Where does $-b/2a$ come from?

Question

Ive came across the idea to find the minimum/maximum of a parabola , I keep running into the formula $-b/2a$ i wanted to know where does this actually come from ? I know that the parabola has to be opening upwards to have a min value , and on the other hand the parabola has to be opening downward to have a maximum value.A few questions I have would be , is $-b/2a$ derived from a basic quadratic representation: $ax^2 + bx + c$? Not only just quadratic functions , i want to know the logic between all polynomial functions and their lowest point(how are they found , using given coefficient's a,b,c ......).

Answer 1

$y = x^2$ is your basic parabola. It's "tip" is at $(0,0)$.

We can make it wider or narrower or even turn it upside down but scaling it by $a$ so $y = ax^2$. It's the same shape but narrower (if $|a|< 1$) or wider (if $|a\ > 1$) and maybe pointing downward (if $a < 0$). The tip is still at $(0,0)$.

You can shift the tip up or down by adding a $e$. $y=ax^2 + e$. It's the same shape but now the tip is at $(0,e)$.

And finally we can shift the tip left or right by subtracting $d$ from the $x$. So $y = a(x - d)^2 + e$ is the same shape but now the tip is at $(d,e)$.

And so all quadratic equations can be written as $y=a(x-d)^2 + e$ and that tells us that the tip is at $(d,e)$.

But if we exapand that out we get $y = ax^2 -2adx + (d^2 +e)$. As a polynomial this is $y = ax^2 + bx + c$ where $a = a; b = -2ad;$ and $c = (d^2 + e)$.

Or in other word $d = -\frac b{2a}$ and $e = c -d^2 =c - \frac {b^2}{4a^2} \frac {4ac - b^2}{4a^2}$.

So if $y =f(x) = ax^2 + bx + c = a(x +\frac b{2a})^2 +c -\frac {b^2}{4a^2}=a(x -d)^2 + e$ is a quadratic equation it's graph is a parabola with a tip at $(d,e) = (-\frac b{2a}, \frac {4ac - b^2}{4a^2})$.

Now, admittedly we usually figure this out by going in the other direction.

We start with $y = ax^2 + bx + c$ and we complete the square to get

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

Answer 2

It comes from completing the square. You write $$ax^2+bx+c=a\left(x^2+\frac ba x +\frac ca\right)=a\left(\left(x+\frac b{2a}\right)^2+\frac ca-\frac {b^2}{4a^2}\right)$$ and the maximum or minimum is at $x=-\frac b{2a}$ depending on the sign of $a$

Answer 3

You can go calculus way. You might know that in the case of parabola, the slope of tangent at the point where the parabola attains maximum or minimum is zero.

Hence for a general parabola $f(x)=ax^2+bx+c$, the slope of tangent to this curve is given by its derivative.

Hence $f'(x)=2ax+b$

But we know that at the point of minimum say $x=x_0$ the slope of tangent is 0 which means $f'(x_0)=0$

Hence we have $$f'(x_0)=2ax_0+b=0\Rightarrow x_0=\frac {-b}{2a}$$

Q. E. D

Answer 4

The solution of

$$ a x^2 + b x +c = 0 $$

has two parts $ -b/ 2a ,\, \sqrt{b^2-4ac}/2a $ ( which can be found by completing the square)

and by also by differentiation ( but this can be shown only after starting calculus class).

To find maximum or minimum position on x-axis you with a simple procedure you set its derivative of $ a x^2 + b x + c $ to zero.

That way you get $ 2ax+b=0,\, x= -b/2a$.

A graph reveals this general x-position always for its vertex location whether the coefficient is positive or negative. The corrosponding $y$ position is $ c-b^2/4a$

Answer 5

Here is a geometric explanation:

• Write $y = ax^2+bx+c$ as $ y = x(ax+b) + c$
• $y = x(ax+b)$ is a parabola with zeroes at $x_1 = -\frac{b}{a}$ and $x_2 = 0$
• Because of the symmetry of the parabola, the parabola attains its vertex exactly in the middle between $x_1$ and $x_2$: $$x_E = \frac{x_1+x_2}{2} = \frac{-\frac{b}{a}+0}{2}= -\frac{b}{2a}$$ Now, note that $x_E$ does not change, if we shift the parabola by $c$ $$y = x(ax+b) \stackrel{shift\,by\,c}{\longrightarrow}y = x(ax+b) + c$$