A quadratic polynomial uses constants, multiples of the input, and a nonzero multiple of the squared input. Think of enlarging a square carpet: the old square stays in one corner, two strips appear beside it, and a corner fills the gap. Multiplying out (x + h)2 records all four pieces: x2, xh, xh, and h2. That is why x2 + h2 loses something. Subtracting the old output removes the old square, leaving the pieces due to the input change. If every remaining term contains h, write each as h times something and divide out that common multiplier. A number in front multiplies every piece. The final rate can depend on where you start and how far you step.
For positive lengths, the square has two xh strips; the algebraic identity remains valid for negative x or h.
Reminder
Dividing powers.h2 ÷ h = (h · h) ÷ h = h for h ≠ 0. One of the two h factors is removed.
Distributing a multiplier. 3(x2 + 2xh + h2) = 3x2 + 6xh + 3h2. The 3 reaches every term.
Subtracting a whole expression. −(3x2 − 4x + 1) = −3x2 + 4x − 1. The outside −1 changes all three signs.
Why it works. Multiplying (x + h)(x + h) gives x2 + xh + hx + h2 = x2 + 2xh + h2. Subtracting x2 leaves h(2x + h), so dividing by h gives 2x + h. For a fully expanded polynomial, the terms independent of h reproduce the original f(x) and cancel when it is subtracted. This is a polynomial check, not a universal description of all functions. The rate varies on a curve: for 3x2 − 4x + 1, steps of 1 beginning at 1 and at 2 give different rates, 5 and 11.
Rule(x + h)2 = x2 + 2xh + h2. For f(x) = x2, the difference quotient is 2x + h, with h ≠ 0. Multiply any outside coefficient into every expanded term, subtract all of f(x), then factor the whole numerator before canceling h.
The same idea, five ways
Say it
Say 'multiply out the square, subtract the old output, factor the step, then divide'.
Write it
The extra square pieces produce the output change, and dividing by the input step gives the average rate.
In math
(x + h)2 = x2 + 2xh + h2
For f(x) = x2: f(x + h) − f(x)h = 2x + h, h ≠ 0
[x, x + h] if h > 0; [x + h, x] if h < 0
Like
A square carpet grows by two strips and a corner, then its added area is compared with the side-length change.
See itFor positive lengths, the square has two xh strips; the algebraic identity remains valid for negative x or h.
The same idea, other ways
Four pieces in the square
The old x2 square, two xh strips, and h2 corner account for the entire new square. The picture describes positive lengths; the algebraic identity still holds when x or h is negative.
For positive lengths, the square has two xh strips; the algebraic identity remains valid for negative x or h.
Follow the matching pairs
For 3x2 − 4x + 1, subtracting the original output creates +3x2 with −3x2, −4x with +4x, and +1 with −1. Each pair adds to 0, leaving 6xh + 3h2 − 4h.
+3x2 and −3x2 cancel
−4x and +4x cancel
+1 and −1 cancel
Subtracting the complete starting output removes its three matching pieces.
Give the letters numerical meaning
For the harder function, from 1 to 2 the rate is 5 − 01 = 5. From 2 to 3 it is 16 − 51 = 11. The same step h = 1 gives a different rate because the curve is steeper in a different place.
The curve represents 3x2 − 4x + 1. From 1 to 2, the dashed ruler has slope 5 = 6(1) + 3(1) − 4; it is not part of f.
.1A coefficient outside the square
Multiplying a square by a number means multiplying its entire area by that number. It does not mean moving that number inside the square and squaring it too.
5(x + h)2 = 5x2 + 10xh + 5h2.
The outside coefficient acts after the square is computed.
For x = 1 and h = 2, the dashed ruler through two outputs of 5x2 has slope 20.
Worked exampleRung 2: a coefficient multiplies every square piece
This asks for the same square rate when every output is multiplied by 5. Find the difference quotient of f(x) = 5x2.
For x = 1 and h = 2, the dashed ruler through two outputs of 5x2 has slope 20.
5(x + h)2
= 5x2 + 10xh + 5h2
rate = 10x + 5h
Square first; the outside coefficient multiplies every resulting piece.For x = 1 and h = 2, the dashed ruler through two outputs of 5x2 has slope 20.
f(x + h) = 5(x + h)2 = 5(x2 + 2xh + h2).Multiply out the square before applying its outside coefficient 5.
f(x + h) = 5x2 + 10xh + 5h2.The 5 multiplies all three terms, including the middle term.
