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This is one of those questions that I know intuitively, but find it hard to prove mathematically.

Problem

The idea is to try and prove that a second degree polynomial function, $f(x) = ax^2 + bx + c$, always has an extremum.

My thoughts

My thought is to try an explain that given $a > 0$, then $\lim\limits_{x\to\pm\infty}f(x) \to \infty$, so if a non-infinite value exists, it must have a bottom point between the extremes.

Likewise for $a<0$ but with a top point between two $-\infty$ extremes.

But I can't see how to prove this rigorously.

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\begin{eqnarray*} ax^2+bx+c = a \underbrace{\left(x+\frac{b}{2a} \right)^2}_{ \text{extremum when } x= -\frac{b}{2a}} +c - \frac{b^2}{4a}. \end{eqnarray*}

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Given $f(x)=ax^2+bx+c$ multiply by the constant $4a$ to give $$4af(x)=4a^2x^2+4abx+4ac=(2ax+b)^2+(4ac-b^2)$$

The right-hand side as a square plus a constant has a minimum value when the square is zero ie $x=-\frac b{2a}$

This gives a minimum value for $f(x)$ if $a$ is positive and a maximum value if $a$ is negative. If $a=0$ the expression is not quadratic.

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Whilst your ideas can be extended into full proofs, given that we are working with quadratics, there is a much easier way:

We write,

\begin{align} f(x) &= ax^2 + bx + c\\ &= a(x+\frac{b}{2a})^2 + (c - \frac{b^2}{4 a})\\ &\ge (c - \frac{b^2}{4 a}) \, \quad \text{ (here we assume $a > 0$*)} \end{align}

with equality if and only if the squared term is equal to $0$ - that is, when $ x = - \frac{b}{2a}$.


*Proof is identical in the case $a<0$, but with $\ge$ replaced with $\le$

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If you have access to calculus, then you can note that the derivative of $ax^2+bx+c$ is the function $2ax+b$, which equals zero at precisely one place: $x=-\frac{b}{2a}$. When a real valued function defined on $\Bbb R$ has only one local extreme, it's a global extreme. To see what kind, we can check the second derivative, which equals $2a$ everywhere. When $a>0$, that's positive, so the extremum is a minimum, and when $a<0$, the second derivative is negative, so it's a maximum.

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