why is the vertex of a parabola located at x=-b/2a?

Question

please, any suggestions?
mucho gracias=]

Answer 1

The standard form of a parabolic equation is
y = a(x - h)^2 + k
where (h, k) is the vertex. (h is the x-value, which you gave by x = -b/2a)

If it's in the form
y = ax^2 + bx + c
you would use completing the square to get it to the other form.

y = ax^2 + bx + c
y = a[x^2 + (b/a)x] + c
y = a[x^2 + (b/a)x + (b/2a)^2] + c + a(b/2a)^2
y = a[x + (b/2a)]^2 + c + (b^2/4a)

So....
h = -b/2a
(the x-value of the vertex)
and
k = c + (b^2/4a)
(the y-value of the vertex)

Answer 2

It comes from completing the square.

y = ax^2+bx+c
y = a(x^2+bx/a) + c
= a(x^2+bx/a +b^2/4a) + c-b^2/4
= a(x+b/2a)^2 + (c-b^2/4)

There are several ways of explaining why it's the vertex. First off, the vertex splits the parabola into 2 equal sections. It's the only point that yields a unique y value. The only unique y value comes when (x+b/2a)^2 = 0, which means x = -b/2a.

Second, if a is positive, it's concave up. In this case, the vertex yields the minimum y value. Since it's (x+b/2a) is squared, it's minimum value is 0 when x = -b/2a. If a is negative, the parable is concave down. It comes out the same but you use maximum value instead of minimum.

Lastly, if you've studied calculus, you can take the derivative and set it equal to 0 to find the extrema values. You'll get x = -b/2a.

Answer 3

It's a Calculus thing. If you take the derivative of a parabola in 'standard' form ax² + bx + c = y you get the 'slope' of a tangent to the parabola. If you set that equal to 0 (where the tangent is horizontal, which is at the vertex) and solve for x you get x = -b/2a.

You'll get there soon enough ☺

Doug

Answer 4

that tells you what the x-coordinate is. then you put the number you got for X in the original problem and find what Y is.

standard form- Ax+By=C

im doing the same thing in algebra

Answer 5

'cos
dy/dx = 0 when x=-b/2a