How do you find a vertex in a Parabola equation?

Question

So say it's y=(x-5)(x-3)

Answer 1

Multiply it out to get x^2 - 8x + 15

The x-coordinate of the vertex is x = -b/(2a), or 8/2 = 4.

Plug this in for x to get y.

Answer 2

Change the equation to something that looks like the usual equation for a parabola: Y = aX + b
You can do this by completing the square:
y=(x-5)(x-3)=x^2 -8x +15 + 0
I put the zero in the last equation because you need to add (and subtract) something to get the zero:
y = x^2 -8x +16 -1 = (x-4)^2 - 1
where I added 1 to the 15 but subtracted the 1 (-1) to get it in the form:
y=ax^2+b where
a=1
x-->x-4
b=-1
Now we know that the vertex is where the x-->x-4 term is zero, or at x=4 and y = ax^2+b = (1)(0)-1 = -1, or
at (4,-1). I hope that this helps

Answer 3

The first derivative gives you the slope of a line tangent to the curve. So y=(x-5)(x-3)=x^2-5x-3x+15=x^2-8x+15. The slope of the tangent line at a maximum or minimum is zero (it's horizontal). The first derivative is dy=2xdx-8dx, dy/dx=2x-8, if dy/dx is zero then 2x=8 and x=8/2=4. y=[(4)-(5)][(4)-(3)]=
(-1)x(1)=-1.
If x=-1, y=(-1-5)(-1-3)=(-6)(-4)=24

Somebody else check this, please. It's been a long day and I'm kind of loopy now.

Answer 4

let's say that your quadratic equation is
ax^2 + bx +c.
then your vertex vould be -b/2a.
hope it's helpful :D

by the way, ^2 means exponent 2. basically squared -__-x