Quadratic Approximation: Formula, Error, Examples

What Is Quadratic Approximation?

Quadratic approximation is the estimate of a function $f$ near a point $x = a$ by the unique parabola that has the same value, the same first derivative and the same second derivative as $f$ at $a$. It is the tangent line from the tangent line equation plus one correction term built from the second derivative, and it is the degree 2 member of the family of Taylor polynomials.

The formula is

$$Q(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2, \qquad x \text{ near } a.$$

Every symbol has one job:

Two conditions matter. The function must be twice differentiable at $a$, so that $f''(a)$ exists, and $x$ must be close to $a$. "Close" has no fixed size; it depends on how fast the third derivative is changing, which the error section below makes precise.

How Do You Find A Quadratic Approximation Step By Step?

Finding a quadratic approximation takes four steps, and the order matters only in the last one.

1. Pick the base point. Choose $a$ close to the input you care about and where $f(a)$ is easy to compute exactly. For $\sqrt{9.5}$, choose $a = 9$, not $a = 10$. Actually, $a = 9.61$ (which is $3.1^2$) sits even closer, but $a = 9$ keeps the arithmetic in whole numbers and the error is already tiny.

2. Compute three numbers. Evaluate $f(a)$, $f'(a)$ and $f''(a)$.

3. Build $Q(x)$. Substitute the three numbers into $Q(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2$.

4. Evaluate at the step. Put in the actual $x$, so the step is $h = x - a$, and compute.

Written with the step $h$, the same formula reads

$$f(a + h) \approx f(a) + f'(a),h + \frac{f''(a)}{2},h^2.$$

This form shows the structure at a glance: a constant, a term proportional to $h$, and a term proportional to $h^2$. When $h$ is small, $h^2$ is much smaller, which is exactly why the new term is a correction rather than a rewrite.

Where Does The 1/2 In The Quadratic Approximation Formula Come From?

The $\frac{1}{2}$ is there so that the parabola has the same second derivative as $f$. Without it, the parabola would bend twice as hard as the curve.

Test the formula on a function that is already a parabola. Take $f(x) = c_0 + c_1 x + c_2 x^2$ with base point $a = 0$:

$$f(0) = c_0$$

$$f'(x) = c_1 + 2c_2 x \quad\Longrightarrow\quad f'(0) = c_1$$

$$f''(x) = 2c_2 \quad\Longrightarrow\quad c_2 = \frac{f''(0)}{2}$$

So the coefficient of $x^2$ is half the second derivative. That 2 appears because differentiating $x^2$ twice gives 2, and the $\frac{1}{2}$ cancels it. A good approximation of a parabola should be the parabola itself, and with the $\frac{1}{2}$ in place it is.

Check it the other way. Differentiate $Q(x)$ twice:

$$Q'(x) = f'(a) + f''(a)(x - a)$$

$$Q''(x) = f''(a)$$

At $x = a$, the values give $Q(a) = f(a)$, $Q'(a) = f'(a)$ and $Q''(a) = f''(a)$. Three matching conditions pin down exactly one parabola, and that parabola is $Q$. (For a degree $n$ polynomial the same argument produces a denominator of $n$ factorial, which is why the Taylor coefficients carry 2, 6, 24 and so on.)

How Is Quadratic Approximation Different From Linear Approximation?

Linear approximation keeps the value and the slope; quadratic approximation also keeps the bend. Linear approximation is $L(x) = f(a) + f'(a)(x - a)$, so

$$Q(x) = L(x) + \frac{f''(a)}{2}(x - a)^2.$$

The added term is positive when $f''(a) > 0$ (the curve bends up, so the tangent line sits below it) and negative when $f''(a) < 0$. The quadratic term therefore corrects the direction in which the tangent line is wrong.

The table compares both estimates on four functions, each with a step of 0.5 or less. Every value was recomputed in SymPy and rounded to 4 decimal places.

In every row the quadratic error is roughly 8 to 300 times smaller. The cosine row is the most telling: the tangent line at 0 is the flat line $y = 1$, which knows nothing about the curve falling away, while $1 - \frac{x^2}{2}$ captures it.

