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Translating Parabolas and Completing the Square

Date: 23 May 1995 19:29:57 -0400
From: Eva Jason
Subject: Translating Parabolas from the Origin.


The subject in my textbook is "parabolas translated from the origin" 
and standard equations. I was zipping right along until I came to 
the following question:

State the co-ordinates of the vertex, and sketch the graph 
of the following: y^2+8y+8x-8=0

I know I have to make this equation look like the stardard - 
either (x-h)^2 = 4p(y-k) or (y-k)^2 = 4p(x-h) - but I just 
can't make it look like either one.  Can you tell me what 
information I might be missing?

Thanks again for any help!
Eva Jason

Date: 24 May 1995 00:23:04 -0400
From: Dr. Sydney
Subject: Translating Parabolas from the Origin.

Dear Eva,

Hello again!  I'm glad we were able to help. 

You have taken the right first step in approaching this problem - 
you want to make the equation given to you fit into one of the two 
general equations, either the one with x to the second power and 
y to the first power or the one with x to the first power and 
y to the second power. Since the given equation, y^2 + 8y + 8x - 8 = 0, 
has x to the first power and y to the second power, we want to make 
it look like the general equation, (y-k)^2 = 4p(x-h), since that is 
the general equation that also has x to the first power and y to 
the second power. Does that make sense?

So, now we need only find out what h and k are.  First, let's get the 
x's and y's on different sides of the equation.  Put the constant 
with the x (you can put it with the y if you want, but things will be 
easier if you put it with the x). So, your rewritten equation is:

y^2 + 8y = -8x + 8

Now this is looking much closer to the general form, right?  
The left-hand side of the equation needs to be changed, though, and 
we change it by using a process called "completing the square."  
When you complete the square of an equation, say y^2 + ay (for any 
constant, a), you take the coefficient of the y term, divide by 2, 
and square it.  If you add this constant, (a/2)^2, to the original 
equation, you get y^2 + ay + (a/2)^2.  This factors into the equation, 
(y + a/2)^2.  This will make more sense in context, so let's complete 
the square on this problem.

To get the above equation so that we have something that looks like 
(y-k)^2 on the left, we complete the square on the left.  We don't 
want to change the equation's meaning, so anything we do on the left 
we must also do on the right.  So to complete the square, we take 
the coefficient of the y term, 8, divide by 2 (so we get 4), and square 
it, so we get 16.  Now we add 16 to both sides of the equation.  Our 
new equation is:

y^2 + 8y + 16 = -8x + 24

The left-hand side of the equation factors:

y^2 + 8y + 16 = (y + 4)^2

Now your equation is:

(y+4)^2 = -8x + 24

With only a few more steps that I'll let you do, you can put this 
equation into the general form, and then you will know what the vertex is.

I hope this makes sense to you. This method of completing the square is 
just a methodical way of finding out what number you should add to each 
side of the equation so the side with the variable to the second power 
factors into a square.  At the stage where you have the equation, 
y^2 + 8y = -8x +8, you know you have to add a constant to both sides 
of the equation so the left side will become a perfect square.  
Completing the square is just a nice method for figuring out what that 
constant should be.

I hope this helps.  Feel free to write back with any other questions. 
Good luck!
--Sydney, "Dr. Math"
Associated Topics:
High School Equations, Graphs, Translations

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