Can you find the maximum or minimum of an equation without calculus?

Question

Without using calculus is it possible to find provably and exactly the maximum value or the minimum value of a quadratic equation

$$ y:=ax^2+bx+c $$

(and also without completing the square)?

I'd love to know the answer.

Answer 1

The vertex of $y=x^2$ is $(0, 0)$.

The vertex of $y = Ax^2$ is $(0, 0)$.

The vertex of $y = A(x - k)^2$ is just shifted right $k$, so it is $(k, 0)$.

The vertex of $y = A(x - k)^2 + j$ is just shifted up $j$, so it is $(k, j)$.

Multiply that out, you get $y = Ax^2 - 2Akx + Ak^2 + j$. So you get

$$b = -2ak \tag{1}$$ $$c = ak^2 + j \tag{2}$$

Solve (1) for $k$ and plug it into (2), then solve for $j$,you get:

$$k = \frac{-b}{2a}$$ $$c = a\left(\frac{-b}{2a}\right)^2 + j \implies j = \frac{4ac - b^2}{4a}$$

So the vertex occurs at $(j, k) = \left(\frac{-b}{2a}, \frac{4ac - b^2}{4a}\right)$.

Answer 2

We assume (for the sake of discovery; for this purpose it is good enough if this is just an inspired guess) that the curve $y = ax^2 + bx + c$ is symmetric around a vertical axis. It's obvious this is true when $b = 0$, and if we have plotted $y = ax^2 + bx + c$ for various other values of $a$, $b$, and $c$, we may observe enough appearance of symmetry to suppose that it might be true in general. So it's reasonable to say: supposing it were true, what would that tell us about the minimum/maximum value of the polynomial?

We find the points on this curve of the form $(x,c)$ as follows: \begin{align} y &= c. \\ c &= ax^2 + bx + c. \\ 0 &= ax^2 + bx = (ax + b)x. \end{align} Hence if $(x,c)$ is on the curve, then either $ax + b = 0$ or $x = 0$. Setting $x_1 = -\dfrac ba$ and $x_2 = 0$, we can plug in these two values for $x$ and confirm that indeed the two points $\left(-\frac ba, c\right)$ and $(0, c)$ are on the curve.

Using the assumption that the curve is symmetric around a vertical axis, the vertical axis would have to be halfway between $\left(-\frac ba, c\right)$ and $(0, c)$, that is, it is the line $x = -\dfrac b{2a}$. If there is a global maximum or minimum, it is a reasonable guess that it would be on this line, so let's see what we have at $x_0 = -\dfrac b{2a}$. Plugging this into the equation and doing the algebra to find the point $(x_0, y_0)$ on the curve, \begin{align} y_0 &= a\left(-\frac b{2a}\right)^2 + b\left(-\frac b{2a}\right) + c \\ &= c - \frac{b^2}{4a}. \end{align}

So that's our candidate for the maximum or minimum value.

To prove this is correct, consider any value of $x$ other than $-\dfrac b{2a}$. Any such value can be expressed by its difference from $-\dfrac b{2a}$, that is, we let $$ x = -\frac b{2a} + t$$ where $t \neq 0$. Now plug this value into the equation and do the algebra: \begin{align} y &= a\left(-\frac b{2a} + t\right)^2 + b\left(-\frac b{2a} + t\right) + c \\[.5ex] &= at^2 + c - \frac{b^2}{4a}. \tag 1 \end{align}

If $a$ is positive, $at^2$ is positive, hence $y > c - \dfrac{b^2}{4a} = y_0$ for every point $(x,y)$ on the curve such that $x \neq x_0$, and therefore $y_0 = c - \dfrac{b^2}{4a}$ is a minimum. But if $a$ is negative, $at^2$ is negative, and similar reasoning says that $y_0 = c - \dfrac{b^2}{4a}$ is a maximum.

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Note that the proof made no assumption about the symmetry of the curve. On the contrary, the equation $y = at^2 + c - \dfrac{b^2}{4a}$ can be used to prove that the curve is symmetric.

