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Help on Parabolas vertex form please =)
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Pinkpantherkat

by Pinkpantherkat at June 12, 2011

x=-y^(2)-2

and another one is

x= 1/2y(2) + 7y + 65/2

We are suppose to put the equation in vertex form. I'm slightly confused, i can get it into vertex form but then isn't the equation equal to x, so don't you have to change it to make it equal to y?

 

Thanks Steve204 for all your help getting me through Algebra 2Honors Sophomore year! I can't wait for summer. I'm taking Precal A next year to make it easier. At my school we learn Algebra in 8th grade so we can skip it Freshman year if we have good enough grades. And the classes at my school are different so we have B Classes, A Classes, and Honors Classes for the advanced kids. I had a bad fundamental in math which makes this wicked hard for me, (my 6th grade teacher couldn't pass the math test she made and gave to us!) So I thank you so much for helping me Steve 204! I am really grateful! =)

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Pink - Thanks for the kind words … I’m glad I was able to help.  The Vertex Form of a parabola can be written two ways:The first one is an x^2 parabola and it points upward or downward.The second equation is a y^2 parabola and it points right or left.Both of your examples are y^2 parabolas.This is already in Vertex Form.The vertex (h,k) = (0,-2)The graph looks like this:For the second equation you need to complete the square to get it in Vertex Form:, whew! Finally done.This is Vertex Form.(h,k) = (8,-7) and here is what it looks like:Have a GREAT SUMMER!
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Steve204 Steve204 June 12, 2011

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Oops! The correct vertex (h,k) of your first equation is:(h,k) = (-2, 0)I had it reversed.When you have a y^2 parabola you need to be very careful writing down (h,k) since is it opposite of what it would be for an x^2 parabola.
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Steve204 Steve204 June 12, 2011

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