Output change: (5x2 + 10xh + 5h2) − (5x2) = 10xh + 5h2.Subtract the whole original output; 5x2 cancels with −5x2.
10xh + 5h2 = h(10x + 5h).Write 10xh = h · 10x and 5h2 = h · 5h.
h(10x + 5h)h = 10x + 5h, with h ≠ 0.Cancel the nonzero multiplier of the whole numerator.
Answer
10x + 5h, with h ≠ 0.
Check With x = 1 and h = 2, f(3) = 45 and f(1) = 5. The direct rate is 45 − 52 = 402 = 20, and 10(1) + 5(2) = 20.
.2A square plus a linear term
Each appearance of x needs the new input. The square gives 2xh + h2 after subtraction; the extra linear term gives another 2h.
For f(x) = x2 + 2x, rate = 2x + h + 2, with h ≠ 0.
For x = 2 and h = 3, the dashed ruler joins outputs 8 and 35 with slope 9.
Worked exampleDifference quotient of f(x) = 3x2 − 6x and a secant slope
Let f(x) = 3x2 − 6x. The graph of y = 3x2 − 6x passes through (0, 0) and (2, 0), and its lowest point is (1, −3). (a) Find and simplify the difference quotient f(x + h) − f(x)h for h ≠ 0. (b) Use your result with x = 1 and h = 1 to find the slope of the secant line through (1, −3) and (2, 0).
The parabola y = 3x2 − 6x = 3(x − 1)2 − 3 with lowest point (1, −3) and the point (2, 0). The secant line through these two points has slope 3.
f(x + h) = 3(x + h)2 − 6(x + h) = 3(x2 + 2xh + h2) − 6x − 6h = 3x2 + 6xh + 3h2 − 6x − 6hEvery x is replaced by (x + h). The square expands as x2 + 2xh + h2, and the outside multipliers 3 and −6 must reach every term inside their parentheses.
f(x + h) − f(x) = (3x2 + 6xh + 3h2 − 6x − 6h) − (3x2 − 6x) = 3x2 + 6xh + 3h2 − 6x − 6h − 3x2 + 6xAll of f(x) is kept in parentheses after the minus, so both of its terms change sign.
Combine terms: 3x2 − 3x2 = 0 and −6x + 6x = 0, leaving 6xh + 3h2 − 6hThe terms that do not contain h cancel, as they always do in a difference quotient.
6xh + 3h2 − 6hh = h(6x + 3h − 6)h = 6x + 3h − 6, for h ≠ 0Every remaining term contains h, so h factors out of the whole numerator. It cancels with the denominator only when h ≠ 0.
For part (b), substitute x = 1 and h = 1: 6(1) + 3(1) − 6 = 3With x = 1 and h = 1 the difference quotient is the slope between the inputs 1 and 2, and these are the points (1, −3) and (2, 0).
Answer
(a) f(x + h) − f(x)h = 6x + 3h − 6, for h ≠ 0. (b) The secant slope through (1, −3) and (2, 0) is 3.
Check Compute directly: f(1) = 3 − 6 = −3 and f(2) = 12 − 12 = 0, so 0 − (−3)2 − 1 = 3, which matches. A second check uses x = 2 and h = 1: f(3) = 27 − 18 = 9, so 9 − 01 = 9, and the formula gives 12 + 3 − 6 = 9.
Work to write
f(x + h) = 3x2 + 6xh + 3h2 − 6x − 6h
f(x + h) − f(x) = 6xh + 3h2 − 6h
h(6x + 3h − 6)h = 6x + 3h − 6, h ≠ 0
x = 1, h = 1: 6 + 3 − 6 = 3
Secant slope = 3
(a) f(x + h) − f(x)h = 6x + 3h − 6, for h ≠ 0. (b) The secant slope through (1, −3) and (2, 0) is 3.
.3A three-term quadratic
A three-term function needs every substitution and every subtraction sign. The constant term disappears in the difference, while the linear term leaves a multiple of h.
