Inflection points theorem for quadratic functions.

[zketrouble]

Hey all,

I was bored as heck with my internet down and started scribbling some random math stuff. I'm sure what I've come up with is old news to most but I found some interesting properties in the following proof and I'm curious what my fellow physicsforums users can come up with to see what the causes are for the relationships between my proof and the age-old quadratic formula.

An inflection point occurs when the slope of a function equals zero. So for quadratic equations (and all other equations) of the form f'(x) = ax^2 + bx + c, f'(x) = 0 at inflection points.

d/dx ax^2 + bx + c = 0
2ax + b = 0
2ax = -b
x = -b/(2a).

It's interesting (though not at all surprising), that the terms in this are similar to what is seen in the quadratic formula -b +/- [(b^2 - 4ac)^1/2]/2a

Both expression have a -b and both have a 1/2a. Its not all that surprising that the two expressions would be related, but where is the direct link between the quadratic formula and the expression I came up with?

 

[HallsofIvy]

Hey all,

I was bored as heck with my internet down and started scribbling some random math stuff. I'm sure what I've come up with is old news to most but I found some interesting properties in the following proof and I'm curious what my fellow physicsforums users can come up with to see what the causes are for the relationships between my proof and the age-old quadratic formula.

An inflection point occurs when the slope of a function equals zero. So for quadratic equations (and all other equations) of the form f(x) = ax^2 + bx + c, f'(x) = 0 at inflection points.

This is not true, although it might be a typo. A critical point for a function is where the derivative is 0 (or does not exist). An inflection point is where the derivative changes from increasing to decreasing and so occurs where the second derivative is 0 (or does not exist).

d/dx ax^2 + bx + c = 0
2ax + b = 0
2ax = -b
x = -b/(2a).

It's interesting (though not at all surprising), that the terms in this are similar to what is seen in the quadratic formula -b +/- [(b^2 - 4ac)^1/2]/2a

Both expression have a -b and both have a 1/2a. Its not all that surprising that the two expressions would be related, but where is the direct link between the quadratic formula and the expression I came up with?

Yes, for a parabola, the derivative is 0 only at the vertex- and you can find that by completing the square.
$$ax^2+ bx+ c= a(x^2+ (b/a)x+ c/a)= a(x^2+ (b/a)x+ (b^2/4a^2)- b^2/4a^2+ c/a)$$
$$= a((x+ b/2a)^2+ c/a- b^2/4a^2)= a((x+ b/2a)^2- \frac{b^2- 4ac}{4a^2}$$

Now, if a> 0, that is the number $(b^2- 4ac)/(4a^2)$ plus something that is positive unless $x+ b/2a= 0$ and so has the value $-(b^2- 4ac)/(4a^2)$ as its minimum value at $x= -b/2a$. If a< 0, it's turned over and that point is a maximum.

Of course, the quadratic formula comes from applying "completing the square" to $ax^2+ bx+ c= 0$ so you get similar results.

 

[tiny-tim]

hey zketrouble!

(have a ± and try using the X² icon just above the Reply box )

… Both expression have a -b and both have a 1/2a. Its not all that surprising that the two expressions would be related, but where is the direct link between the quadratic formula and the expression I came up with?

ax² + bx + c = 0

-> a[(x + b/2a)² + (4ac - b²)/4a²] = 0

 

[zketrouble]

This is not true, although it might be a typo. A critical point for a function is where the derivative is 0 (or does not exist). An inflection point is where the derivative changes from increasing to decreasing and so occurs where the second derivative is 0 (or does not exist).

Thanks for the correction, I always get the two mixed up :p

 

[atomthick]

The roots of the quadratic equation are found at equal distance of the point where minimum/maximum value is obtained for ax^2 + bx +c meaning.

As you well computed for x = -b/2a you get the minimum/maximum value of ax^ + bx + c. This point is in the middle of the two roots therefore you have to substract a value D to find the first root and to add a value D to find the second root.

x1 = -b/2a - D
x2 = -b/2a + D

 

[zketrouble]

Thanks everybody! I was almost tearing my hair out in wondering the connection.