Everything in this readiness chapter has been leading here. The difference quotient is the one formula every idea in the next chapter is built directly on top of, calculus's central idea of instantaneous change eventually included. So this section is less "one more topic to refresh" and more "make sure the launchpad is solid."
h just means "how wide is the interval you're averaging over." It's one letter instead of subtracting two endpoints every time. Simplifying the difference quotient means clearing that h out of the denominator before you'd ever plug a number in. That matters because h is about to become the denominator of an expression whose behavior as h shrinks toward 0 is exactly what the next chapter studies. That expression needs to survive substituting h = 0 without instantly dividing by zero. Every algebra step in this section exists to make that substitution safe.
The difference quotient of a function f at x is \frac{f(x+h)-f(x)}{h}, the same average-rate-of-change quantity College Algebra called \frac{f(b)-f(a)}{b-a}. It's just written with b = x+h and a = x, so the interval's width, h, appears explicitly as its own symbol instead of being buried in two endpoints. Simplifying it always follows the same three moves. Expand f(x+h). Subtract f(x) and collect like terms so every remaining term has a factor of h. Then divide that factor of h out. What's left is a formula in x and h with no h left in the denominator. That is the form the next chapter needs. The next chapter's central question is what happens as h shrinks all the way to 0, and that question can't be answered from the original \frac{f(x+h)-f(x)}{h} form at all. Plugging in h=0 there gives \frac{0}{0}, undefined, before the algebra clears that factor of h out of the denominator.
Formula
Average Rate of Change
Recalled from College Algebra · Average rate of change — first taught there.
The average rate of change of f on [a, b] is \frac{f(b) - f(a)}{b - a}, the slope of the secant line through (a, f(a)) and (b, f(b)).
Formula
The Difference Quotient
Recalled from College Algebra · Complex fractions and the difference quotient — first taught there.
For a function f and h \neq 0, \frac{f(x+h) - f(x)}{h}. Geometrically, this is the slope of the line through (x, f(x)) and (x+h, f(x+h)), the algebraic starting point for the derivative in calculus.
Shrink the interval
Drag Q toward P — watch the secant's slope (the difference quotient) settle down as h shrinks toward 0.
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Try it on a steeper curve
Drag Q toward P here too — the settling behavior is the same, even though the curve is steeper.
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Worked example
Simplifying a difference quotient
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1
Simplify \dfrac{f(x+h)-f(x)}{h} for f(x) = x^2 + 3x.
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2
Expand f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h.
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3every remaining term has a factor of h, and that's not a coincidence
Subtract f(x) = x^2 + 3x: \big(x^2+2xh+h^2+3x+3h\big) - \big(x^2+3x\big) = 2xh + h^2 + 3h.
every remaining term has a factor of h, and that's not a coincidence
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4
Factor out h: \dfrac{h(2x+h+3)}{h}, then cancel it (valid since h \neq 0 in a difference quotient): 2x + h + 3.
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Build the secant line
Drag two points to construct the secant line through (1, f(1)) and (3, f(3)) for f(x) = x^2 + 3x -- the same secant this section's difference quotient measures.
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Where this goes next
The very next section
Set f(x) = x^2 + 3x and let h shrink toward 0 in the answer just found, 2x+h+3. It approaches 2x+3, with no algebra left to do and no division by zero in sight. That number, 2x+3, is the exact quantity the next chapter names and studies in full. It opens by asking what "let h shrink toward 0" means precisely enough to compute with.
Check your understanding (counts toward your progress)
Why must every remaining term in the numerator carry a factor of h before dividing?
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This isn't a stylistic convention -- it's a direct algebraic consequence of f(x+h)-f(x) vanishing at h=0.
It matters a great deal -- dividing by h is only valid once every term genuinely contains an h factor to cancel, not before.
This property holds for any function's difference quotient, not just quadratics -- it follows from f(x+h)-f(x)=0 whenever h=0, regardless of f's shape.
Why
Setting h=0 makes f(x+h)-f(x) collapse to f(x)-f(x)=0 for any f, which is only possible if every term in the expanded numerator carries at least one factor of h.
Simplify the difference quotient \dfrac{f(x+h)-f(x)}{h} for f(x) = -4x^2-3x.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
f(x+h)-f(x) = -4(2xh+h^2)-3h = h(-8x+-4h-3). Dividing by h leaves -8x+-4h-3, with no h left in the denominator.
Correct answer
Also accepted:
Why
Practice: Simplifying the difference quotient (3 problems · doesn’t affect your progress)
For f(x) = 3x^2, simplify the difference quotient, then evaluate it at x=2, h=0.1.
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Why
f(x+h)-f(x)=3(x+h)^2-3x^2=6xh+3h^2=h(6x+3h), so the simplified quotient is 6x+3h. At x=2, h=0.1: 6(2)+3(0.1)=12+0.3=12.3.
Correct answer
Also accepted:
Why
The simplified difference quotient for f(x)=3x^2 is 6x+3h -- substitute x=2 and h=0.1 into that.
In diffq-2's result, 2x+h-4, what expression does it approach as h\to0?
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Setting h=0 in 2x+h-4 leaves 2x-4 -- the value the difference quotient settles toward as the interval shrinks to nothing.
Correct answer
Also accepted:
Why
Substitute h=0 directly into the simplified difference quotient 2x+h-4.
Compare the difference quotient of a linear function f(x)=mx+b to that of a quadratic. Which one comes out with no h left in it, even before taking any limit?
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A quadratic's simplified difference quotient always keeps an h term (like 2x+h+5 in diffq-1) -- it only vanishes once you separately let h\to0.
Only the linear case drops h entirely on its own -- the quadratic case keeps an h term until a limit is taken.
The quadratic case is the one that keeps h in its simplified form -- diffq-1 and diffq-2 above both still carry an h term.
Why
For f(x)=mx+b: f(x+h)-f(x)=m(x+h)+b-(mx+b)=mh, so the quotient is \dfrac{mh}{h}=m -- constant, with h already gone. A quadratic's quotient, like diffq-1's 2x+h+5, still carries an h term until a limit is taken.
Simplify f(x+h)-f(x) over h for f(x)=mx+b directly, the same way diffq-1 and diffq-2 did for a quadratic.
The linear function's difference quotient simplifies to exactly m, with no h remaining at all
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Simplify the difference quotient \dfrac{f(x+h)-f(x)}{h} for f(x)=\dfrac{3}{x} (assume all quantities are defined).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
f(x+h)-f(x)=\frac{3}{x+h}-\frac3x=\frac{3x-3(x+h)}{x(x+h)}=\frac{-3h}{x(x+h)}. Dividing by h cancels the h: -\frac{3}{x(x+h)}.
Correct answer
Also accepted:
Why
Combine f(x+h) and f(x) over a common denominator, x(x+h), before subtracting.
From College Algebra
Drag the widget's points so the secant runs from x=1 to x=1.5, then read off the average rate of change of f(x) = 1/x over that interval.
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What this task checks
Structure checked automatically. Whether your ideas are convincing isn't graded, so reread with the checklist.
The compiler says
A model answer
Why
\dfrac{f(1.5)-f(1)}{0.5} = \dfrac{\frac{1}{1.5}-1}{0.5} = \dfrac{-\frac13}{0.5} \approx -0.6667 — matching the simplified formula -\dfrac{1}{x(x+h)} = -\dfrac{1}{1(1.5)} \approx -0.6667 derived earlier in this section.
Correct answer
Also accepted:
Why
The average rate of change is \dfrac{f(x+h)-f(x)}{h} — the secant's rise over run.