For f(x) = 3x2 − 4x + 1, rate = 6x + 3h − 4, with h ≠ 0.
6xh = h · 6x; 3h2 = h · 3h; −4h = h · (−4).
The curve represents 3x2 − 4x + 1. From 1 to 2, the dashed ruler has slope 5 = 6(1) + 3(1) − 4; it is not part of f.
Worked exampleDifference quotient with a number in front of x2
This asks for the average-rate formula for a quadratic with three terms. Find f(x + h) − f(x)h for f(x) = 3x2 − 4x + 1.
The curve represents 3x2 − 4x + 1. From 1 to 2, the dashed ruler has slope 5 = 6(1) + 3(1) − 4; it is not part of f.
f(x + h) = 3(x + h)2 − 4(x + h) + 1.The new input replaces every x position, including the linear term.
f(x + h) = 3(x2 + 2xh + h2) − 4x − 4h + 1.Multiply out the square first, then let −4 multiply both terms of x + h.
Output change: (3x2 + 6xh + 3h2 − 4x − 4h + 1) − (3x2 − 4x + 1).The whole original output stays in parentheses until its minus sign is distributed.
= 3x2 + 6xh + 3h2 − 4x − 4h + 1 − 3x2 + 4x − 1.The outside −1 changes +3x2 to −3x2, −4x to +4x, and +1 to −1.
= 6xh + 3h2 − 4h.The explicit matching pairs are +3x2 with −3x2, −4x with +4x, and +1 with −1.
6xh = h · 6x, 3h2 = h · 3h, and −4h = h · (−4). Therefore the output change is h(6x + 3h − 4).Each remaining term contains the same multiplier h.
h(6x + 3h − 4)h = 6x + 3h − 4, with h ≠ 0.A nonzero h multiplying the whole top cancels with h on the bottom.
Answer
6x + 3h − 4, with h ≠ 0.
Check Use x = 2 and h = 3. f(5) = 75 − 20 + 1 = 56 and f(2) = 12 − 8 + 1 = 5, so 56 − 53 = 513 = 17. The result gives 6(2) + 3(3) − 4 = 17. The wrong expression 6x + 3h2 − 4 gives 35, so this check catches the power error.
.4A negative squared term
The minus outside x2 negates the entire square after it has been multiplied out. Subtracting the starting negative square then adds a matching positive x2.
−(x + h)2 = −x2 − 2xh − h2.
For −x2 + 6x, rate = −2x − h + 6, with h ≠ 0.
For x = 2 and h = 3, the endpoints give average slope −1 even though the curve turns between them.
Worked exampleRung 5: a negative square coefficient
This asks for the rate formula when a minus sign is outside the square. Find the difference quotient of f(x) = −x2 + 6x.
For x = 2 and h = 3, the endpoints give average slope −1 even though the curve turns between them.
−(x + h)2
= −x2 − 2xh − h2
rate = −2x − h + 6
The outside minus changes every square term's sign.For x = 2 and h = 3, the endpoints give average slope −1 even though the curve turns between them.
f(x + h) = −(x + h)2 + 6(x + h).The new input stays inside the square; the outside minus remains outside.
= −(x2 + 2xh + h2) + 6x + 6h = −x2 − 2xh − h2 + 6x + 6h.The outside minus reaches all three square terms, while 6 multiplies both input pieces.
Output change: (−x2 − 2xh − h2 + 6x + 6h) − (−x2 + 6x).Subtract the whole original output in parentheses.
= −x2 − 2xh − h2 + 6x + 6h + x2 − 6x.Multiplying the starting expression by −1 changes −x2 to +x2 and +6x to −6x.
= −2xh − h2 + 6h = h(−2x − h + 6).The −x2 and +x2 cancel; 6x and −6x cancel. The remaining factors after taking out h are −2x, −h, and 6.
h(−2x − h + 6)h = −2x − h + 6, with h ≠ 0.The entire numerator has a nonzero factor h that divides out.
Answer
−2x − h + 6, with h ≠ 0.
Check At x = 2 and h = 3, f(5) = −25 + 30 = 5 and f(2) = −4 + 12 = 8. The direct rate 5 − 83 = −1 matches −2(2) − 3 + 6 = −1.
