When completing the square:

Question

5x^2 + 10x - 20
5 ( x^2 + 2x) -20 = 0
5 ( x + 1)^2 -15 = 0

Did I do that right? If so I have another question. When it's in this form, -15 is the y-coordinate of the vertex and 2 the x coordinate of the vertex? If so isnt the x cooridinate of the vertex -2 because you flip the sign?

Answer 1

Hmmmm.

5 ( x^2 + 2x) -20 = 0
add 20 to each side.
5 ( x^2 + 2x)=20

divide each side by 5
x^2+ 2x =4
add one to each side to complete the square

x^2+2x+1=5 factor, and subtract five from each side
(x+1)^1-5=0
I think this is different from yours. Check it.

I don't know how you are getting a vertex. Is there an equation at says y is equal to something? I don't understand your questions.

Answer 2

5 x^2 + 10 x - 20 = y
first divide everything by 5, the coef of x

x^2 + 2 x - 4 = y/5
now add 4 to both sides

x^2 + 2 x = y/5 +4
now take half the coef of x, square it, and add to both sides
(2/2)^2 =1
so
x^2 + 2 x + 1 = y/5 + 5
NOW the left is
(x+1)^2 = y/5 + 1

to the right and left of x = -1 where the left is 0
so the x coordinate of the vertex is x = -1

What is y then?
y/5 + 1 = 0
y = -1(5) = -5
so the vertex is at
(-1,-5)

Answer 3

This is how I did this:
5x^2+10x=20 <--- I left that out, sorry
5x^2+10x-20=0
5(x^2+2x)-20=0
5(x^2+2x-1+1)-20+5
5(x+1)^2-15=0

So I was thinking that (-1,-15) was the vertex because it is in the form y=a(x-h)^2+k

??
Does this make sense?

Answer 4

If you just have an equation in x with no y, you just have two points where the left is equal to the right, no parabola, no vertex.