The price is one more derivative and one more term. For hand estimates, quadratic approximation is usually the sweet spot; going to degree 3 or 4 adds accuracy but the arithmetic quickly stops being something you would do on paper.

What Does Quadratic Approximation Look Like On A Graph?

On a graph, the quadratic approximation is the best-fit parabola at the base point: it touches the curve, leaves at the same angle and curls the same way. The tangent line only touches and leaves at the same angle.

A classic homework request is to find the local quadratic approximation of $\cos x$ at $x = 0$. The three numbers are $f(0) = 1$, $f'(0) = -\sin 0 = 0$ and $f''(0) = -\cos 0 = -1$, so

$$\cos x \approx 1 - \frac{x^2}{2}.$$

The graph makes the difference obvious. The tangent line $y = 1$ runs flat across the top of the cosine hump, while the parabola $1 - \frac{x^2}{2}$ opens downward and hugs the hump for a good distance on both sides.

The sign of $f''(a)$ also decides what the base point is. With $f'(a) = 0$, a positive $f''(a)$ makes $Q$ an upward parabola, so $a$ is a local minimum; a negative one makes it a downward parabola, so $a$ is a local maximum. That is the second derivative test seen through the quadratic approximation: the test reads the shape of the best-fit parabola.

What Are The Standard Quadratic Approximations Near Zero?

Near $a = 0$ the formula simplifies to $f(x) \approx f(0) + f'(0),x + \frac{f''(0)}{2}x^2$, and a handful of results come up so often that they are worth knowing on sight. Each one is the start of the function's Maclaurin series, cut after the $x^2$ term.

The last row is the start of the binomial series, and it contains several others as special cases ($r = \frac{1}{2}$ gives the square-root row, $r = -1$ with $x$ replaced by $-x$ gives the $\frac{1}{1 - x}$ row).

Notice that $\sin x$ and $\tan x$ get no $x^2$ term. Both are odd functions, so every even derivative at 0 is zero; their quadratic approximation equals their linear approximation, and the first real correction is cubic.

How Accurate Is Quadratic Approximation?

The error of a quadratic approximation shrinks like the cube of the step: halve the distance from the base point and the error drops by a factor of about 8. That is the precise sense in which it beats the tangent line, whose error shrinks only like the square.

Taylor's theorem gives the exact statement. If $f$ has a continuous third derivative between $a$ and $x$, then

$$f(x) - Q(x) = \frac{f'''(c)}{6}(x - a)^3$$

for some number $c$ between $a$ and $x$. This is the Lagrange form of the remainder. You rarely know $c$, but if $|f'''(t)| \leq M$ for every $t$ between $a$ and $x$, then

$$|f(x) - Q(x)| \leq \frac{M}{6},|x - a|^3.$$

Apply it to $e^{0.1}$ with $a = 0$. Every derivative of $e^x$ is $e^x$, which is at most $e^{0.1}$ on $[0, 0.1]$, so

$$|e^{0.1} - 1.105| \leq \frac{e^{0.1}}{6}(0.1)^3 \approx 0.000184.$$

The actual error is $0.000171$, inside the bound, as it must be.

The cube law is easy to see in a table for $e^x$ at $a = 0$:

From $h = 0.1$ to $h = 0.05$ the linear error falls by about 4 (the square law) and the quadratic error by about 8 (the cube law).

A related question students post: if you use $\sin x \approx x - \frac{x^3}{6} + \frac{x^5}{120}$ for $0 \leq x \leq 1.5$, what is the maximum error? The same reasoning applies one level up. That polynomial is also the degree 6 approximation (the $x^6$ coefficient is zero), so the remainder involves the seventh derivative, which is bounded by 1:

$$|\text{error}| \leq \frac{(1.5)^7}{5040} \approx 0.0034.$$

The actual worst error, at $x = 1.5$, is $0.0033$.

There is one honest warning. Far from the base point, the quadratic approximation can be worse than the linear one. For $\ln x$ at $a = 1$, estimating $\ln 3$ gives $L(3) = 2$ but $Q(3) = 2 - \frac{1}{2}(2)^2 = 0$, against a true value of $1.0986$. The extra term is only guaranteed to help when the step is small.