Strategy: step by step
1. Plug in: replace every x by (x + h), multiply out the square, and apply each outside multiplier.
2. Subtract: keep all of f(x) in parentheses after a minus, then flip every term's sign.
3. Simplify: combine matching polynomial terms; the terms independent of h cancel.
4. Cancel h: write every remaining term as h times something, factor h from the whole numerator, and cancel only for h ≠ 0.
Strategy
The quadratic difference quotient in four moves
1
Is there a number in front of the square?
YesSquare the whole input first, then multiply every resulting term by that number.
NoUse x2 + 2xh + h2.
↓
2
Is there a minus outside the square?
YesAfter expanding, change every square term's sign.
NoKeep the signs from the square expansion.
↓
3
Have all original-output signs been reversed in the subtraction?
YesCombine the matching terms.
NoRewrite −(f(x)) before removing the parentheses.
↓
4
In this fully expanded polynomial, does a term without h remain?
YesRecheck the expansion and subtraction; the copy of f(x) should cancel.
NoWrite each term as h times something.
↓
5
Does the whole numerator now have a factor h?
YesCancel it for h ≠ 0 and interpret the average rate.
NoRecheck the algebra rather than canceling selected pieces.
Plug in and multiply the square into four products.
Subtract the complete original formula with parentheses.
Simplify by grouping like terms.
Factor a whole nonzero h and keep the original restriction.
Worked exampleDifference quotient for a garden plot with a compost corner
A rectangular garden plot is x meters wide and 2x meters long. A 2 square-meter corner is set aside for a compost bin, so the planted area is A(x) = 2x2 − 2 square meters. The graph of y = 2x2 − 2 passes through (2, 6) and (3, 16), and its lowest point is (0, −2). (a) Find and simplify the difference quotient A(x + h) − A(x)h for h ≠ 0. (b) Use your result with x = 2 and h = 1 to find the average rate of change of the planted area as the width grows from 2 m to 3 m.
Graph of the planted area A(x) = 2x2 − 2, with lowest point (0, −2) and the points (2, 6) and (3, 16) that mark the interval used in part (b).
A(x + h) = 2(x + h)2 − 2 = 2(x2 + 2xh + h2) − 2 = 2x2 + 4xh + 2h2 − 2Replace every x by (x + h), expand with (x + h)2 = x2 + 2xh + h2, and multiply the outside coefficient 2 into every expanded term.
A(x + h) − A(x) = (2x2 + 4xh + 2h2 − 2) − (2x2 − 2) = 2x2 + 4xh + 2h2 − 2 − 2x2 + 2Keep all of A(x) in parentheses after the minus, then flip the sign of each of its terms. The −2 becomes +2.
2x2 − 2x2 = 0 and −2 + 2 = 0, so the numerator is 4xh + 2h2.Combine matching terms. The terms that do not contain h cancel.
4xh + 2h2h = h(4x + 2h)h = 4x + 2h, for h ≠ 0Every remaining term contains h. Factor h from the whole numerator, then cancel it, which is allowed only because h ≠ 0.
With x = 2 and h = 1: 4(2) + 2(1) = 8 + 2 = 10Substitute into the simplified difference quotient. The result is the average rate of change of A on [2, 3].
Answer
(a) A(x + h) − A(x)h = 4x + 2h, for h ≠ 0. (b) 10 square meters per meter. The planted area grows by 10 m2 for each meter of added width, on average, from 2 m to 3 m.
Check From the graph points, A(2) = 2(4) − 2 = 6 and A(3) = 2(9) − 2 = 16. So 16 − 63 − 2 = 10, which matches 4(2) + 2(1) = 10.
(a) A(x + h) − A(x)h = 4x + 2h, for h ≠ 0. (b) 10 square meters per meter. The planted area grows by 10 m2 for each meter of added width, on average, from 2 m to 3 m.
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 for the area of a strip of five square tiles
A decorative border is a straight strip of 5 identical square tiles placed side by side. If each tile has side length x inches, the total area of the strip is A(x) = 5x2 square inches. (a) Find and simplify the difference quotient A(x + h) − A(x)h for h ≠ 0. (b) Use your result to find the average rate of change of the area when the side length grows from x = 4 inches to x = 4.2 inches, so h = 0.2.
