FWHM for a quadratic

This post contains a derives a result I needed recently. The derivation is simple but a little tedious, so I wanted to save it in case I need it again.

Full width half maximum

A common way to measure the width of a function peak in a function f(x) is to find the place x0 where f takes on its maximum, and find two points, x-1 to the left and x1 to the right, where f drops to half its peak value, i.e.

f±1 = f(x0) / 2.

The width of the peak is then defined to be the distance between these two points:

FWHM = x1x-1

where FWHM stands for “full width half maximum.” I’ve mentioned FWHM a few times before, such as here.

It’s also useful sometimes to find the full width at k times the maximum for values of k other than 1/2 and so we’ll solve the more general problem.

Quadratic case

Now suppose f is a quadratic function

f(x) = ax² + bx + c.

where a is not zero. We want to find the FWHM of f, and more generally find the distance between two points where f takes on values k times its maximum (or minimum).

Taking the derivative of f shows that the vertex of the parabola occurs when

2ax + b = 0

and so

x0 = –b/(2a).


f(x0) = c – b²/(4a).

Now we have to find two solutions to

f(x) = k f(x0)

which means

ax² + bx + ck(c – b²/(4a)) = 0.

This is a quadratic equation with constant term

c‘ = ck(cb²/(4a))

and so from the quadratic formula, the difference between the two roots is

√( b² – 4ac‘ ) / |a| = √( b² – 4a(1-k)c – kb² ) / |a|

When k = 1/2, this reduces to

FWHM = √(b²/2 – 2ac) / |a|.


Let’s try this on a couple examples to see if this checks out.

Maximum example


f(x) = 20 – (x – 2)² = –x² + 4x +16

Clearly the maximum is 20 and occurs at x = 2. The quadratic formula shows the two places where f takes half its maximum value are

x = 2 ± √10

and so the FWHM equals 2√10.

If we use the formula for FWHM above we get

√( 4²/(2) + 32) = √40 = 2√10.

Minimum example

Let’s do another example, this time looking for where a convex parabola takes on twice its minimum value. So here we set k = 2 and so the expression

√( b² – 4ac‘ ) / |a| = √( b² – 4a(1-k)c – kb² ) / |a|

above reduces to

√(4acb²) / |a|.


f(x) = 3x² + 2x + 1

Then the minimum of f occurs at -1/3 and the minimum equals 2/3. We want to find where f equals 4/3. The quadratic formula shows this occurs at

(-1 ± √2)/3

and so the distance between these two points is 2√2 / 3.

If we plug a = 3, b = 2, and c = 1 into

√(4acb²) / |a|

we get the same result

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