Question

I was making a formula to get the critical point of a quadratic without calculus because i thought it would be funny, but it only works if you follow those two conditions outside of the formula. Is there a way to incorporate those into the formula?

Answer 1

Btw, this simplifies to -b/2a

Funny thing is, my 7th grade teacher had us memorize this formula upon learning quadratics lol

Answer 2

So what you’ve done here is take the average of the roots (aka find the midpoint of the roots) which is where the vertex is located.

I’m not understanding those conditions you created. If the quadratic has 2 roots, the vertex is between them and your formula is fine. If it has 1 root you will find it twice from the formula, add it to itself then divide by 2 yielding it again (but in that case the vertex is on the x axis where the root is so you don’t need to use this formula). If the quadratic formula yields no valid real roots then it doesn’t cross the x axis, not sure why you then think you can change the sign of a or set x=0?

Answer 3

The easiest way to do this is completing the square.

ax^2 + bx + c = a[(X+b/2a)^2 -(b^2 -4ac)/4a^2].

This has a minimum when X=-b/2a (i.e. when X+b/2a=0).

At this point, y=c-b^2/4a

Answer 4

Also, is there a way to incorporate the whole process of putting the x value back into the original quadratic to get the Y into the formula besides from just doing


a * formula^2 + b * formula + c

because this wont work if the quadratic has subtraction

Answer 5

First - just place module on -4ab
Second - don’t know, need to know when and why that doesn’t work so x = 0.

Answer 6

So, I'm not quite sure I understand what's going on here:

>but it only works if you follow those two conditions outside of the formula

You have the two roots (-b+√(b^(2)\-4ac))/(2a) and (-b+√(b^(2)\-4ac))/(2a), and you add them up.  When you add them like you've shown, you get x = (-2b)/(2a).  Dividing this by 2 gives you x = -b/(2a).

x = -b/(2a) is an extremely well-known formula for the location of a parabola's vertex, and to my knowledge doesn't fail or have restrictive conditions.  How are you finding that these additional steps are necessary?

Answer 7

Praise OP for thinking for themselves and exploring a question they found interesting. Boo anyone who discourages this kind of exploration just because its not new and has a much simpler solution.

Answer 8

This is just the well known -b/2a formula with many extra unnecessary steps. You're not "inventing" any math just because you call things by a longer more complicated name, and you'll definitely have no lack in inventing math in a field such as algebra that is complete and known to humanity for maaany years.

Answer 9

If else lol

Answer 10

why would this not work when roots don’t exist? even if complex roots, the discriminant cancels out and gives -b/2a ALWAYS.

EDIT: okay i realise it isn’t intuitive for you to take the midpoint of two complex numbers.

So i’ll introduce a bit of conic sections. A parabola can always be represented in the form

(x-h)\^2 = 4a(y-k).

vertex of the parabola are (h,k). and focus is (h,k+ a). This is only for a vertical parabola, no rotated or horizontal parabolas work like this. Also, you havent taken them into account so this satisfies the normal quadratic equation well. Try proving how you can go from ax\^2 + bx + c to here and how vertex and focus are found. You might have trouble if you don’t know exactly what a parabola is. A parabola is the locus of a point, the ratio of whose distance from a fixed line and a fixed point is 1. This ratio is called the eccentricity.

Now let’s talk a little about complex numbers. Think of them as vectors, and the real number part of the complex number is always  on the x axis or the real line. You can almost think of it as being multiplied with the ihat unit vector. And think of iota or sqrt(-1) as the jhat unit vector. In the end the roots or a parabola with D < 0 will end up looking like m +/- i(n). m is the real part and n is the imaginary scalar. Now the midpoint of the two points that these vectors make them land at is still (m,0). Here an important distinction however is that the origin of a complex plane is not necessarily origin of the cartesian plane in which you plot the parabola.

IF THAT RAMBLE DOESN’T MAKE SENSE, read this. Remember how at the critical point we can say that the parabola is symmetric about it. Why not try doing that. Put in x = 0 in a parabola with no real roots. You’ll know the y intercept of it. Now let’s say the vertex has the coordinates (h,k). Then if you move h distance along the x axis backwards you reach x = 0. And if you move h distance forwards, you reach x = 2h. However, both at x = 0 and x + 2h the parabola has the same y value. So if you know the value of y at x = 0, you know the value of k which is just half of it. What about h? Well at x = 0, ax\^2 + bx + c = c.  Now if you solve for ax\^2 + bx + c = c, you get ax\^2 = -bx. So x = 0 or x = -b/a. Here 2h = -b/a. So h is -b/2a.

I put the best proof in the end because im dumb.