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Degree: Physics (Masters) - Manchester University

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You have surely met a teacher who is bad at teaching, or maybe one who understands less than the students. I did. So, back at school, I organised tutoring sessions with small groups of younger students who wanted to actually learn. I'm here to find the best way for you to understand the subject. And hopefully, show you that everything boils down to common sense, if you just look closely.

Don't hesitate to send me a free meet the tutor request! I will do my best to help.

About me: I am a physics student at the University of Manchester. I have been the top student, I've attended "the best highschool in Poland", won olympiads, done a couple university courses while at school, played with lasers and atoms in a lab, been the captain of a yacht... And in free time, I learn Japanese and programming. I love maths, physics, all things analytical, learning and exploring new areas.

#### Subjects offered

SubjectQualificationPrices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

#### Qualifications

MathsA-levelA2100%
English LanguageA-levelA2100%
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### How to transform graphs of functions?

So you've encountered something like: g(x) = | 4 f( |x + 1| - 3 ) + 5 | What a mess! And you are only given the graph of f(x). How do we go about sketching g(x)? As in any problem in mathematics, let's break it all down to a list of simple steps. We start with a graph of f(x). 1. Let's crea...

So you've encountered something like:

g(x) = | 4 f( |x + 1| - 3 ) + 5 |

What a mess! And you are only given the graph of f(x). How do we go about sketching g(x)?

As in any problem in mathematics, let's break it all down to a list of simple steps. We start with a graph of f(x).

1. Let's create a function f1(x) = f(x+1). You might already know that adding a number to the argument of a function is equivalent to 'sliding' its graph by that number to the left (it's quite simple, after all, you make an x=5 give f(5+1)=f(6) - which is to the right of f(5) we call that translation by a vector)

2. f1(x) is not quite like g(x) yet. Let's make f2(x) = f1( |x| ) = f( |x+1| ) . How does that affect the graph? As we know |x| = x, for positive x's and |x| = -x, for negative x's. So to sketch f2(x), we take the part of the graph of f1(x) which is to the right of the y-axis (where x is positive), and on the left half, we need to draw a mirror image of the right half. We're one step closer to g(x).

3. Next we need to include the number -3 in the argument. Let f3(x) = f2(x-3). Which means another shift along the x-axis. This time we move the graph of f2(x) to theright, by 'amount' 3.

4. Now we are 'outside' f3(x) - we are not going to play with the argument anymore. Which means that now we will change the graph in the y-axis only. The remaining steps include multiplying the function by 4. This means that everywhere on the graph, the function is 4 times farther away from the x-axis (from 0). So sketch f4(x) = 4*f3(x), 4 times taller and steeper.

5. Then there is the 5 left to add. If we add a number to the function, it means that it goes up in the y direction by 5, everywhere. So let's shift the graph upwards. f5(x) = f4(x) + 5

6. We are nearly done now. The last thing to do is the absolute value. Once again, for all positive values, it stays as is, and for all negative values it gets a minus sign. Thus everywhere the function f5(x) has values smaller than 0, we need to 'flip' it upside down (symmetry with respect to the x-axis). f6(x) = |f5(x)|

g(x) = f6(x)

And that is it. You now have an accurate depiction of g(x). It wasn't that complicated after all. You just needed to make incremental changes and step by step move 'from the inside out'.

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3 years ago

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