Graph of the strip's area A(x) = 5x2, marking the point (4, 80): tiles 4 inches on a side give a total area of 80 square inches.
Plug in: A(x + h) = 5(x + h)2 = 5(x2 + 2xh + h2) = 5x2 + 10xh + 5h2.Every x is replaced by (x + h). The square expands as (x + h)2 = x2 + 2xh + h2, and the outside coefficient 5 must multiply every expanded term.
Subtract: A(x + h) − A(x) = (5x2 + 10xh + 5h2) − (5x2) = 5x2 + 10xh + 5h2 − 5x2.All of A(x) goes in parentheses after the minus sign, and the sign of each term inside is then flipped.
Simplify: 5x2 − 5x2 = 0, so the numerator is 10xh + 5h2.Matching terms are combined. The terms with no h cancel, and every remaining term contains h.
Cancel h: 10xh + 5h2h = h(10x + 5h)h = 10x + 5h, for h ≠ 0.Write each term as h times something and factor h from the whole numerator. The common factor h may be canceled only because h ≠ 0.
Evaluate at x = 4 and h = 0.2: 10(4) + 5(0.2) = 40 + 1 = 41.The simplified difference quotient is the average rate of change from x to x + h, so the numbers are substituted directly.
Answer
(a) A(x + h) − A(x)h = 10x + 5h, for h ≠ 0. (b) 41 square inches of area per inch of side length.
Check Compute the change in area directly. A(4.2) = 5(4.2)2 = 5(17.64) = 88.2, and A(4) = 5(16) = 80. Then 88.2 − 800.2 = 8.20.2 = 41, which matches 10(4) + 5(0.2) = 41.
(a) A(x + h) − A(x)h = 10x + 5h, for h ≠ 0. (b) 41 square inches of area per inch of side length.
Rung 2Rung 2: a coefficient multiplies every square piece
This asks for the same square rate when every output is multiplied by 5. Find the difference quotient of f(x) = 5x2.
For x = 1 and h = 2, the dashed ruler through two outputs of 5x2 has slope 20.
5(x + h)2
= 5x2 + 10xh + 5h2
rate = 10x + 5h
Square first; the outside coefficient multiplies every resulting piece.For x = 1 and h = 2, the dashed ruler through two outputs of 5x2 has slope 20.
f(x + h) = 5(x + h)2 = 5(x2 + 2xh + h2).Multiply out the square before applying its outside coefficient 5.
f(x + h) = 5x2 + 10xh + 5h2.The 5 multiplies all three terms, including the middle term.
Output change: (5x2 + 10xh + 5h2) − (5x2) = 10xh + 5h2.Subtract the whole original output; 5x2 cancels with −5x2.
10xh + 5h2 = h(10x + 5h).Write 10xh = h · 10x and 5h2 = h · 5h.
h(10x + 5h)h = 10x + 5h, with h ≠ 0.Cancel the nonzero multiplier of the whole numerator.
Answer
10x + 5h, with h ≠ 0.
Check With x = 1 and h = 2, f(3) = 45 and f(1) = 5. The direct rate is 45 − 52 = 402 = 20, and 10(1) + 5(2) = 20.
Rung 3Difference quotient of f(x) = 2x2 − 4x
Let f(x) = 2x2 − 4x. The graph of y = 2x2 − 4x passes through (0, 0) and (3, 6), and its lowest point is (1, −2). (a) Find and simplify the difference quotient f(x + h) − f(x)h for h ≠ 0. (b) Use your result with x = 0 and h = 3 to find the slope of the secant line through (0, 0) and (3, 6).
Parabola y = 2x2 − 4x = 2(x − 1)2 − 2 with vertex (1, −2). The secant line through (0, 0) and (3, 6) has slope 2.
Replace every x by (x + h): f(x + h) = 2(x + h)2 − 4(x + h).f(x + h) means the input x + h goes into each place x appears, in both the square term and the linear term.
Expand the square: (x + h)2 = x2 + 2xh + h2, so 2(x + h)2 = 2x2 + 4xh + 2h2.The outside coefficient 2 multiplies every term of the expanded square, not just the first one.
Distribute the −4: −4(x + h) = −4x − 4h. So f(x + h) = 2x2 + 4xh + 2h2 − 4x − 4h.The multiplier −4 applies to both x and h.
