*Hi I'm Bia.*I study Maths with a module in Spanish at the University of Warwick!

I absolutely **love** maths. So whether you just want a better grade or you're looking to study it at University level, I'm always here to help. I'll go through **many examples**, as from my previous tutoring experience, I find that practice makes perfect!

In my Spanish sessions I'll start by introducing an idiomatic phrase e.g 'to be totally amazed' is 'flipar en colores' which is literally- 'to flip colours', as I believe that these will not only** improve your essay writing/ oral**, but will prepare you for travelling to Spanish- speaking countries. I can also assist you with any grammatical difficulties.

I believe that everyone learns in a unique way and am ready to tailor to your needs. I'm **patient** and will** motivate** you to realise your potential!

*I look forward to meeting you!*

*Hi I'm Bia.*I study Maths with a module in Spanish at the University of Warwick!

I absolutely **love** maths. So whether you just want a better grade or you're looking to study it at University level, I'm always here to help. I'll go through **many examples**, as from my previous tutoring experience, I find that practice makes perfect!

In my Spanish sessions I'll start by introducing an idiomatic phrase e.g 'to be totally amazed' is 'flipar en colores' which is literally- 'to flip colours', as I believe that these will not only** improve your essay writing/ oral**, but will prepare you for travelling to Spanish- speaking countries. I can also assist you with any grammatical difficulties.

I believe that everyone learns in a unique way and am ready to tailor to your needs. I'm **patient** and will** motivate** you to realise your potential!

*I look forward to meeting you!*

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5from 36 customer reviews

David (Parent from Leatherhead)

September 29 2017

very good

David (Parent from Leatherhead)

September 25 2017

very helpful and understanding

Ann maria (Student)

August 16 2016

great :)

Afshan (Parent from London)

October 18 2015

Another beneficial session.

The asymptotes of a curve are the lines or curves that approach a given curve very closely without actually touching it. They make curve sketching much easier. You will probably only have asymptotes that are lines in which case you will have either horizontal asymptotes or vertical asymptotes.

__Horizontal asymptotes__

To find the horizontal asymptotes, we want to see the behaviour of the curve as x tends to infinity. So we look at the order of both the numerator and denominator:

**Case 1**: If the numerator's order is greater than the denominator's (e.g if the function is f(x) = x^{3}/(2x^{2}+1) since x^{3 }has order 3 and x^{2} has order 2) then there will be no horizontal asymptotes.

**Case 2**: If the numerator's order is less than the denominator's (e.g if the function is f(x) = 1/x) then the horizontal asymptote will be y=0

**Case 3**: If the numerator's order is the same as the denominator's, e.g the function f(x) = (2x^{2 }- 4) / (x^{2} - 1), then we look at the highest order term for both sides and look at the coefficient of it. In other words, we ignore all other terms besides 2x^{2} and x^{2} so 2x^{2}/x^{2} = 2. So in this case the asymptote is y=2.

But why does this work? For the horizontal asymptotes we are effectively looking at what our curve will do as x tends to infinity. So in Case 1, when the numerator is bigger than the denominator, as x tends to infinity, y tends to infinity so there is no asymptote. In Case 2, as x tends to infinitiy, we find that y tends to zero. In case 3, as x tends to infinity, all of the smaller terms become insignificant. Let's take a closer look. In order to fully understand, we can substitute x=1000 (or any large number) to roughly see what happens.

**Case 1:** f(x) = x^{3}/(2x^{2}+1). Setting x=1000, we get f(1000)= 1000^{3}/(2(1000)^{2}+1) = 500 and if we set x= 10000 then f(x)=5000 so there is no horizontal asymptote here.

**Case 2:** f(x) = 1/x. Setting x = 1000 we get f(1000)= 0.001 which is close to zero. Similarly, f(10000) = 0.0001 so as we increase x, f(x) tends to zero. Which is why the asymtote is y=0

**Case 3:** f(x) = (2x^{2 }- 4) / (x^{2} -1). f(1000) = (2(1000)^{2 }- 4) / ((1000)^{2 } -1)= 1.999998 which is roughly 2. If we increase x even more then f(10000)= 1.99999998 which is even closer to 2. Which is why the asymptote is y=2

__Vertical asymptotes:__

Here we are effectively looking at the behaviour of the curve as f(x)=y tends to infinity. The only times there will be vertical asymptotes is when the order of the denominator of the fraction is greater than the numerator's. Since f(x) will tend to infinity, the denominator of our curve will tend to zero. For example, let's consider the curve f(x) = (2x^{2 }- 4) / (x^{2} -1)

Here we simply set the denominator x^{2 } -1=0 to find x=1 or x= -1 and these are the vertical asymptotes.

The asymptotes of a curve are the lines or curves that approach a given curve very closely without actually touching it. They make curve sketching much easier. You will probably only have asymptotes that are lines in which case you will have either horizontal asymptotes or vertical asymptotes.

__Horizontal asymptotes__

To find the horizontal asymptotes, we want to see the behaviour of the curve as x tends to infinity. So we look at the order of both the numerator and denominator:

**Case 1**: If the numerator's order is greater than the denominator's (e.g if the function is f(x) = x^{3}/(2x^{2}+1) since x^{3 }has order 3 and x^{2} has order 2) then there will be no horizontal asymptotes.

**Case 2**: If the numerator's order is less than the denominator's (e.g if the function is f(x) = 1/x) then the horizontal asymptote will be y=0

**Case 3**: If the numerator's order is the same as the denominator's, e.g the function f(x) = (2x^{2 }- 4) / (x^{2} - 1), then we look at the highest order term for both sides and look at the coefficient of it. In other words, we ignore all other terms besides 2x^{2} and x^{2} so 2x^{2}/x^{2} = 2. So in this case the asymptote is y=2.

But why does this work? For the horizontal asymptotes we are effectively looking at what our curve will do as x tends to infinity. So in Case 1, when the numerator is bigger than the denominator, as x tends to infinity, y tends to infinity so there is no asymptote. In Case 2, as x tends to infinitiy, we find that y tends to zero. In case 3, as x tends to infinity, all of the smaller terms become insignificant. Let's take a closer look. In order to fully understand, we can substitute x=1000 (or any large number) to roughly see what happens.

**Case 1:** f(x) = x^{3}/(2x^{2}+1). Setting x=1000, we get f(1000)= 1000^{3}/(2(1000)^{2}+1) = 500 and if we set x= 10000 then f(x)=5000 so there is no horizontal asymptote here.

**Case 2:** f(x) = 1/x. Setting x = 1000 we get f(1000)= 0.001 which is close to zero. Similarly, f(10000) = 0.0001 so as we increase x, f(x) tends to zero. Which is why the asymtote is y=0

**Case 3:** f(x) = (2x^{2 }- 4) / (x^{2} -1). f(1000) = (2(1000)^{2 }- 4) / ((1000)^{2 } -1)= 1.999998 which is roughly 2. If we increase x even more then f(10000)= 1.99999998 which is even closer to 2. Which is why the asymptote is y=2

__Vertical asymptotes:__

Here we are effectively looking at the behaviour of the curve as f(x)=y tends to infinity. The only times there will be vertical asymptotes is when the order of the denominator of the fraction is greater than the numerator's. Since f(x) will tend to infinity, the denominator of our curve will tend to zero. For example, let's consider the curve f(x) = (2x^{2 }- 4) / (x^{2} -1)

Here we simply set the denominator x^{2 } -1=0 to find x=1 or x= -1 and these are the vertical asymptotes.