How Do You Find The Quadratic Approximation Of A Function Of Two Variables?

For a function $f(x, y)$ near $(a, b)$, the quadratic approximation keeps the value, both first partial derivatives and all three second partial derivatives. With $h = x - a$ and $k = y - b$:

$$Q(x, y) = f + f_x,h + f_y,k + \frac{1}{2}\left(f_{xx},h^2 + 2f_{xy},hk + f_{yy},k^2\right),$$

where every derivative is evaluated at $(a, b)$. The middle term carries a 2 because $f_{xy}$ and $f_{yx}$ are equal and both contribute. That 2 is the piece even experienced users of the formula second-guess, and writing the two mixed terms out separately once is the surest way to remember it. Geometrically, the tangent plane becomes a curved surface (a bowl, a dome or a saddle).

A common request is the quadratic approximation of $f(x, y) = e^x \cos y$ near the origin. The values at $(0, 0)$ are:

$$f = 1, \quad f_x = 1, \quad f_y = 0$$

$$f_{xx} = 1, \quad f_{xy} = 0, \quad f_{yy} = -1$$

So

$$Q(x, y) = 1 + x + \frac{x^2}{2} - \frac{y^2}{2}.$$

At $(0.1, 0.1)$ this gives $1.1$, against the true $e^{0.1}\cos 0.1 = 1.0996$. The surface is a saddle: it curves up in the $x$ direction and down in the $y$ direction.

Another posted question asks for the best quadratic approximation of $\sqrt{1 + 7x + y}$ near $(0, 0)$. The derivatives at the origin are $f = 1$, $f_x = \frac{7}{2}$, $f_y = \frac{1}{2}$, $f_{xx} = -\frac{49}{4}$, $f_{xy} = -\frac{7}{4}$ and $f_{yy} = -\frac{1}{4}$, so

$$Q(x, y) = 1 + \frac{7x}{2} + \frac{y}{2} - \frac{49x^2}{8} - \frac{7xy}{4} - \frac{y^2}{8}.$$

At $(0.01, 0.02)$ it gives $1.0440$, matching the true value $1.0440$ to 4 decimal places.

The three second partials form a 2 by 2 table called the Hessian matrix, which is where multivariable optimization picks up the story.

Examples Of Quadratic Approximation

Example 1

Use the quadratic approximation of $e^x$ at $a = 0$ to estimate $e^{0.1}$.

$$f(0) = 1, \quad f'(0) = 1, \quad f''(0) = 1$$

$$Q(x) = 1 + x + \frac{x^2}{2}$$

$$Q(0.1) = 1 + 0.1 + 0.005$$

$$Q(0.1) = 1.105$$

The true value is $e^{0.1} = 1.1052$, so the error is about $0.0002$.

Final answer: $e^{0.1} \approx 1.105$

Example 2

Estimate $\sqrt{9.5}$ using a quadratic approximation at $a = 9$.

A first attempt often looks like this. With $f(x) = \sqrt{x}$:

$$f(9) = 3, \quad f'(x) = \frac{1}{2\sqrt{x}}, \quad f''(x) = -\frac{1}{4x^{3/2}}$$

$$f'(9) = \frac{1}{6}, \quad f''(9) = -\frac{1}{108}$$

Now the second derivative goes straight in as the coefficient, forgetting the $\frac{1}{2}$:

$$3 + \frac{1}{6}(0.5) - \frac{1}{108}(0.5)^2 = 3.0810$$

Compare with the plain tangent line, $L(9.5) = 3 + \frac{1}{6}(0.5) = 3.0833$, and the true value $\sqrt{9.5} = 3.0822$. The "improved" estimate is off by $0.0012$, which is worse than the tangent line's $0.0011$. Adding information should not make an estimate worse, so the correction is twice too big.

The rescue is the $\frac{1}{2}$:

$$Q(9.5) = 3 + \frac{1}{6}(0.5) + \frac{1}{2}\left(-\frac{1}{108}\right)(0.5)^2$$

$$Q(9.5) = 3 + 0.083333 - 0.001157$$

$$Q(9.5) = 3.0822$$

Final answer: $\sqrt{9.5} \approx 3.0822$, correct to 4 decimal places.

