Sketching Hyperbolas! (1 Viewer)

[fatima96]

Hey people!
So like yeah I forgot how to sketch hyperbolas like these ones :

 

[bladeys]

Change x+3/x+1 to 1 + [2/(x+1)] and graph from there

 

[Alkanes]

Change x+3/x+1 to 1 + [2/(x+1)] and graph from there

There is no polynomial division in 2U.

I'm not sure of 2U way, but you should know that there is a restriction on x = -1.

 

[bladeys]

There is no polynomial division in 2U.

I'm not sure of 2U way, but you should know that there is a restriction on x = -1.

Oh sorry,
Find the vertical asymptote which is x = -1
Find the horizontal asymptote which is y =1 (basically just find the limit of the function)
Find the y intercept which is 3
and then graph the usual shape of a hyperbola from there

 

[Fawun]

Hey people!
So like yeah I forgot how to sketch hyperbolas like these ones :

You make the denominator equal to 0 so x+1 = 0 so x = -1 (that's your vertical asymptote)
The horizontal asymptote is 1 because of the coefficients of x or you can find the limit of it.

Then you find the x and y intercepts by subbing in y = 0 and x = 0

and then graph it

I think.

 

[asianese]

The 2U way is easier IMO since you still need to find intercepts and everything with the function 1+2/(x+1), whereas if you do it straight off with x+3/(x+1) you get it.

 

[twinklegal19]

There is no polynomial division in 2U.

I'm not sure of 2U way, but you should know that there is a restriction on x = -1.

You don't need polynomial division for this simple transformation.

 

[D94]

Change x+3/x+1 to 1 + [2/(x+1)] and graph from there

Question, how is this more superior than to just sub in x = 0, y = 0, x -> ±infinity, and finding the vertical asymptote?

You will still need to do what I just said to the 2/(x+1) term. The only real advantage is that you'll see the horizontal asymptote immediately (although, just subbing in large numbers is still an effective method).

 

[asianese]

Question, how is this more superior than to just sub in x = 0, y = 0, x -> ±infinity, and finding the vertical asymptote?

You will still need to do what I just said to the 2/(x+1) term. The only real advantage is that you'll see the horizontal asymptote immediately (although, just subbing in large numbers is still an effective method).

^ Exactly what I wanted to say.

 

[twinklegal19]

Question, how is this more superior than to just sub in x = 0, y = 0, x -> ±infinity, and finding the vertical asymptote?

You will still need to do what I just said to the 2/(x+1) term. The only real advantage is that you'll see the horizontal asymptote immediately (although, just subbing in large numbers is still an effective method).

It's not really, just an alternative method for people to recognise it as simply a hyperbola shifted up one units and two to the left.