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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Rose34 on March 08, 2020, 08:23:29 pm

Title: Hyperbola graphing?
Post by: Rose34 on March 08, 2020, 08:23:29 pm

How to graph the hyperbola of y=1/x-1?

Normally when it is like y=-1/2x+5 I know how to graph it but when the last value(5 or any number) is missing I do not know how to graph it

Thanks in advance!

Title: Re: Hyperbola graphing?
Post by: Ionic Doc on March 08, 2020, 09:17:55 pm

Quote from: Rose34 on March 08, 2020, 08:23:29 pm

How to graph the hyperbola of y=1/x-1?

Normally when it is like y=-1/2x+5 I know how to graph it but when the last value(5 or any number) is missing I do not know how to graph it

Thanks in advance!

Hey Rose,

x-1   simply means a horizontal shift on the x-axis to the right by 1 unit.
You usually equate the bottom part of the fraction to 0 and solve for x, to find the vertical asymptote.
So your vertical asymptote will be at x=1

For your other example of y=-1/2x+5
Your vertical asymptote will be at

2x+5=0
Therefore x= -5/2

Back to your original question, of y=1/x-1
The horizontal asymptote would be 0.

Edit: Forgot to mention, if you want to find your y-intercept, then let x=0
So y=1/0-1
y = -1
Therfore your y-intercept is (0,-1)

For your x-intercept you let y=0.
Therefore in this graph, there  is no x-intercept

Knowing this you can graph your hyperbola, it should look something like:
(https://i.imgur.com/L080UWN.png)
(make sure your hyperbolae never touch the vertical and horizontal asymptotes)

I hope this helped, also use this website called desmos to play around with graphs and see what they would look like, it really helps (here's the link:  https://www.desmos.com/calculator     

Title: Re: Hyperbola graphing?
Post by: Rose34 on March 08, 2020, 09:41:58 pm

Quote from: Ionic Doc on March 08, 2020, 09:17:55 pm

Hey Rose,

x-1   simply means a horizontal shift on the x-axis to the right by 1 unit.
You usually equate the bottom part of the fraction to 0 and solve for x, to find the vertical asymptote.
So your vertical asymptote will be at x=1

For your other example of y=-1/2x+5
Your vertical asymptote will be at

2x+5=0
Therefore x= -5/2

Back to your original question, of y=1/x-1
The horizontal asymptote would be 0.

Knowing this you can graph your hyperbola, it should look something like:
(https://i.imgur.com/L080UWN.png)
(make sure your hyperbolae never touch the vertical and horizontal asymptotes)

I hope this helped, also use this website called desmos to play around with graphs and see what they would look like, it really helps (here's the link:  https://www.desmos.com/calculator   

Thank you very much!!