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
I think this is different from yours. Check it.
I dont know how you are getting a vertex. Is there an equation at says y is equal to something? I dont understand your questions.
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
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
This is how I did this:
5x^2+10x=20 <--- I left that out, sorry
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?
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.
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