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I'm tutoring a Grade 11 Math student in BC, Canada, and we're going over parabolas. He's having difficulty with finding the vertex of a parabola - not how to find it, but WHY it works. And I'm having a hard time finding a good way to explain it.

With Calculus I can explain this in a heartbeat - we're just finding where the derivative is zero. But this student definitely has no knowledge of calculus, and teaching him derivatives is probably out of the equation (math pun, har har).

Is there a non-calculus way to derive the solution for the x coordinate of the vertex of a parabola (y is easy to explain graphically). A hand-wavy or purely graphical depiction is fine, so long as it seems to 'make sense' without any knowledge of calculus.

Cheers!

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    $\begingroup$ Why not use transformations? The simplest parabola is $f(x)=x^{2}$ which has vertex at (0,0), so shifting it horizontally by h units, and vertically by k units, should shift its vertex by the same amount. $\endgroup$ – MintChocolateIceCream Oct 2 '15 at 16:07
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    $\begingroup$ @mintchocolateicream you should make that an answer and I'd up vote. $\endgroup$ – Karl Oct 2 '15 at 19:57
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    $\begingroup$ You can use a symmetric argument for the vertex to be located between the roots; from the quadratic equations, these roots are at: $x = \frac{-b}{2a} \pm \text{blah}$, hence the $x$-coordinate between the roots is at $x = \frac{-b}{2a}$. $\endgroup$ – Benjamin Dickman Oct 2 '15 at 20:35
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    $\begingroup$ In England, we use "Completing the Square" for this $\endgroup$ – nickjamesuk Oct 2 '15 at 23:28
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    $\begingroup$ As an addition to @MintChocolateIceCream 's remark, in Belgium we usually consider both $y=ax^2+bx+c$ and $y=a(x-p)^2+q$ as 'standard' forms for the equation of a parabola. Finding the vertex from the 2nd equation is absolutely easy, so you should only find the correspondence between the 2 formula's (and indeed, $p=-\frac{b}{2a}$). $\endgroup$ – long tom Oct 3 '15 at 13:24
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Take your equation and complete the square to get it into the form $y=a(x - h)^2 + k$. Assume first that $a$ is positive. That means that $a(x - h)^2$ is positive so your $y$ values are always $k$ plus some positive number. The original equation has its minimum when the "something", $a(x - h)^2$, is equal to 0 which occurs when $x = h$. That makes the minimum value $y=a(h - h)^2 + k = k$.

For the case where $a$ is negative, the argument is the same only you're looking for the maximum value.

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    $\begingroup$ IMHO, this should really be the accepted answer. Completing the square is a basic skill, and really allows you to understand everything about quadratics. $\endgroup$ – Steven Gubkin Oct 5 '15 at 2:30
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    $\begingroup$ I also believe this would serve as an ideal answer -- it was mentioned in the comments, but probably it is best to record here in full. (Perhaps, for completeness, it would be worthwhile to include the actual algebraic manipulations?) $\endgroup$ – Benjamin Dickman Oct 6 '15 at 3:13
  • $\begingroup$ Personally I find this answer very satisfying. One thing to note though is that my tutee didn't find the process of completing the square very intuitive, and so the symmetric arguments outlined above and below were also very valuable. $\endgroup$ – Brandon Oct 16 '15 at 0:54
  • $\begingroup$ Until the student is comfortable with symbolic thinking, mweiss' solution (or one, like it, see youtube.com/watch?v=Z5hVo4q5xZI) is so much easier. $\endgroup$ – Sue VanHattum Dec 3 '16 at 10:59
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Curiously enough, I wrote a short article about this that has just been published in The Mathematics Teacher (available here, although you may need subscriber access, or access to a library with such access, in order to get the full text). The advantage of this approach is that it requires very little algebra -- and in particular does not require students to complete the square or use the quadratic formula. Instead, the approach relies on three basic ideas:

  1. Parabolas are symmetric
  2. Adding a constant term to a quadratic function shifts its graph vertically
  3. Quadratic equations with no constant term are very easy to solve by factoring

Here is the approach, in detail:

Suppose you want to find the vertex of $y=5x^2 - 20x + 12$. Ask yourself: Could I solve an easier, related problem? One such easier, related problem comes by dropping the constant term entirely. So let's ask ourselves:

Where is the vertex of $y = 5x^2 - 20x$?

Well, we know that parabolas are symmetric. A consequence of that is that if a parabola has two $x$-intercepts, those will be located equally spaced on opposite sides of the line of symmetry. Or, to put it another way, if you can find the two $x$-intercepts, the line of symmetry will be halfway between them (i.e. at their average).

Fortunately, the $x$-intercepts of $y = 5x^2 - 20x$ are easy to find: We just solve $$5x^2-20x = 0$$ $$5x(x-4) = 0$$ $$x = 0,4$$

and that means that the line of symmetry is at $x = 2$.

Now, how does that help us solve the original problem? Here is the crucial idea:

Restoring the $ + 12 $ term to the original equation shifts the parabola vertically but does not change its line of symmetry.

