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I'm working on a lesson plan for my students regarding completing the square for a parabola, and I've done the following:

$$\begin{align}ax^2+bx+c &= a\left(x^2+{b\over a}x\right) + c \\ & = a\left(x^2+{b\over a}x+{b^2\over 4a^2}-{b^2\over 4a^2}\right)+c \\& = a\left(x^2+{b\over a}x+{b^2\over4a^2}\right) - a{b^2\over4a^2} + c \tag{$*$}\\ &= a\left(x^2-2hx+h^2\right)-ah^2+c \tag{Let $h=-{b\over 2a}$}\\ &=a(x-h)^2 + c-ah^2 \\ &= a(x-h)^2+k.\tag{Let $k=c-ah^2$} \end{align}$$

This gives us the standard form of a parabola we all know and love to pick out the vertex of the parabola, but how am I going to justify the selection of $\displaystyle h = -{b\over2a}$ over $\displaystyle h={b\over 2a}$? The selection is to give us the axis of symmetry $\displaystyle x=-{b\over2a}$, of course, but how do I explain why the negative gives us the axis of symmetry? I found a demonstration here, but for my purposes it seems like a chicken-and-egg situation: I haven't discussed the quadratic formula yet, which I will be proving the quadratic formula as part of the lecture, but I need to discuss completing the square first. So it seems kind of silly to use the quadratic formula to prove a result that will be used to prove the quadratic formula.

I know that by definition, the axis of symmetry is the line where any two $x$ values of the quadratic equation $ax^2+bx+c$ have the same $y$ value. But I'm stuck with trying to find these $x$-values otherwise. Do I need to reorganize my lesson plan?

Community
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Decaf-Math
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5 Answers5

8

The axis of symmetry is such that $f(h-x)=f(h+x).$

Hence

$$a(h-x)^2+b(h-x)+c=a(h+x)^2+b(h+x)+c$$

or

$$4ahx+2bx=0.$$

As this must hold for all $x$,

$$2ah+b=0.$$

(The extra solution $x=0$ trivially comes from $f(h)=f(h)$ which holds for any $h$.)

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    I really like this answer. For those coming here in the future, it really helps to draw a picture. From my experience learning algebra, there's a convention to write "$x = -{b\over2a}$". However, we need to set aside this habit in order to completely understand the power of this answer. Rather, we're trying to get $h=-{b\over 2a}$ instead. In this case, $x$ is the distance away from $h$ on the $x$-axis, and these ($h-x$ and $h+x$) $x$-values are what we plug into the quadratic functions to yield the desired result. (Just an explanation to help others who might think like me) :) Thanks so much! – Decaf-Math Mar 02 '17 at 11:30
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First show that the parabola $y=ax^2$ is symmetrical about the $y-$axis ($x=0$).

Displacing it by $c$ vertically gives $y=ax^2+c$ which is also symmetrical about $x=0$.

Moving it by $h$ to the left gives $y=a(x-h)^2+c$ which is symmetrical about $x-h=0$ or $x=h$.

From the equation given

$$ax^2+bx+c=a\left(x+\frac b{2a}\right)^2+c-\frac {b^2}{4a}$$ which is symmetrical about $x+\dfrac b{2a}=0$ or $x=-\dfrac b{2a}$.

Hypergeometricx
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2

Regardless of what choices are involved, completing the square leads to

$$ a\left(x + \frac{b}{2a} \right)^2 + \left(c - \frac{b^2}{4a} \right) $$

Since you were trying to arrive at a formula of the form

$$ a (x-h)^2 + k $$

The choice of $h = -\frac{b}{2a}$ is nothing more than pattern matching

$$ x - h = x + \frac{b}{2a} $$

0

The stationary point occurs at $x=-\frac{b}{2a}$.

Therefore demonstrate that $$f(-\frac{b}{2a}+\epsilon)= f(-\frac{b}{2a}-\epsilon)$$ for all $\epsilon$.

Vincent
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David Quinn
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$$ ax^2+bx+c = a(x- \alpha)(x- \beta) $$

First derivative of $ y = ax^2+bx+c,\quad 2ax+b = 0 \rightarrow \Sigma =\dfrac{-b}{a}$

Looking at sum and difference of roots $x_1,x_2$ is equally advantageous.

$$ \Sigma = (\alpha + \beta)= \frac{-b}{a}; \quad \Delta =\, |(\alpha - \beta)| = \frac{\sqrt{b^2-4 a c}}{a}$$

where

$$ { \alpha}=\frac{-b+\sqrt{b^2-4 a c}}{2 a} ; \quad { \beta}=\frac{-b -\sqrt{b^2-4 a c}}{2 a}; $$

Narasimham
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