If we take this a little further, we can even derive the standard quadratic formula from it. The roots of the equation $y = ax^2 + bx + c$ are the values of $x$ such that $y = 0$. But as we know from Equation $(1)$, above, if we make the substitution $x = -\dfrac b{2a} + t$, that means \begin{align} 0 = y &= ax^2 + bx + c \\ &= at^2 + c - \frac{b^2}{4a}. \end{align} A little algebra (isolate the $at^2$ term on one side and divide by $a$) gives us $$ t^2 = \frac{b^2}{4a^2} - \frac ca. \tag 2 $$ In general, if $p^2 = q$ then $p = \pm \sqrt q$, so Equation $(2)$ tells us that \begin{align} t &= \pm \sqrt{\frac{b^2}{4a^2} - \frac ca} \\ &= \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}\\ &= \pm \frac{\sqrt{b^2 - 4ac}}{\lvert 2a \rvert}\\ &= \pm \frac{\sqrt{b^2 - 4ac}}{2a}, \end{align} and recalling that we set $x = -\dfrac b{2a} + t$, \begin{align} x &= -\frac b{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\ &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}, \end{align} which is precisely the usual quadratic formula.

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Is the reasoning above actually just an example of "completing the square," or is it sufficiently different from the usual method of "completing the square" that it can be considered a different method? The equation $x = -\dfrac b{2a} + t$ is equivalent to $t = x + \dfrac b{2a}$; the method of completing the square involves expanding $\left(x + \dfrac b{2a}\right)^2$; and in fact we do see $t^2$ figuring prominently in the equations above.

Certainly we could be inspired to try completing the square after noticing how neatly the equation $ax^2 + bx + c = at^2 + c - \dfrac{b^2}{4a}$ simplified the problem; but we never actually expanded the binomial $\left(x + \dfrac b{2a}\right)^2$, and we never subtracted the original polynomial from it to find the amount we needed to "complete" the square. Instead, the quantity $c - \dfrac{b^2}{4a}$ just "appeared" in the original equation as the result of a direct substitution.

I think that may be about as different from "completing the square" as a purely algebraic method can get.

Answer 3

One approach for finding the maximum value of $y$ for $y=ax^2+bx+c$ would be to see how large $y$ can be before the equation has no solution for $x$.

First rearrange the equation into a standard form:

$ax^2+bx+c-y=0$

Now solving for $x$ in terms of $y$ using the quadratic formula gives:

$x= \frac{-b\pm \sqrt{b^2-4a(c-y)}}{2a}$

This will have a solution as long as $b^2-4a(c-y) \geq 0$

You can rearrange this inequality to get the maximum value of $y$ in terms of $a,b,c$. See if you get the same answer as the calculus approach gives. Remember that $a$ must be negative in order for there to be a maximum.

Answer 4

This is almost the same as completing the square but .. for giggles.

If $a = 0$ we know $y = xb + c$ will get "extreme" and "extreme" positive and negative values of $x$ so no max or minimum is possible.

So if $ax^2 + bx + c = a(x^2 + x b/a)+c := a(x^2 + b'x) + c$ So finding the max/min is simply a matter of finding the max/min of $x^2 + b'x$ and multiplying by $a$ and adding $c$.

In other words .... wolog $a = 1$ and $c = 0$.

Now we know $x^2 + bx$ has only a min as $x^2$ is positive and as $|x|$ increases the $x^2$ term "overpowers" the $bx$ term. Or if $x > |b|/2$ then $(x+ h)^2 + b(x + h) = x^2 + bx +h(2x + b) + h^2 > 0$ so the expression has no max value.

So we want to find the minimum of $x^ + b'x = x(x + b)$. Let $y := x - b'/2$ then $x(x + b')=(y -b'/2)(y + b'/2)= y^2 - (b'^2/4)$. As $y^2 \ge 0$ the min will occur when $y = 0$ or in other words, $x= b'/2 = b/2a$

So the max/min of $ax^2 + bx + c$ occurs at $x = b/2a$ and the max/min value is $b^2/4 + b^2/2a + c$

Okay, that really was the same thing as completing the square but it didn't feel like it so what the @@@@.