So for the problem we are interested in, $y = 5x^2 - 20x + 12$, the line of symmetry is also at $x = 2$. Since the vertex is on the line of symmetry, its $x$-coordinate is $2$.

Finally, we can find the $y$-coordinate by simply plugging $x=2$ into the formula for the parabola: $ y= 5 \cdot 2^2 - 20(2) + 12 = -8$. So the vertex is at $(2, -8)$.

Having gone through this for a single example, now let's consider the general problem. If we want to find the vertex of $ y = ax^2 + bx + c $, let's ask ourselves first the simpler problem obtained by dropping the constant term:

Where is the vertex of $y=ax^2 + bx$?

The answer is: it will have an $x$-coordinate halfway between the $x$-intercepts of the parabola. We can find those easily: $$ax^2 + bx = 0$$ $$x(ax + b) = 0$$ $$x = 0, -b/a $$

Therefore the line of symmetry is at $x = -\frac{b}{2a}$. Because adding or subtracting a constant term just shifts a parabola vertically, this is also the line of symmetry for the parabola we are actually interested in, and therefore is also the $x$-coordinate of its vertex, too.

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    $\begingroup$ You may have seen this, for it is hidden in a comment to the OP, but this is the method described (in less detail) in MSE 709. $\endgroup$ – Benjamin Dickman Oct 4 '15 at 19:36
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    $\begingroup$ @BenjaminDickman Ah, nice. I had not seen that. $\endgroup$ – mweiss Oct 4 '15 at 20:32
  • $\begingroup$ Just read this -- and am teaching quadratic functions right now in Algebra 2! Fun reading; I especially enjoyed the $a = rise/run^2$ part. $\endgroup$ – Benjamin Dickman Dec 1 '16 at 22:26
  • $\begingroup$ @BenjaminDickman Thanks for reminding me to update this. :) $\endgroup$ – mweiss Dec 2 '16 at 2:39
  • $\begingroup$ There are some great illustrations of this sort of idea by James Tanton. (He keeps the original and asks for two points with the same y-coordinate. The work is virtually the same.) youtube.com/watch?v=Z5hVo4q5xZI $\endgroup$ – Sue VanHattum Dec 3 '16 at 10:56
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Consider $f(x)=x^2$ then $f(x-a)+b=(x-a)^2+b$ and geometrically corresponds to translating the graph of $y=f(x)$ by a vector $[a,b]^T$

In particular the vertex $(0,0) \mapsto (a,b)$

Cases such as $kf(x)$ were omitted for simplicity but follow similar reasoning.

Hope this helps.

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    $\begingroup$ Thanks! I'll combine this with the symmetric argument that Benjamin outlined above. $\endgroup$ – Brandon Oct 3 '15 at 17:21
  • $\begingroup$ I think the transformation argument is a little abstract for some beginners, so I normally start by asking the minimum value of (x-a)^2 and what we need x to be to get there. It's then a simple jump to the minimum of (x-a)^2+b. $\endgroup$ – AndrewC Oct 4 '15 at 14:15
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This response is similar to some of what has already been written, but I thought it might be helpful to connect finding the vertex's $x$-coordinate with figuring out the roots given by the quadratic equation.

Finding the $x$-coordinate of the vertex for $x \mapsto wx^2$ shouldn't be too tough: it is at $x = 0$. Observe also that the $x$-coordinate is not affected by shifting the curve up by $z$, i.e., for $x \mapsto wx^2 + z$.

The difficulty with a curve like $x \mapsto ax^2 + bx + c$ is the presence of that pesky $b$ term; so let us shift $x$ by some distance so as to get a new curve where the coefficient of $x^1$ is $0$. Call the distance $d$.

Now we have (after expanding and combining like terms):

$$a(x+d)^2 + b(x+d) + c = ax^2 + (2ad + b)x + ad^2 + bd + c$$

To achieve our goal means ensuring $2ad + b = 0$, i.e., $d = -b/2a$.

So we end up with a curve${^1}$ whose vertex's $x$-coordinate is $0$; returning to our original curve means shifting back by $-b/2a$ and the vertex's $x$-coordinate shifts along with it to the desired $x = -b/2a$.

And so the original question is answered: the $x$-coordinate of the vertex is at $x = -b/2a$.

$1$. The curve we end up with after using $d = -b/2a$ is:

$$x \mapsto ax^2 + 0 + \frac{b^2}{4a} - \frac{b^2}{2a} + c = ax^2 - \frac{b^2}{4a} + c$$

For what $x$ values does this return $0$? Setting this equal to $0$ and multiplying through by $4a$:

$$4a^2 x^2 - b^2 + 4ac = 0 \Longrightarrow x^2 = \frac{b^2 - 4ac}{4a^2} \Longrightarrow x = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$$

Shifting back by the $-b/2a$ now gives the roots found in the quadratic equation:

$$x = \frac{-b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}$$

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