2
$\begingroup$

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.

$\endgroup$

closed as off-topic by Spencer, JonMark Perry, Silvia Ghinassi, Claude Leibovici, user26857 Feb 6 '16 at 8:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Spencer, JonMark Perry, Silvia Ghinassi, Claude Leibovici, user26857
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Also here is meta.math.stackexchange.com/questions/5020/… for formatting your functions properly. $\endgroup$ – Arbuja Jan 30 '16 at 20:35
  • $\begingroup$ Odd question... No calculus, no completion of the square... I guess asking the teacher should work. $\endgroup$ – user228113 Jan 30 '16 at 21:16
  • 2
    $\begingroup$ Without completing the square, or without calculus? $\endgroup$ – Mark Bennet Jan 30 '16 at 21:16
  • $\begingroup$ This is like asking how to win a martial arts tournament while unconscious. $\endgroup$ – Matt Samuel Jan 31 '16 at 0:13
  • $\begingroup$ That said, I would guess the ancient Greeks knew how to do this, and I think completing the square was discovered less than a thousand years ago. $\endgroup$ – Matt Samuel Jan 31 '16 at 0:33
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ So it works out the values in the shifts of the maxima or minima at (0,0) , in the specific quadratic, to deduce the actual maxima or minima in any quadratic. I think this is a good answer to the question I asked. $\endgroup$ – Karlie Kloss Feb 7 '16 at 13:22
  • $\begingroup$ This is one of the best answer I have come across $\endgroup$ – Kirthi Raman Oct 15 '16 at 22:44
1
$\begingroup$

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.


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.


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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Yes a variation of this idea can be used to find the minimum too $\endgroup$ – Karlie Kloss Jan 30 '16 at 20:59
  • 5
    $\begingroup$ @Karlie Kloss Technically speaking this solution is also not without completion of squares because you are still using the quadratic formula and how do you get that??? $\endgroup$ – Anurag A Jan 30 '16 at 21:09
  • $\begingroup$ @KarlieKloss Just because you don't see something spelled out in its full detail doesn't mean it is "not used." The question then is, what is the proof of the quadratic formula that does not use any form of completing the square? $\endgroup$ – David K Feb 1 '16 at 14:58
  • $\begingroup$ Yes, t think now that is a better question to ask. So this method answers the question if there is a proof of the quadratic formula that does not use any form of completing the square. $\endgroup$ – Karlie Kloss Feb 1 '16 at 16:46
0
$\begingroup$

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 @@@@.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.