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.
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:
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