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November 20, 2014

November 20, 2014

Posted by **Lena** on Wednesday, January 23, 2008 at 9:55pm.

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?

- Math -
**bobpursley**, Wednesday, January 23, 2008 at 10:09pmHmmmm.

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 dont know how you are getting a vertex. Is there an equation at says y is equal to something? I dont understand your questions.

- Math -
**Damon**, Wednesday, January 23, 2008 at 10:15pm5 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)

- Math -
**Lena**, Wednesday, January 23, 2008 at 10:22pmThis 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?

- Math -
- Math -
**Damon**, Thursday, January 24, 2008 at 6:46amIf 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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