Graph of cosh x (hyperbolic cosine, the catenary)

Updated 20 October 2026 · Values computed by the ihateDesmos engine

The hyperbolic cosine, cosh x = (eˣ + e⁻ˣ)/2, is the average of eˣ and e⁻ˣ. Its graph is a U-shape with its lowest point at (0, 1). A chain or cable hanging under its own weight takes exactly this shape, the catenary.

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The graph of

y=cosh(x)

for −4 ≤ x ≤ 4. Dots mark its minimum.

Key features

Domain: All real numbers
Range: y ≥ 1
Definition: cosh x = (eˣ + e⁻ˣ)/2
Minimum: (0, 1)
Symmetry: Even: cosh(−x) = cosh x
Concavity: Concave up everywhere
y-intercept: None
Derivative: sinh(x)
Antiderivative: sinh(x)+C

Derivative and antiderivative from the calculator’s algebra system. Domain, range and asymptotes are stated for real x.

What the graph of cosh x looks like

At the bottom, cosh x ≈ 1 + x²/2, so it looks like a parabola. Further out it grows exponentially, and from a distance the difference is obvious: a catenary has steeper sides than any parabola eventually.

Galileo thought a hanging chain formed a parabola; in 1691 Leibniz, Huygens and Johann Bernoulli independently showed it is a catenary, y = a cosh(x/a). Turned upside down, it is the ideal shape for an arch in pure compression.

cosh x is always at least 1, and cosh²x − sinh²x = 1, which is why (cosh t, sinh t) traces a hyperbola.

Points of interest

Found numerically to about twelve digits in the window −4 ≤ x ≤ 4, then shown in exact form where one exists.

x-intercepts (roots): None in this window
Local maxima: None in this window
Local minima: (0, 1)
Inflection points: None in this window

Table of values

A few points to plot, with exact values where they are simple.

Values of cosh x
x	y = cosh x
−3	10.0677
−2	3.7622
−1	1.54308
0	1
1	1.54308
2	3.7622
3	10.0677

How to graph cosh x by hand

1
Plot the minimum (0, 1).
2
Plot (±1, 1.54), (±2, 3.76) and (±3, 10.07).
3
Draw a smooth U through them, symmetric about the y-axis.

Transformations

How the graph moves when you change the formula. The original is cosh x in blue.

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y = 1 + x^2/2 The parabola that matches cosh x at the bottom; the catenary pulls away from it.
y = 2cosh(x/2) A wider catenary, a cosh(x/a) with a = 2.
y = -cosh(x) + 5 An upside-down catenary arch.