ) − 2 − 2h + 5 = −1 − 2h − $h^{2}$ − 2 − 2h + 5 = 2 − 4h − $h^{2}$ .Every x is replaced by the whole input (1 + h). The square is expanded before the leading minus sign is distributed.
Subtract: f(1 + h) − f(1) = (2 − 4h − $h^{2}$ ) − (2) = 2 − 4h − $h^{2}$ − 2.Writing −(f(1)) with parentheses makes the minus sign reach the whole starting output.
Simplify the numerator: 2 − 2 − 4h − $h^{2}$ = −4h − $h^{2}$ .The terms that do not depend on h cancel, as they must for a fully expanded polynomial.
Factor and cancel h: $−4 h − h squared over h$ = $h(−4 − h) over h$ = −4 − h, for h ≠ 0.h multiplies the entire numerator. It is a nonzero factor, so it can be divided out. The restriction h ≠ 0 stays attached to the result.
For (b), substitute h = 2: −4 − 2 = −6. The inputs are 1 and 1 + 2 = 3, so the interval is [1, 3].With h > 0, the difference quotient is the average rate of change on [x, x + h] = [1, 3].

Answer

(a) = −4 − h, for h ≠ 0. (b) With h = 2, the average rate of change of f on [1, 3] is −6. On average, f falls 6 units for each 1-unit increase in x.

Check Compute directly from the given points: f(3) = −9 − 6 + 5 = −10 and f(1) = 2. Then = = −6, which matches −4 − 2 = −6.

Work to write

f(1) = 2
f(1 + h) = −(1 + h $) squared$ − 2(1 + h) + 5 = 2 − 4h − $h^{2}$
f(1 + h) − f(1) = (2 − 4h − $h^{2}$ ) − (2) = −4h − $h^{2}$
$−4 h − h squared over h$ = $h(−4 − h) over h$ = −4 − h, h ≠ 0
h = 2: −4 − 2 = −6
Interval: [1, 3]; average rate of change = −6

(a) $f(1 + h) − f(1) over h$ = −4 − h, for h ≠ 0. (b) With h = 2, the average rate of change of f on [1, 3] is −6. On average, f falls 6 units for each 1-unit increase in x.

Rung 6Difference quotient with a negative step on y = + 2x − 3

Let f(x) = $x^{2}$ + 2x − 3. Its graph is the upward-opening parabola y = (x + 1 $) squared$ − 4, with vertex (−1, −4). It passes through (−1, −4) and (2, 5). (a) Find and simplify the difference quotient $f(x + h) − f(x) over h$ , with h ≠ 0. (b) Use your result with x = 2 and h = −3 to find the average rate of change of f on the matching interval, and name that interval.

The parabola y = + 2x − 3 = (x + 1 − 4, with vertex (−1, −4) and the point (2, 5). A step of h = −3 from x = 2 reaches x = −1, so the secant through these two points has slope 3.

f(x + h) = (x + h $) squared$ + 2(x + h) − 3 = $x^{2}$ + 2xh + $h^{2}$ + 2x + 2h − 3Plug in: every x becomes the whole input (x + h). Then (x + h $) squared$ = $x^{2}$ + 2xh + $h^{2}$ , and 2(x + h) = 2x + 2h.
f(x + h) − f(x) = $x^{2}$ + 2xh + $h^{2}$ + 2x + 2h − 3 − ( $x^{2}$ + 2x − 3) = $x^{2}$ + 2xh + $h^{2}$ + 2x + 2h − 3 − $x^{2}$ − 2x + 3Subtract: f(x) goes in parentheses, so the minus reaches every one of its terms. This turns −3 into +3.
f(x + h) − f(x) = 2xh + $h^{2}$ + 2hSimplify: $x^{2}$ − $x^{2}$ = 0, 2x − 2x = 0 and −3 + 3 = 0. The terms independent of h cancel, as they must for a fully expanded polynomial.
$2 xh + h squared + 2 h over h$ = $h(2 x + h + 2) over h$ = 2x + h + 2, with h ≠ 0Cancel h: h is a factor of the entire numerator and h ≠ 0, so it divides out. The restriction h ≠ 0 still holds.
With x = 2 and h = −3: 2(2) + (−3) + 2 = 4 − 3 + 2 = 3Substitute into the simplified quotient. Here −3 ≠ 0, so this value of h is allowed.
x + h = 2 + (−3) = −1, so the interval is [−1, 2]The step h is negative, so x + h lies to the left of x. Sorting the endpoints gives [x + h, x] = [−1, 2].

Answer

(a) = 2x + h + 2, for h ≠ 0. (b) The average rate of change of f on [−1, 2] is 3.

Check Use the graph's points directly. f(2) = 4 + 4 − 3 = 5 and f(−1) = 1 − 2 − 3 = −4. Then = = = 3. In left-to-right order, = = 3. Both match the formula's value.

Work to write

f(x + h) = $x^{2}$ + 2xh + $h^{2}$ + 2x + 2h − 3
f(x + h) − f(x) = $x^{2}$ + 2xh + $h^{2}$ + 2x + 2h − 3 − ( $x^{2}$ + 2x − 3) = 2xh + $h^{2}$ + 2h
$2 xh + h squared + 2 h over h$ = 2x + h + 2, h ≠ 0
x = 2, h = −3: 2(2) + (−3) + 2 = 3
x + h = −1, so the interval is [−1, 2]
Average rate of change on [−1, 2] = 3

(a) $f(x + h) − f(x) over h$ = 2x + h + 2, for h ≠ 0. (b) The average rate of change of f on [−1, 2] is 3.

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

✗ Not this: f(x + h) = f · x + f · h.

f is the function's name, not a number multiplying its input.

✓ Instead: For f(x) = , f(x + h) = (x + h.

✗ Not this: f(x + h) = f(x) + f(h).

With f(x) = , x = 3 and h = 1, f(4) = 16, but f(3) + f(1) = 9 + 1 = 10.

✓ Instead: Feed the whole new input x + h into the formula.

✗ Not this: f(x + h) = f(x) + h.

For the same square example, f(4) = 16, but f(3) + 1 = 10. Adding to an input happens before the rule runs.

✓ Instead: For g(x) = 6 − 2x, g(x + h) = 6 − 2x − 2h.

✗ Not this: A leftover 12 − 2h is the correct output change for g(x) = 6 − 2x.

The starting +6 was added rather than subtracted. The proper subtraction removes the copy 6 − 2x.

✓ Instead: (6 − 2x − 2h) − (6 − 2x) = −2h.

✗ Not this: h must always be positive, and [x, x + h] is ordered for every h.

If x = 4 and h = −3, x + h = 1 lies to the left of 4.

✓ Instead: The quotient allows h < 0; use the sorted interval [1, 4]. Only h = 0 is always forbidden.

Tips and tricks

Remember the name: difference means subtraction, quotient means division.
Here x names the starting input, not an unknown you automatically solve for. h names the signed input change.
The four moves are plug in, subtract, simplify, cancel a valid whole factor.
For a fully expanded polynomial numerator, every term independent of h cancels. Do not apply that wording blindly to square roots or other unexpanded functions.
Check algebra with x = 2 and h = 3; values 0 and 1 often hide missing factors and wrong powers.
The assigned algebra skill is an interval ending at a letter or having step h. This is the same endpoint-rate idea from section 1.1.

Trap. Adding h to the old output, treating f as a multiplier, or allowing a zero input step. Substitute x + h into the whole function formula and keep h ≠ 0.