Hello!

I'm Alex and I'm currently a **masters student in mathematics** at the university of Bristol. I've always had a real love for maths and I hope my tutorials can help you share that love.

I'm a friendly and patient tutor have a wide range of teaching experience: ranging from one-on-one maths tutoring to classroom assistance to my current job as a gymnastics coach.

I'm offering tuiton in **A-Level Maths and Further Maths**, as well as **GCSE Maths and Statistics**, for all exam boards.

__Sessions__

As a tutor, I prefer to take and **informal and relaxed approach**, so that as a student you get more out of our sessions. We'll focus on **whatever it is that you're finding tough**, whether that's exam technique, a particularly tricky method or topics from your syllabus you aren't sure of.

I will work with you to come up with a strategy that helps you get the most of of our sessions. A lot can be covered in 55 minutes and I want to help you be prepared for both your exams and whatever future study in maths you are planning.

__What next?__

If you have any questions then please drop me a message or book a** free 'Meet the Tutor' session. **Let me know what you're struggling with and what exam board you're studying and we can arrange tutoring that fits in with your schedule.

Hello!

I'm Alex and I'm currently a **masters student in mathematics** at the university of Bristol. I've always had a real love for maths and I hope my tutorials can help you share that love.

I'm a friendly and patient tutor have a wide range of teaching experience: ranging from one-on-one maths tutoring to classroom assistance to my current job as a gymnastics coach.

I'm offering tuiton in **A-Level Maths and Further Maths**, as well as **GCSE Maths and Statistics**, for all exam boards.

__Sessions__

As a tutor, I prefer to take and **informal and relaxed approach**, so that as a student you get more out of our sessions. We'll focus on **whatever it is that you're finding tough**, whether that's exam technique, a particularly tricky method or topics from your syllabus you aren't sure of.

I will work with you to come up with a strategy that helps you get the most of of our sessions. A lot can be covered in 55 minutes and I want to help you be prepared for both your exams and whatever future study in maths you are planning.

__What next?__

If you have any questions then please drop me a message or book a** free 'Meet the Tutor' session. **Let me know what you're struggling with and what exam board you're studying and we can arrange tutoring that fits in with your schedule.

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May 3 2016

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Paul (Parent)

May 5 2016

This is an example of a rational function. So to sketch it, we need to know three things:

1) where it crosses the x and y axis

2) Where its turning points are, if it has any

3) Where its asymptotes are, if it has any.

Let's start with 1):

The graph crosses the y axis when x=0. So if x=0, then y=4/1 = 4. So the graph crosses the y axis at (0,4)

The graph crosses the x axis when y=0, which means the numerator of the fraction (x^{2} + 4*x + 4)/(3*x + 2) = 0. We see that the numerator can be factorised into (x + 2)^{2}, which means that the numberator only equals zero when x = -2. So the graph crosses the x axis at (-2,0)

Now onto 2):

Using the quotient rule, dy/dx = (3*x^{2} + 4*x - 4)/(3*x + 2)^{2}

Now this is zero when the numerator is zero. We can factorise the numberator into (x+2)*(3*x - 2) to see that dy/dx is zero when x = -2 or x = 2/3. We could have also used the quadratic formula to work this out.

When x = -2, y = 0 as we already established. So this is where we cross the x axis and a turning point at the same time.

When x = 2/3, y = (2/3 + 2)^{2}/(3*(2/3) + 1) = 64/27

by differentiating again to find dy/dx we can classify each stationary point: (-2,0) is a maximum and (2/3,64/27) is a minimum.

Now for part 3)

An asymptote is when the graph "shoots off to inifinity". So it occurs when the denominator of the function is zero. Here that only occurs when (3*x+1) = 0, or when x = -1/3.

So we have the parts of our graph we need to draw it. The curve comes in from the bottom right, touches the x axis at (-2,0) to change direction, shoots off to negative infinity along the line x=-1/3 (but never crossing it). Emerges from positive infinity the other side of the line x=-1/3, turns at the point (2/3,64/27) and swoops off the top right corner.

This is an example of a rational function. So to sketch it, we need to know three things:

1) where it crosses the x and y axis

2) Where its turning points are, if it has any

3) Where its asymptotes are, if it has any.

Let's start with 1):

The graph crosses the y axis when x=0. So if x=0, then y=4/1 = 4. So the graph crosses the y axis at (0,4)

The graph crosses the x axis when y=0, which means the numerator of the fraction (x^{2} + 4*x + 4)/(3*x + 2) = 0. We see that the numerator can be factorised into (x + 2)^{2}, which means that the numberator only equals zero when x = -2. So the graph crosses the x axis at (-2,0)

Now onto 2):

Using the quotient rule, dy/dx = (3*x^{2} + 4*x - 4)/(3*x + 2)^{2}

Now this is zero when the numerator is zero. We can factorise the numberator into (x+2)*(3*x - 2) to see that dy/dx is zero when x = -2 or x = 2/3. We could have also used the quadratic formula to work this out.

When x = -2, y = 0 as we already established. So this is where we cross the x axis and a turning point at the same time.

When x = 2/3, y = (2/3 + 2)^{2}/(3*(2/3) + 1) = 64/27

by differentiating again to find dy/dx we can classify each stationary point: (-2,0) is a maximum and (2/3,64/27) is a minimum.

Now for part 3)

An asymptote is when the graph "shoots off to inifinity". So it occurs when the denominator of the function is zero. Here that only occurs when (3*x+1) = 0, or when x = -1/3.

So we have the parts of our graph we need to draw it. The curve comes in from the bottom right, touches the x axis at (-2,0) to change direction, shoots off to negative infinity along the line x=-1/3 (but never crossing it). Emerges from positive infinity the other side of the line x=-1/3, turns at the point (2/3,64/27) and swoops off the top right corner.

There are 12 + 8 = 20 beads in the bag in total.

If sasha draws a bead at random then she has 8 chances out of 20 possible beads that the bead is blue. So the probability that the bead is blue is 8/20 = 2/5

There are 12 + 8 = 20 beads in the bag in total.

If sasha draws a bead at random then she has 8 chances out of 20 possible beads that the bead is blue. So the probability that the bead is blue is 8/20 = 2/5