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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 two differentiable functions $f,g$ on an interval $I$, their Wronskian is the functionThe order matters: exchanging $f$ and $g$ reverses the sign. For $n$ sufficiently differentiable functions, the Wronskian is the determinant $\det (f_j^{(i-1)}(x){)}_{i,j=1}^n$.^[ref]

The Wronskian measures whether the vectors $(f(x),f^{\prime }(x))$ and $(g(x),g^{\prime }(x))$ supply independent value-and-derivative data at that point. For real-valued functions, its absolute value is the area of the parallelogram they span. If a constant linear combination $af+bg$ is zero throughout $I$, then $a{f}^{\prime }+b{g}^{\prime }=0$ too. A nonzero Wronskian at even one point makes this pair of linear equations invertible and forces $a=b=0$. Thus a nonzero Wronskian somewhere always proves linear independence of the functions on $I$.

For an explicit example, the exponential functionssolve the same homogeneous linear differential equation $y^{\prime \prime }-y^{\prime }-2y=0$ on the whole real line: substitution gives $1+1-2=0$ and $4-2-2=0$, respectively. Their Wronskian isFor the initial value problem $y(0)=2$, $y^{\prime }(0)=-1$, the linear combination $y=c_1{y}_1+c_2{y}_2$ requires $c_1+c_2=2$ and $-c_1+2c_2=-1$. Hence $c_1=5/3$, $c_2=1/3$, and $y=\frac{5}{3}{e}^{-x}+\frac{1}{3}{e}^{2x}$. The nonzero Wronskian makes the two initial conditions independently adjustable.

The stronger converse requires a common equation. Suppose $y_1,y_2$ solve $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ with continuous functions $p,q$ on one interval $I$. Then the functions are linearly independent if and only if their Wronskian is nonzero at one point, equivalently at every point. If $W(x_0)=0$, choose constants $a,b$, not both zero, making $z=a{y}_1+b{y}_2$ satisfy $z(x_0)=z^{\prime }(x_0)=0$. The superposition principle makes $z$ a solution of a differential equation for the same equation, and the uniqueness theorem for ordinary differential equations forces $z$ to be identically zero. This proves linear dependence.^[ref]

For arbitrary differentiable functions, an identically zero Wronskian need not imply linear dependence. Consider, on $I=(-2,2)$,Both functions have three continuous derivatives. On each side of zero, $g$ is a constant multiple of $f$, so $W(f,g)=0$; at zero their values and first derivatives are all zero, so the Wronskian is zero there too. Nevertheless $af+bg=0$ evaluated at $x=-1$ gives $a-b=0$, while evaluation at $x=1$ gives $a+b=0$. Hence $a=b=0$: these functions are linearly independent on the whole interval. There is no single constant multiplier valid on both sides of the common zero.

These functions cannot both solve any equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$ with continuous functions $p,q$ throughout $(-2,2)$: even $f$ alone would contradict uniqueness because $f(0)=f^{\prime }(0)=0$ but $f$ is not identically zero. They do solve the singular equation $x^2{y}^{\prime \prime }-3x{y}^{\prime }=0$, whose leading coefficient vanishes at zero. Dividing by $x^2$ gives $p=-3/x$ only away from zero; Abel's identity cannot be applied across that point. The example pinpoints why checking the equation and its interval is essential before using a zero Wronskian to infer linear dependence.