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The figure illustrates Pierre de Fermat's use of adequality to construct a tangent to a parabola. Reading the local exam image is decisive: $D$ is the vertex of a parabola, $DC$ lies on its axis of symmetry, $B$ lies on the parabola, and the proposed tangent $BE$ meets the extended axis at $E$. The horizontal through the auxiliary point $O$ meets the axis at $I$; $O$ lies on $BE$, outside the parabola. In particular, $O$ must not be treated as a second point of the parabola. The target is the subtangent $CE$. The diagram and this construction occur in the tangent discussion associated with Varia opera mathematica by Pierre de Fermat. The 1679 book was posthumous: its date does not date the invention of the method.^[ref][ref]

Write $d=CD>0$, $a=CE>0$, and $e=CI$, with $I$ between $C$ and $D$. These are lengths, and the increment $e$ is distinct from the labelled point $E$. Then $DI=d-e$ and $EI=a-e$. The parabola's characteristic relation makes the square of an ordinate proportional to its distance from the vertex of a parabola. At height $I$, the actual ordinate of the parabola is shorter than $OI$. Hencewhere the equality follows from similar triangles $BCE$ and $OIE$.

The method of adequality replaces the comparison by a relation used to extract the condition of contact:Expanding and cancelling $d{a}^2$ gives $-2dae+d{e}^2\sim -a^{2}e$. Divide by the nonzero increment $e$, then suppress the remaining term containing $e$. The result is $a^2=2da$, so, since $a>0$,Thus $DE=DC$: the intersection with the axis lies as far above the vertex of a parabola as $C$ lies below it. This is a concrete construction, not just a rule for computing a numerical slope.

A modern check clarifies why the calculation succeeds without pretending that Pierre de Fermat used a modern limit. Put $D=(0,0)$, let $y$ increase down the axis, and write the parabola as $x^2=ky$, $k>0$. For $B=(b,d)$, with $b^2=kd$, the proposed tangent through $E=(0,-d)$ is $y=2bx/k-d$. Substitution givesIt meets the parabola with a double root at $B$, and for every other point its $y$ coordinate is less than that of the parabola. This also verifies the strict comparison used for $O$.

The significance is a systematic algebraic condition for geometric contact. Subtracting neighbouring expressions, dividing by an increment, and discarding its remaining terms anticipates differentiation. But the historical question is precisely what licenses the last step. The symbol $\sim$ above marks adequality, not ordinary equality at a finite displacement. Treating the construction as an already stated epsilon-delta definition of a limit would conceal the methodological innovation and its unresolved explanatory burden.

At a point $B$ of a planar curve, let the ordinate through $B$ meet the chosen horizontal coordinate axis at $C$, and let the tangent line at $B$ meet that axis at $E$. The axis segment $CE$ is the subtangent; this construction requires the indicated finite intersection and depends on the chosen axes.

For the standard parabola $y^2=4ax$, $a>0$, take $B=(x_0,y_0)$ with $y_0\ne 0$. The tangent line is $y_{0}y=2a(x+x_0)$, so $E=(-x_0,0)$ and $C=(x_0,0)$. Therefore $CE=2x_0=2CD$, where $D=(0,0)$ is the vertex of a parabola. Here tangent refers to a straight line, rather than the trigonometric tangent function.

Write the quadratic polynomial asIts function graph is a parabola with vertex of a parabola $(h,k)$. Equal offsets from $h$ give equal values, since $p(h+t)=p(h-t)=a{t}^2+k$. If $a>0$, the minimum is $k$; if $a<0$, the maximum is $k$. In both cases it is attained only at $x=h$.

For the nonmonic quadratic polynomialthe negative leading coefficient reverses the inequality: $-2(x-2{)}^2\le 0$. Its maximum is $3$ at $x=2$. Its coefficients are different from those of the positive example, but the same form exposes the shape of both function graphs.

Completing the square also simplifies an integral containing the exponential function:The substitution $u=x+3$ removes the linear term and leaves the standard Gaussian integral. The same shift that locates the minimum makes this integral recognizable.

For a parabola described by the function graph $y=a(x-h{)}^2+k$, with nonzero $a$ in the real numbers, its vertex of a parabola is the point $(h,k)$. This point gives its minimum or maximum value of $y$ in these coordinates.