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

When completing the square:
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 - , Wednesday, January 23, 2008 at 10:09pm

    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.

  • Math - , Wednesday, January 23, 2008 at 10:15pm

    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)

  • Math - , Wednesday, January 23, 2008 at 10:22pm

    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?

  • Math - , Thursday, January 24, 2008 at 6:46am

    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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