Hi, can someone help me with finding the range of this function: “(x+2) / (x-1)”

My teacher suggested substituting in very large values of x and very small values to see what y values the graph tends towards but I’m not sure if this applies for this example. Is there a fool-proof method of working out the range of any function?

(Substituting in large values of x (100,000) gives me 1.00003 so the asymptote can’t be at y = 1)

Is it relevant to know what translations have been applied to the function? Thanks!

My teacher suggested substituting in very large values of x and very small values to see what y values the graph tends towards but I’m not sure if this applies for this example. Is there a fool-proof method of working out the range of any function?

(Substituting in large values of x (100,000) gives me 1.00003 so the asymptote can’t be at y = 1)

Is it relevant to know what translations have been applied to the function? Thanks!

Original post by subbhy

Hi, can someone help me with finding the range of this function: “(x+2) / (x-1)”

My teacher suggested substituting in very large values of x and very small values to see what y values the graph tends towards but I’m not sure if this applies for this example. Is there a fool-proof method of working out the range of any function?

(Substituting in large values of x (100,000) gives me 1.00003 so the asymptote can’t be at y = 1)

Is it relevant to know what translations have been applied to the function? Thanks!

My teacher suggested substituting in very large values of x and very small values to see what y values the graph tends towards but I’m not sure if this applies for this example. Is there a fool-proof method of working out the range of any function?

(Substituting in large values of x (100,000) gives me 1.00003 so the asymptote can’t be at y = 1)

Is it relevant to know what translations have been applied to the function? Thanks!

Its obv related to a reciprocol and its worth making the fraction proper so

(x+2)/(x-1) = (x-1 + 3)/(x-1) = 1 + 3/(x-1)

The -1 in the translation of x makes no difference to the range. So think about the range of 1/x and how this then differs.

Original post by mqb2766

Its obv related to a reciprocol and its worth making the fraction proper so

(x+2)/(x-1) = (x-1 + 3)/(x-1) = 1 + 3/(x-1)

The -1 in the translation of x makes no difference to the range. So think about the range of 1/x and how this then differs.

(x+2)/(x-1) = (x-1 + 3)/(x-1) = 1 + 3/(x-1)

The -1 in the translation of x makes no difference to the range. So think about the range of 1/x and how this then differs.

Great thank you

Is there a general way of approaching questions like this?

Original post by subbhy

Great thank you

Is there a general way of approaching questions like this?

Is there a general way of approaching questions like this?

I always divide by x on top and bottom of fraction to get something like:

y=(1 + 2/x) / (1 - 1/x)

as x-->infinity, 2/x and 1/x tend to zero

so as x-->infinity, y-->1

so there must be an asymptote at y=1

(edited 3 months ago)

Original post by subbhy

Great thank you

Is there a general way of approaching questions like this?

Is there a general way of approaching questions like this?

Not really but you should really understand how transformations (translations, scaling, ...) affect elementary functions so that covers a reasonable number of quesitons like this. Tbh, a good thing to do would be to simply sketch functions (even when the question doesnt ask for it) and if you can do that, the range, domain, ... is fairly trivial. Here youd have a vertical asymptote at x=1, horizontal asymptote at y=1 and the usuall reciprocal curves ((scaled by 3) in the translated Q1 and Q3.

An alternative way to get the horizontal asymptote (point to exclude in the range) is to rearrange

y = x+2 / x-1

so

x = ...

Youd have a y-1 on the denom which gives the horiz asymptote as there is no x value which corresponds to y=1, so the inverse map.

(edited 3 months ago)

Original post by mqb2766

Not really but you should really understand how transformations (translations, scaling, ...) affect elementary functions so that covers a reasonable number of quesitons like this. Tbh, a good thing to do would be to simply sketch functions (even when the question doesnt ask for it) and if you can do that, the range, domain, ... is fairly trivial. Here youd have a vertical asymptote at x=1, horizontal asymptote at y=1 and the usuall reciprocal curves ((scaled by 3) in the translated Q1 and Q3.

An alternative way to get the horizontal asymptote (point to exclude in the range) is to rearrange

y = x+2 / x-1

so

x = ...

Youd have a y-1 on the denom which gives the horiz asymptote.

An alternative way to get the horizontal asymptote (point to exclude in the range) is to rearrange

y = x+2 / x-1

so

x = ...

Youd have a y-1 on the denom which gives the horiz asymptote.

Great thanks! I was looking to sketch the curve but didn’t think of rewriting it as a proper fraction.