Subtract all of f(x) in parentheses: f(x + h) − f(x) = (2x2 + 4xh + 2h2 − 4x − 4h) − (2x2 − 4x) = 2x2 + 4xh + 2h2 − 4x − 4h − 2x2 + 4x.The minus sign applies to every term of f(x), so −4x becomes +4x.
Combine like terms: 2x2 − 2x2 = 0 and −4x + 4x = 0, leaving 4xh + 2h2 − 4h.The terms that do not contain h cancel. Every remaining term should contain h.
Factor h from the whole numerator: 4xh + 2h2 − 4h = h(4x + 2h − 4).Each term has a factor of h. Only a factor of the entire numerator can be canceled.
Divide: h(4x + 2h − 4)h = 4x + 2h − 4, for h ≠ 0.h divided by h equals 1 only when h is not zero.
For part (b), substitute x = 0 and h = 3: 4(0) + 2(3) − 4 = 0 + 6 − 4 = 2.The difference quotient with x = 0 and h = 3 is the slope from the point at x = 0 to the point at x = 0 + 3 = 3.
Answer
(a) f(x + h) − f(x)h = 4x + 2h − 4, for h ≠ 0. (b) The secant slope through (0, 0) and (3, 6) is 2.
Check Compute the slope directly. f(0) = 2(0)2 − 4(0) = 0 and f(3) = 2(9) − 12 = 6. Then 6 − 03 − 0 = 63 = 2, which matches 4(0) + 2(3) − 4 = 2. A second test with x = 1 and h = 1: f(2) = 8 − 8 = 0 and f(1) = 2 − 4 = −2, so 0 − (−2)1 = 2. The formula gives 4(1) + 2(1) − 4 = 2. ✓
Output change: (3x2 + 6xh + 3h2 − 4x − 4h + 1) − (3x2 − 4x + 1).The whole original output stays in parentheses until its minus sign is distributed.
= 3x2 + 6xh + 3h2 − 4x − 4h + 1 − 3x2 + 4x − 1.The outside −1 changes +3x2 to −3x2, −4x to +4x, and +1 to −1.
= 6xh + 3h2 − 4h.The explicit matching pairs are +3x2 with −3x2, −4x with +4x, and +1 with −1.
6xh = h · 6x, 3h2 = h · 3h, and −4h = h · (−4). Therefore the output change is h(6x + 3h − 4).Each remaining term contains the same multiplier h.
h(6x + 3h − 4)h = 6x + 3h − 4, with h ≠ 0.A nonzero h multiplying the whole top cancels with h on the bottom.
Answer
6x + 3h − 4, with h ≠ 0.
Check Use x = 2 and h = 3. f(5) = 75 − 20 + 1 = 56 and f(2) = 12 − 8 + 1 = 5, so 56 − 53 = 513 = 17. The result gives 6(2) + 3(3) − 4 = 17. The wrong expression 6x + 3h2 − 4 gives 35, so this check catches the power error.
Rung 5Rung 5: a negative square coefficient
This asks for the rate formula when a minus sign is outside the square. Find the difference quotient of f(x) = −x2 + 6x.
For x = 2 and h = 3, the endpoints give average slope −1 even though the curve turns between them.
−(x + h)2
= −x2 − 2xh − h2
rate = −2x − h + 6
The outside minus changes every square term's sign.For x = 2 and h = 3, the endpoints give average slope −1 even though the curve turns between them.
f(x + h) = −(x + h)2 + 6(x + h).The new input stays inside the square; the outside minus remains outside.
= −(x2 + 2xh + h2) + 6x + 6h = −x2 − 2xh − h2 + 6x + 6h.The outside minus reaches all three square terms, while 6 multiplies both input pieces.
Output change: (−x2 − 2xh − h2 + 6x + 6h) − (−x2 + 6x).Subtract the whole original output in parentheses.
= −x2 − 2xh − h2 + 6x + 6h + x2 − 6x.Multiplying the starting expression by −1 changes −x2 to +x2 and +6x to −6x.