Example 3

Find the quadratic approximation of $f(x) = x^{2/3}$ near $a = 1$, and use it to estimate $(1.1)^{2/3}$.

$$f(1) = 1$$

$$f'(x) = \frac{2}{3}x^{-1/3} \quad\Longrightarrow\quad f'(1) = \frac{2}{3}$$

$$f''(x) = -\frac{2}{9}x^{-4/3} \quad\Longrightarrow\quad f''(1) = -\frac{2}{9}$$

$$Q(x) = 1 + \frac{2}{3}(x - 1) - \frac{1}{9}(x - 1)^2$$

$$Q(1.1) = 1 + 0.066667 - 0.001111$$

$$Q(1.1) = 1.0656$$

The true value is $1.0656$; the linear estimate $1.0667$ is off by $0.0011$.

Final answer: $Q(x) = 1 + \frac{2}{3}(x - 1) - \frac{1}{9}(x - 1)^2$, and $(1.1)^{2/3} \approx 1.0656$

Example 4

Estimate $\sin 31^\circ$ with a quadratic approximation centered at $30^\circ$.

The derivative formulas for sine hold only in radians, so convert the step first. One degree is $\frac{\pi}{180} \approx 0.017453$ radians, so $a = \frac{\pi}{6}$ and $h = 0.017453$.

$$f(a) = \sin\frac{\pi}{6} = \frac{1}{2}$$

$$f'(a) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$$

$$f''(a) = -\sin\frac{\pi}{6} = -\frac{1}{2}$$

$$Q = \frac{1}{2} + \frac{\sqrt{3}}{2}(0.017453) - \frac{1}{4}(0.017453)^2$$

$$Q = 0.5 + 0.015115 - 0.000076$$

$$Q = 0.515039$$

The true value is $\sin 31^\circ = 0.515038$. The quadratic error is about one millionth; the linear estimate $0.515115$ is off by about $0.000077$. The first instinct here is to use $h = 1$ (one degree) in the formula, which produces nonsense like $0.5 + 0.866 - 0.25 = 1.116$, a sine value above 1.

Final answer: $\sin 31^\circ \approx 0.5150$

Example 5

Find the quadratic approximation of $\tan x$ at $a = 0$ and explain why it equals the linear approximation.

$$f(0) = \tan 0 = 0$$

$$f'(x) = \sec^2 x \quad\Longrightarrow\quad f'(0) = 1$$

$$f''(x) = 2\sec^2 x \tan x \quad\Longrightarrow\quad f''(0) = 0$$

$$Q(x) = 0 + x + 0 \cdot x^2 = x$$

Because $f''(0) = 0$, the parabola degenerates into the tangent line itself. The error is still cubic, because $\tan x = x + \frac{x^3}{3} + \cdots$, so it is about $\frac{x^3}{3}$. At $x = 0.1$, $Q = 0.1$ and $\tan 0.1 = 0.1003$.

Final answer: $Q(x) = x$; a zero second derivative at the base point removes the quadratic term.

Example 6

Use the quadratic approximation of $f(x) = x^4 - 4x$ at $a = 2$ to take one step toward the minimum of $f$.

$$f(2) = 8, \quad f'(x) = 4x^3 - 4 \Longrightarrow f'(2) = 28$$

$$f''(x) = 12x^2 \Longrightarrow f''(2) = 48$$

$$Q(x) = 8 + 28(x - 2) + 24(x - 2)^2$$

The parabola $Q$ is easy to minimize: set $Q'(x) = 28 + 48(x - 2) = 0$.

$$x - 2 = -\frac{28}{48}$$

$$x = \frac{17}{12} \approx 1.4167$$

The true minimum of $f$ is at $x = 1$ (where $f'(1) = 0$), and one step moved from 2 to 1.4167. Repeating the step from the new point is exactly Newton's method for optimization: the update $x_{\text{new}} = a - \frac{f'(a)}{f''(a)}$ is the vertex of the quadratic approximation.

