Is there a way of finding the turning point of a quadratic without calculus and without completing the square?

Question

Using completing the square it's trivial to find the minimum of a polynomial such as $x^2+9x+5$. We rearrange to get $(x+\frac{9}{2})^2-\frac{61}{4}$ and can see that when $x=-\frac{9}{2}$ we have a minimum.

Using calculus it's also trivial. The derivative of $x^2+9x+5$ with respect to $x$ is $2x+9$, and the value of $x$ that makes $2x+9=0$ is $-\frac{9}{2}$.

My question is: are there any unusual methods to find the minimum of a quadratic (or, if possible, higher order polynomial)? Something not rooted in algebraic manipulation like completing the square or calculus?

Answer 1

Another way to look at it: If the curve $y=ax^2+bx+c$ has two $x$-intercepts then the max/min happens at their average. But we can slide the parabola up or down without changing the $x$-coordinate of the vertex. So subtract off the constant term:

$$y=ax^2+bx = x(ax+b).$$

The two $x$-intercepts are $0$ and $-b/a$. The vertex then, has $x$-coordinate $(0 + (-b/a))/2 = -b/2a..$

Answer 2

Without completing the square:

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Let,

$$x^2+9x+5=a,\,a\in\mathbb R.$$

Then we have:

$$x^2+9x+(5-a)=0$$

Then note that:

If $x\in\mathbb R$, then $\Delta_x=9^2-4(5-a)\geqslant 0$. This implies that, $\Delta_x=4a+61\geqslant 0$. This yields $a\geqslant-\frac{61}{4}$.

Finally, observe that if $\Delta_x=0$, then the mininum occurs when,

$$x_1=x_2=-\frac 92.$$

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Generalization of the method:

Let,

$$ \begin{align}&ax^2+bx+c=M\\ \implies &ax^2+bx+(c-M)=0\\ &\Delta_x= b^2-4a(c-M)\geqslant 0\\ \implies &4aM≥-(b^2-4ac)\end{align} $$

This tells us,

$$ \begin{align} &M\geqslant -\frac {b^2-4ac}{4a},\;\, \text{if}\; a>0\\ &M\leqslant -\frac {b^2-4ac}{4a},\;\, \text{if}\; a<0\end{align} $$

This also means that the minimum/maximum occurs when:

$$\Delta_x=0\quad\text{or}\quad x=-\frac {b}{2a}.$$

Answer 3

You can "depress" a polynomial $ax^n+bx^{n-1}+\cdots+c$ by substituting $x=y-b/na.$ The new polynomial in $y$ won't have an $x^{n-1}$ term.

If you do this for a quadratic, $ax^2+bx+c$, and let $x = y-b/2a$ the result has no linear term. You get a quadratic of the form $ay^2+D$, which clearly has minimum/maximum at $y = 0$. Put this back in the substitution to get $x = -b/2a.$

Answer 4

There's a rather trivial method of finding the x-value of the extremum of a quadratic. All you have to do is construct a 2x2 matrix with the quadratic as its minimal polynomial, and then take the average of its eigenvalues. Simple! Or you could create a sequence with the quadratic as its generating function, take the limit of the ratios of its successive terms, and also extend it to negative infinity and take the limit of those ratios, then take the average of those two limits.

There are also ways of using Physics formulas to find the vertex. You can derive $v(t) = at+v_0$ rather easily, and then you can use such methods as a geometric argument to derive $x(t)= \frac {at^2}2+v_0t+x_0$. So if $x(t)=x^2+9x+5$, then we can deduce that $a=2,v_0=9,x_0=5$. Then we need $t$ such that $v(t)=0$, i.e. $at+v_0=0$, which gives us $t=-\frac{v_0}a = -\frac 92$.

Answer 5

This is the same as @B.Goddard's answer, and uses the root formula (which assumes completion by square), so this is more of a comment than an answer.

If $r_-$ and $r_+$ are roots of the quadratic, because the quadratic is symmetric, the turning point will happen in the middle between them (the average).

The root formula is $r_\pm = {-b \pm \sqrt{b^2 - 4ac}\over 2a}$, and the average of these is $$ {1\over2}\left({-b \color{red}{+ \sqrt{b^2 - 4ac}}\over 2a} + {-b \color{red}{-\sqrt{b^2 - 4ac}}\over 2a}\right) = -{b\over 2a} $$