You already know that \sqrt{9} = 3 because 3^2 = 9. The square root "undoes" the square. That undoing has a striking visual consequence: the graphs of y = x^2 and y = \sqrt{x} are mirror images of each other, reflected across the line y = x.

Seeing both curves on the same axes is one of the cleanest pictures of what it means for two operations to be inverses. One curve climbs steeply; the other curve grows with deliberate slowness. Fold the picture along the diagonal y = x and the two curves land exactly on top of each other.

The two curves, side by side

The square function y = x^2 turns small inputs into tiny outputs and blows up fast: 2^2 = 4, 3^2 = 9, 10^2 = 100. The square root function y = \sqrt{x} does the reverse — big inputs produce modest outputs: \sqrt{100} = 10, \sqrt{9} = 3, \sqrt{4} = 2.

Drag the slider (or drag directly on the canvas) to move $x$. The orange dot rides on $y = \sqrt{x}$; the green dot rides on $y = x^2$. The thin dashed segment between them crosses the diagonal $y = x$ at its midpoint, and both dots sit the same distance from the fold line. Tick animate fold to rotate the parabola across the diagonal and watch it land exactly on the square-root curve.

Look at how the two curves cross the line y = x: they both pass through the origin (0, 0) and through the point (1, 1). Those are the only two real fixed points of either function, because x^2 = x and \sqrt{x} = x both have solutions x = 0 and x = 1.

Why they are mirror images

Two functions f and g are inverses of each other when g(f(x)) = x and f(g(x)) = x — applying one undoes the other. For f(x) = x^2 and g(x) = \sqrt{x} with x \geq 0:

g(f(x)) = \sqrt{x^2} = x \quad \text{and} \quad f(g(x)) = (\sqrt{x})^2 = x

The general rule is that the graph of any function and its inverse are reflections of each other across the line y = x. Here is the reason.

If (a, b) is on the graph of f, that means f(a) = b. But if f and g are inverses, then g(b) = a — so the point (b, a) is on the graph of g. Swapping the coordinates of every point (a, b) \leftrightarrow (b, a) is exactly what reflection across the line y = x does geometrically. So the graph of g is the mirror image of the graph of f across that diagonal, automatically.

Why swapping coordinates is a reflection across y = x: the line y = x treats the two axes symmetrically. A point like (2, 4) sits above the line (because y > x); its mirror image (4, 2) sits below (because y < x). The reflection move "swap x and y" is the algebraic version of "flip across the diagonal." Every inverse-function pair is linked this way, not just square and square root.

A few specific mirror pairs

Check three points to feel the reflection:

Point on y = x^2 Mirror on y = \sqrt{x}
(1, 1) (1, 1) — on the diagonal, maps to itself
(2, 4) (4, 2)
(3, 9) (9, 3)
(0.5, 0.25) (0.25, 0.5)

Each row reads the same equation two ways. 3^2 = 9 says the parabola contains (3, 9); \sqrt{9} = 3 says the square-root curve contains (9, 3). Same fact, different graphs.

Why the parabola is restricted to x \geq 0 in this picture

The function y = x^2 accepts negative inputs too: (-3)^2 = 9, so the point (-3, 9) is also on the parabola. But the square-root function \sqrt{x} is only defined for x \geq 0 in the real numbers, and its output is always non-negative (by the principal-root convention from Roots and Radicals). So the clean mirror-image relationship only works in the first quadrant, where both curves live.

If you let the parabola extend to negative x, there is no matching square-root curve below the x-axis to reflect into — because \sqrt{x} is not defined for negative x, and its outputs are never negative. The mirror-image picture is a first-quadrant affair.

This is also why x^2 doesn't quite have a "full" inverse over all real numbers — it is not one-to-one, because 3^2 = (-3)^2 = 9 both give 9. The square-root function inverts only the right half of the parabola, the part with x \geq 0. To invert the left half, you would need the negative-root function y = -\sqrt{x}, whose graph is the reflection of the left half of the parabola across y = x.

The growth-rate contrast

Quantitatively, x^2 eventually dominates every polynomial-of-smaller-degree, while \sqrt{x} is dominated by every such polynomial. A doubling in x quadruples x^2 (factor of 4), but multiplies \sqrt{x} by only \sqrt{2} \approx 1.414 (factor of less than 1.5).

  • At x = 100: x^2 = 10{,}000 and \sqrt{x} = 10. Five orders of magnitude apart.
  • At x = 10{,}000: x^2 = 10^8 and \sqrt{x} = 100. Six orders apart.
  • At x = 10^6: x^2 = 10^{12} and \sqrt{x} = 1000. Nine orders apart.

The two curves spread apart as fast as any pair of inverses can: one explodes, the other tames. That is the visual signature of a function and its inverse — they pull in exactly opposite directions, symmetrically across the diagonal.

The takeaway

y = x^2 and y = \sqrt{x} are the same equation read two different ways — the first says "given x, square it"; the second says "given y, find the x whose square is y." Plotted together, the two readings appear as reflections across the line y = x. That reflection is the universal visual signature of inverse functions: swap the coordinates, and any point on one curve lands on the other.

Related: Roots and Radicals · Exponents and Powers · Algebraic Identities · Quadratic Equations