Final answer: the vertex of $Q$ is $x = \frac{17}{12} \approx 1.4167$

Why Does Quadratic Approximation Matter If The Tangent Line Already Works?

The tangent line fails completely whenever the slope at the base point is zero, and nature is full of those points. The real reason the quadratic term exists: "near a flat point, the bend is the whole story."

• At a flat point, the linear estimate predicts no change. If $f'(a) = 0$, then $L(x) = f(a)$ for every $x$. Any real change near $a$ lives entirely in the $\frac{f''(a)}{2}(x - a)^2$ term.

• Relativity at everyday speeds is a flat point. The time-dilation factor $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$, with $\beta = \frac{v}{c}$, has slope zero at $\beta = 0$, so its tangent line is $\gamma = 1$: no effect. Its quadratic approximation is $\gamma \approx 1 + \frac{\beta^2}{2}$.

• That one term predicts the GPS clock drift. A GPS satellite moves at about 14,000 km/h, according to the Ohio State University relativity and GPS notes. That is $v \approx 3888.9$ m/s, so $\frac{\beta^2}{2} \approx 8.41 \times 10^{-11}$, and over 86,400 seconds in a day this gives $7.27$ microseconds. The same source quotes about 7 microseconds per day for the special-relativity slowdown.

• The same term is classical kinetic energy. Multiply $\gamma \approx 1 + \frac{\beta^2}{2}$ by $mc^2$ and you get $mc^2 + \frac{1}{2}mv^2$. The familiar $\frac{1}{2}mv^2$ is the quadratic approximation of relativistic energy, a point HyperPhysics at Georgia State University derives from the binomial expansion.

Every minimum and maximum in physics, economics and engineering is a flat point too. That is why the cost of a production run, the potential energy of a stretched molecular bond and the error of a machine-learning model near its best fit are all modelled, first, by a parabola.

Where Is Quadratic Approximation Used In Real Life?

Quadratic approximation is the default model wherever a quantity sits near its best, worst or resting value.

• Optimization and machine learning: second-order methods replace the loss function by its quadratic approximation and jump to that parabola's (or paraboloid's) lowest point, as Example 6 showed in one variable.

• Pendulums and springs: for a pendulum, $\cos\theta \approx 1 - \frac{\theta^2}{2}$ turns the potential energy into a parabola, which is what makes small swings behave like simple harmonic motion.

• Chemistry: the energy of a molecule near its stable shape is approximated by a quadratic in the atomic displacements, which gives vibration frequencies.

• Economics: profit near a maximum, or utility near an optimum, is modelled by its second-order expansion to study how sensitive the optimum is.

• Numerical integration: Simpson's rule integrates a parabola through three points instead of a straight line through two, which is why it beats the trapezoidal rule on smooth curves.

• Physics: the low-speed limit of relativity (the GPS and kinetic-energy calculations above) is a quadratic approximation.

Where Do Students Trip Up On Quadratic Approximation?

Three mistakes account for most wrong answers, and each one has a quick check.

Mistake 1: Dropping The 1/2

Where it slips in: Building $Q(x)$ after the derivatives are done, especially when $f''(a)$ is an awkward fraction.

Don't do this: Write $f(a) + f'(a)(x - a) + f''(a)(x - a)^2$. For $\sqrt{9.5}$ at $a = 9$ this gives $3.0810$, worse than the tangent line.

The correct way: Write the $\frac{1}{2}$ before substituting anything: $Q(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2$. Then check that $Q''(a) = f''(a)$.

The memorizer is most at risk here: the formula is held as a string of symbols, so a missing $\frac{1}{2}$ looks no different from a correct line. The habit that fixes it is the parabola test from earlier. If your formula would not reproduce $c_0 + c_1 x + c_2 x^2$ exactly, it is wrong.

Mistake 2: Using Degrees Instead Of Radians

Where it slips in: Any trigonometric estimate, such as $\sin 31^\circ$ or $\cos 62^\circ$.

Don't do this: Use $h = 1$ for a one-degree step. The rule $(\sin x)' = \cos x$ is only true when $x$ is in radians.