= −2xh − h2 + 6h = h(−2x − h + 6).The −x2 and +x2 cancel; 6x and −6x cancel. The remaining factors after taking out h are −2x, −h, and 6.
h(−2x − h + 6)h = −2x − h + 6, with h ≠ 0.The entire numerator has a nonzero factor h that divides out.
Answer
−2x − h + 6, with h ≠ 0.
Check At x = 2 and h = 3, f(5) = −25 + 30 = 5 and f(2) = −4 + 12 = 8. The direct rate 5 − 83 = −1 matches −2(2) − 3 + 6 = −1.
Rung 6Rung 6: a fresh three-term quadratic
This asks for the rate expression when a square coefficient, a negative linear coefficient, and a constant all occur. Find the difference quotient of f(x) = 6x2 − 7x + 2.
For x = 1 and h = 2, the dashed ruler connects outputs 1 and 35 with slope 17.
change = 12xh + 6h2 − 7h
= h(12x + 6h − 7)
rate = 12x + 6h − 7
Every term of this polynomial difference contains h after full simplification.For x = 1 and h = 2, the dashed ruler connects outputs 1 and 35 with slope 17.
f(x + h) = 6(x + h)2 − 7(x + h) + 2.Replace every x by the whole new input.
= 6(x2 + 2xh + h2) − 7x − 7h + 2 = 6x2 + 12xh + 6h2 − 7x − 7h + 2.The square expansion is multiplied term by term by 6, and −7 reaches both input pieces.
Output change: (6x2 + 12xh + 6h2 − 7x − 7h + 2) − (6x2 − 7x + 2).Keep all three original terms grouped for subtraction.
= 12xh + 6h2 − 7h = h(12x + 6h − 7).The matching h-independent polynomial terms cancel. Each remaining term is h times 12x, 6h, or −7.
h(12x + 6h − 7)h = 12x + 6h − 7, with h ≠ 0.Cancel the nonzero multiplier of the whole numerator.
Answer
12x + 6h − 7, with h ≠ 0.
Check At x = 1 and h = 2, f(1) = 6 − 7 + 2 = 1 and f(3) = 54 − 21 + 2 = 35. The direct rate 35 − 12 = 342 = 17 matches 12(1) + 6(2) − 7 = 17.
Rung 7Difference quotient of 4x2 with a negative step
Let f(x) = 4x2. The graph of y = 4x2 passes through (1, 4) and (2, 16). (a) Find and simplify the difference quotient f(x + h) − f(x)h for h ≠ 0. (b) Use your result with x = 2 and the negative step h = −1. This gives the slope of the secant line through (2, 16) and (1, 4).
The parabola y = 4x2 with the points (2, 16) and (1, 4). The step h = −1 moves from x = 2 back to x = 1, and the secant line through these two points has slope 12.
Plug in: f(x + h) = 4(x + h)2 = 4(x2 + 2xh + h2) = 4x2 + 8xh + 4h2.Every x is replaced by (x + h). The square expands as x2 + 2xh + h2, and the outside coefficient 4 multiplies every expanded term.
Subtract: f(x + h) − f(x) = (4x2 + 8xh + 4h2) − (4x2) = 4x2 + 8xh + 4h2 − 4x2.All of f(x) goes in parentheses after the minus sign, so its sign flips.
Simplify: 4x2 − 4x2 = 0, so the numerator is 8xh + 4h2.The terms that do not contain h cancel, which always happens in a difference quotient.
Cancel h: 8xh + 4h2h = h(8x + 4h)h = 8x + 4h, for h ≠ 0.Factor h from the whole numerator first, then cancel it. The cancellation is allowed only because h ≠ 0.
Use the negative step: with x = 2 and h = −1, 8x + 4h = 8(2) + 4(−1) = 16 − 4 = 12.The formula holds for any nonzero h, negative values included. Here x + h = 1, so the result is the slope between x = 2 and x = 1.
Answer
(a) f(x + h) − f(x)h = 8x + 4h, h ≠ 0. (b) With x = 2 and h = −1, the value is 12.
Check Compute directly: f(1) = 4(1)2 = 4 and f(2) = 4(2)2 = 16. Then f(1) − f(2)−1 = 4 − 16−1 = −12−1 = 12. This matches 8(2) + 4(−1) = 12. As a second check, the pattern 2x + h for x2, multiplied by 4, gives 8x + 4h.
