Transformation/how to graph this function

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Tattatty

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Sorry I'm not sure how to type out the question so I just uploaded a photo.
Could someone please help me with this question : Sketch C, stating the coordinates of any vertices and equations of any asymptotes. C is the equation shown in the photo.
I think this is a transformation question?? It looks like a hyperbola equation but the x-3 and y+1 changes it to something else. It's not an ellipse either because there is a minus sign. Plz help me!!! Thank you
 
Harry_the_cat said:
A question first.
How does the graph of y=(x-3)^2 compare to the graph of y=x^2 ?
Click to expand...
There is a translation in the positive X direction by 3 units for y=(x-3)^2? The thing is, I'm not sure if the oblique asymptotes can undergo horizontal or vertical translations. Thx for replying sir
 
A translation affects every point on the plane, so that includes the asymptotes. They just "slide" like everything else.

You might find it easier to think of it as moving the origin. Just draw the hyperbola x^2/2^2 - y^2/3^2 = 1 on a grid, and then label the center as (3, -1), as it has to be after translation to (x-3)^2/2^2 - (y+1)^2/3^2 = 1. Then place the axes appropriately.

Or, if you prefer, plot the point (3, -1), and then draw asymptotes starting at that point with the appropriate slopes. Translation doesn't change slope. Also plot the vertices and the curve itself, as an exact copy of what they would be centered at the origin.
 
Dr.Peterson said:
A translation affects every point on the plane, so that includes the asymptotes. They just "slide" like everything else.

You might find it easier to think of it as moving the origin. Just draw the hyperbola x^2/2^2 - y^2/3^2 = 1 on a grid, and then label the center as (3, -1), as it has to be after translation to (x-3)^2/2^2 - (y+1)^2/3^2 = 1. Then place the axes appropriately.

Or, if you prefer, plot the point (3, -1), and then draw asymptotes starting at that point with the appropriate slopes. Translation doesn't change slope. Also plot the vertices and the curve itself, as an exact copy of what they would be centered at the origin.
Click to expand...
Thx so much for replying!!! Ok I think I understand what you mean now. However, I have just 1 more question, if I draw the basic graph as you said, and the asymptotes are y=±3x/2, what do they become after transformation? This is because the question wants me to write down their equations, so I can't just draw the centre with asymptotes passing through
 
Note that the transformation of the hyperbola replaced x with (x-3) and y with (y+1).

Do the same to the equations of the asymptotes! That will do the translation you need.

Alternatively, you could use what I said about slope. What is the equation of a line through (3, -1) with slope 3/2?

If you use the point-slope form, you'll find that these two methods are essentially the same. That provides an interesting insight on the point-slope form.
 
Dr.Peterson said:
Note that the transformation of the hyperbola replaced x with (x-3) and y with (y+1).

Do the same to the equations of the asymptotes! That will do the translation you need.

Alternatively, you could use what I said about slope. What is the equation of a line through (3, -1) with slope 3/2?

If you use the point-slope form, you'll find that these two methods are essentially the same. That provides an interesting insight on the point-slope form.
Click to expand...
Sir, you are the greatest. Thank you for the great help
 
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