The correct way: Convert the step: $h = \frac{\pi}{180} \approx 0.017453$ per degree. A quick check is size: a one-degree change can never move a sine value by more than about $0.0175$.

Mistake 3: Trusting The Approximation Far From The Base Point

Where it slips in: Choosing a convenient base point that is not close, such as $a = 1$ for $\ln 3$.

Don't do this: Assume $Q$ is always better than $L$. For $\ln 3$, $Q(3) = 0$ while $L(3) = 2$ and the true value is $1.0986$.

The correct way: Pick the nearest point where $f$ is known exactly, and check the step against the error bound $\frac{M}{6}|x - a|^3$ before trusting the answer.

The rusher tends to fall into this one: the formula is applied the moment a base point is spotted, without asking whether $|x - a|$ is small. The real-world version of Mistake 3 is classical kinetic energy, $\frac{1}{2}mv^2$, which is the quadratic approximation of relativistic energy and works only for small $\frac{v}{c}$. For an electron at $0.9c$, the classical formula gives $0.405,mc^2$, while the true kinetic energy is $1.2942,mc^2$, more than three times larger. That is why particle accelerators are designed with the full relativistic formula (HyperPhysics explains the comparison).

Practice Problems On Quadratic Approximation

1. Find the quadratic approximation of $\cos x$ at $a = 0$ and use it to estimate $\cos 0.3$.

2. Find the quadratic approximation of $\sqrt{1 + x}$ at $a = 0$ and estimate $\sqrt{1.2}$.

3. Find the quadratic approximation of $f(x) = \frac{1}{x}$ at $a = 2$ and estimate $\frac{1}{2.1}$.

4. Find the quadratic approximation of $x^3$ at $a = 1$, estimate $(1.1)^3$, and state the exact error.

5. Find the quadratic approximation of $e^{0.2}$ from $a = 0$, then bound its error with the Lagrange remainder.

6. Find the quadratic approximation of $f(x, y) = 9\cos x \cos y$ near $(0, 0)$.

Answer to Question 1: $Q(x) = 1 - \frac{x^2}{2}$, so $Q(0.3) = 0.955$; true value $0.9553$.

Answer to Question 2: $Q(x) = 1 + \frac{x}{2} - \frac{x^2}{8}$, so $Q(0.2) = 1.095$; true value $1.0954$.

Answer to Question 3: $f(2) = \frac{1}{2}$, $f'(2) = -\frac{1}{4}$, $f''(2) = \frac{1}{4}$, so $Q(x) = \frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^2$ and $Q(2.1) = 0.47625$; true value $0.4762$.

Answer to Question 4: $Q(x) = 1 + 3(x - 1) + 3(x - 1)^2$, so $Q(1.1) = 1.33$; true value $1.331$, and the error is exactly $(x - 1)^3 = 0.001$.

Answer to Question 5: $Q(0.2) = 1.22$; the bound is $\frac{e^{0.2}}{6}(0.2)^3 \approx 0.0016$, and the actual error is $0.0014$.

Answer to Question 6: $Q(x, y) = 9 - \frac{9x^2}{2} - \frac{9y^2}{2}$.

What To Remember About Quadratic Approximation

• Quadratic approximation estimates $f(x)$ near $a$ by $f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2$, the parabola matching value, slope and bend.

• The $\frac{1}{2}$ makes the parabola's second derivative equal $f''(a)$, and leaving it out can make the estimate worse than the tangent line.

• The error shrinks like the cube of the step, bounded by $\frac{M}{6}|x - a|^3$.

• At a flat point the tangent line predicts no change, and the quadratic term carries the whole effect.

• Far from the base point, or for trigonometric inputs in degrees, the approximation can fail badly.

Where To Go From Here

Work through practice problems 1 to 6; if the derivatives come out right but the estimate is off, return to the section on where the 1/2 comes from, and if the trigonometric answers look wrong, check the radians step in Example 4.

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Read More

• Taylor series and how the approximations extend to infinitely many terms

• Power series and where they converge

• Maxima and minima of functions

• Applications of derivatives