Original post by yzven

I always divide by x on top and bottom of fraction to get something like:

y=(1 + 2/x) / (1 - 1/x)

as x-->infinity, 2/x and 1/x tend to zero

so as x-->infinity, y-->1

so there must be an asymptote at y=1

y=(1 + 2/x) / (1 - 1/x)

as x-->infinity, 2/x and 1/x tend to zero

so as x-->infinity, y-->1

so there must be an asymptote at y=1

Original post by mqb2766

Not really but you should really understand how transformations (translations, scaling, ...) affect elementary functions so that covers a reasonable number of quesitons like this. Tbh, a good thing to do would be to simply sketch functions (even when the question doesnt ask for it) and if you can do that, the range, domain, ... is fairly trivial. Here youd have a vertical asymptote at x=1, horizontal asymptote at y=1 and the usuall reciprocal curves ((scaled by 3) in the translated Q1 and Q3.

An alternative way to get the horizontal asymptote (point to exclude in the range) is to rearrange

y = x+2 / x-1

so

x = ...

Youd have a y-1 on the denom which gives the horiz asymptote.

An alternative way to get the horizontal asymptote (point to exclude in the range) is to rearrange

y = x+2 / x-1

so

x = ...

Youd have a y-1 on the denom which gives the horiz asymptote.

If there is an asymptote at y = 1, how comes when x = 100 then y = 1.0303….. ?

I’m a bit confused by that

Original post by subbhy

If there is an asymptote at y = 1, how comes when x = 100 then y = 1.0303….. ?

I’m a bit confused by that

I’m a bit confused by that

Try larger values of x and you'll notice that y gets closer to 1 but never reaches it i.e. y=1 is an asymptote.

Or you could think about it like this : as x gets really big, the +2 and -1 become insignificant so you get something close to x/x = 1. You never actually reach 1 because the +2 and -1, even though they are insignificant, keep you away from it.

Original post by subbhy

If there is an asymptote at y = 1, how comes when x = 100 then y = 1.0303….. ?

I’m a bit confused by that

I’m a bit confused by that

Thats pretty much what you should expect from a horizontal aymptote? So as |x| gets large, y approches (but never reaches) the asymptotic value. Again, get in the habit of trying to sketch functions and it will become almost second nature.

Original post by Notnek

Try larger values of x and you'll notice that y gets closer to 1 but never reaches it i.e. y=1 is an asymptote.

Or you could think about it like this : as x gets really big, the +2 and -1 become insignificant so you get something close to x/x = 1. You never actually reach 1 because the +2 and -1, even though they are insignificant, keep you away from it.

Or you could think about it like this : as x gets really big, the +2 and -1 become insignificant so you get something close to x/x = 1. You never actually reach 1 because the +2 and -1, even though they are insignificant, keep you away from it.

Ah! I kept getting myself confused thinking the graph was going upwards so it had already passed 1 by giving me an output of 1.00003. Thanks!

Original post by mqb2766

Thats pretty much what you should expect from a horizontal aymptote? So as |x| gets large, y approches (but never reaches) the asymptotic value. Again, get in the habit of trying to sketch functions and it will become almost second nature.

Thanks: I was getting into a muddle thinking the graph went upwards and so it had already passed y=1 by giving me an output of >1. Thanks for the help!

It does become tricky trying to sketch with the examples my teacher gives 😣

Original post by subbhy

Thanks: I was getting into a muddle thinking the graph went upwards and so it had already passed y=1 by giving me an output of >1. Thanks for the help!

It does become tricky trying to sketch with the examples my teacher gives 😣

It does become tricky trying to sketch with the examples my teacher gives 😣

The more you try, the better youll get. Sketches should really reflect what you understand about the functions properties, so if you have difficulty sketching it, you may find other stuff hard.

Note an alternative (simpler) way to write this function is

(y-1)(x-1)=3

This makes both translations (vertical/horizontal asymptotes) clear. If y=1 or x=1, there is no value for the other term which gives 3 when you multiply by 0.

(edited 3 months ago)

Original post by mqb2766

The more you try, the better youll get. Sketches should really reflect what you understand about the functions properties, so if you have difficulty sketching it, you may find other stuff hard.

Note an alternative (simpler) way to write this function is

(y-1)(x-1)=3

This makes both translations (vertical/horizontal asymptotes) clear.

Note an alternative (simpler) way to write this function is

(y-1)(x-1)=3

This makes both translations (vertical/horizontal asymptotes) clear.

Thanks a lot for the help!

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