# Thread: How to translate x^2-4x = y^2+9y to be in the form x=a(y-k)^2 + h?

1. ## How to translate x^2-4x = y^2+9y to be in the form x=a(y-k)^2 + h?

The equation x^2-4x = y^2+9y is a parabola according desmos.com, but how can it be translated into vertex form for a parabola, x=a(y-k)^2 + h? The equation contains both x^2 and y^2 terms and I don't see how to complete a square for y^2+9y.

2. It isn't a parabola; it just looks sort of like one when you are zoomed in close. Try zooming out to see what it really is: a hyperbola.

Do you know how to complete the square (for BOTH variables) to put a hyperbola into standard form?

3. Originally Posted by Dr.Peterson
It isn't a parabola; it just looks sort of like one when you are zoomed in close. Try zooming out to see what it really is: a hyperbola.

Do you know how to complete the square (for BOTH variables) to put a hyperbola into standard form?
Ah, then the problem occurs when trying to complete the square of the y terms. The first step should be to take the 9 in 9y divide by 2 and then square the result, but this is 9/2 = 4.5 (x 4.5 = 20.25), a fractional answer. Yikes!

4. Originally Posted by mauricev
Ah, then the problem occurs when trying to complete the square of the y terms. The first step should be to take the 9 in 9y divide by 2 and then square the result, but this is 9/2 = 4.5 (x 4.5 = 20.25), a fractional answer. Yikes!
Yes, there will be fractions. And, later, irrational radicals. You can handle that, right?

Note that Desmos showed you a vertex, whose y-coordinate is not very pretty, so you should not be surprised. Ugly doesn't mean wrong! (But if this comes from a class where you have been coddled with nice numbers until now, you might want to check that you copied the problem right. Maybe it's meant to be nicer than this.)

5. Originally Posted by Dr.Peterson
Yes, there will be fractions. And, later, irrational radicals. You can handle that, right?

Note that Desmos showed you a vertex, whose y-coordinate is not very pretty, so you should not be surprised. Ugly doesn't mean wrong! (But if this comes from a class where you have been coddled with nice numbers until now, you might want to check that you copied the problem right. Maybe it's meant to be nicer than this.)
Ha, the question simply calls for identifying which type of graph it is and which way it opens. I was just taking it further because the completing the square step ran into fractions and I hadn't seen that before.

6. Originally Posted by mauricev
Ha, the question simply calls for identifying which type of graph it is and which way it opens. I was just taking it further because the completing the square step ran into fractions and I hadn't seen that before.
So, do you know how to answer that question without having to see a graph? You apparently thought it could be a parabola ... but having both x and y squared makes that impossible.

7. Originally Posted by mauricev
Ha, the question simply calls for identifying which type of graph it is and which way it opens. I was just taking it further because the completing the square step ran into fractions and I hadn't seen that before.
If the orientation of the graph is important to your identification, it will not be completely clear until you finish the work of Completing the Square. CAn't skip it. May as well get good at it now. You'll need it later.

8. Originally Posted by tkhunny
If the orientation of the graph is important to your identification, it will not be completely clear until you finish the work of Completing the Square. CAn't skip it. May as well get good at it now. You'll need it later.
I second this opinion, and state further that it would likely be very wise of you to do as many of these exercises as you can. They can be painful sometimes, and you don't want to freeze on the test. You want your brain to have as much "muscle memory" as possible!

9. Originally Posted by tkhunny
If the orientation of the graph is important to your identification, it will not be completely clear until you finish the work of Completing the Square. CAn't skip it. May as well get good at it now. You'll need it later.
The equation in standard form should be

((y-9/2)^2)/(65/4) - ((x-2)^2)/(65/4) = 1

The -x term is negative, so it opens up and down

10. Well, there you have it. Now, what's all this about it being a parabola and struggling with fractions? Good work.

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