A Level Maths: Algebra And Transformations
Introduction
When it comes to advanced level mathematics there are a number of skills that you will need that you will have discovered at GCSE level. These topics will be revisited and take a little further especially with regards to the problem solving and thinking skills that you will need to adopt.
The Laws of Indices
You were introduced to the laws of indices at GCSE and they are repeated here again.
First you should be aware that in the expression a^m,
where a is the base and m is the power or also known as the index.
For some rules of indices to be able to work the base must be the same.
Multiplication
If two numbers of the same base are multiplied together then the powers are added together.
For instance 2^3 \times 2^7=2^{3+7}=2^{10}
This can be written in general terms: a^m \times a^n=a^{m+n}
Division
If two numbers of the same base are divided then the powers are subtracted.
For instance 2^7 \div 2^3=2^{7-3}=2^4
This can be written in general terms: a^m \div a^n=a^{m-n}
Power to a power
If the power of a base is raised to another power, then the powers are multiplied together.
For instance \left(3^3\right)^4=3^{3 \times 4}=3^{12}
This can be written in general terms: \left(a^m\right)^n=a^{m n}
The Power 0
Anything to the power 0 is always 1.
For instance 125^{\circ}=1
This can be written in general terms: a^0=1
Negative Indices
We know that if we divide the same base to a power, that the powers are subtracted.
So 3^4 \div 3^6=3^{4-6}=3^{-2}
The final answer can also be written as: 3^{-2}=\frac{1}{3^2}
This can be written in general terms: a^{-n}=\frac{1}{a^n}
Fractional Powers
Suppose we have 2^p \times 2^p=2^1
Then when it comes to the powers we can say that p+p=1 (because we add the powers). So 2 p=1 \therefore p=\frac{1}{2}.
What this means is that we have 2^{\frac{1}{2}} and this is the same thing as \sqrt{2}
The same can be said about other roots, such as cube roots and fourth roots. So we can generalise the above to say that \sqrt[n]{a}=a^{\frac{1}{n}}.
Examples
Simplify the following:
\begin{aligned} & a^2 a \\ & a b^3\left(a^2 b\right) \\ & \left(x^2\right)^3 \\ & \left(a b^2\right)^4 \\ & \frac{x^3 y^2}{x y^3} \\ & 125^{\frac{1}{3}} \\ & 16^{\frac{3}{4}} \end{aligned}Solutions
\begin{aligned} & a^2 a=a^{2+1}=a^3 \\ & a b^3\left(a^2 b\right)=a^{1+2} b^{3+1}=a^3 b^4 \\ & \left(x^2\right)^3=x^{2 \times 3}=a^6 \\ & \left(a b^2\right)^4=a^4 b^8 \\ & \frac{x^3 y^2}{x y^3}=x^{3-1} y^{2-3}=x^2 y^{-1}=\frac{x^2}{y} \\ & 125^{\frac{1}{3}}=\sqrt[3]{125}=5 \\ & 16^{\frac{3}{4}}=\left(16^{\frac{1}{4}}\right)^3=2^3=8 \end{aligned}Example
Simplify:
5^{\frac{1}{2}}-5^{\frac{3}{2}}+5^{\frac{5}{2}}Solution
Quite often with questions like these you need an exact answer and you should try to do this without the need for a calculator. Whilst this may seem difficult, our aim should be to find an exact answer and we can do this by re-writing what we have.
You should look for a “common term”. How to do that?
Well \frac{3}{2}=\frac{1}{2}+1 and \frac{5}{2}=\frac{1}{2}+2
So our original question can be written as follows:
5^{\frac{1}{2}}-5^{\frac{3}{2}}+5^{\frac{5}{2}}=5^{\frac{1}{2}}-5^{\frac{1}{2}}(5)+5^{\frac{1}{2}}\left(5^2\right). So we can see that 5^{\frac{1}{2}} is a common term. This gives us:
5^{\frac{1}{2}}\left(1-5+5^2\right)=5^{\frac{1}{2}}(21)=21 \sqrt{5} as the final answer.
Example
Write 125 \sqrt{5} \text { as } 5^k
where k is a constant to be determined
Solution
The answer needs to be in terms of base 5.
125=5 \times 5 \times 5=5^3and \sqrt{5}=5^{\frac{1}{2}}
So 125 \sqrt{5}=5^3 \times 5^{\frac{1}{2}}=5^{\frac{7}{2}} \text {, so } k=\frac{7}{2}
Transformations of Graphs
If we are given a function, f(x) then the curve y=f(x) can be translated, stretched and also translated.
Translations of the form y=f(x)+a
Let us consider the curvey=x^2 which is shown below:
Now what if we have a new equation that we wish to sketch which is y=x^2+2?
What is going to happen to the curve above? You can see from the sketch below that the curve has been moved upwards in the positive y direction by 2 units.
In a similar fashion we could sketch y=x^2-2 and you will see that the curve has been translated downwards in the negative y direction and this can be written as \left(\frac{0}{-2}\right)
In general, for any function f(x) the curve y=f(x) \pm a epresents a translation in the y direction by \left(\frac{0}{ \pm a}\right) units.
Translations of the form y=f(x-a)
Still using the curve y=x^2 what will happen if we now sketch y=(x-2)^2?
What is the curve going to look like now?
The diagram shows that the curve has actually moved two places to the right. And this might seem a little odd given that the sign inside the brackets is negative. It moves to the right because x has got to be larger than 2 in order to get an output.
An easy way to determine the direction is to set the bracket equal to zero. This gives x-2=0 \rightarrow x=2. And because we have a positive value it tells us the curve is translated 2 units in the positive x direction which can be written as \left(\frac{2}{0}\right).
Similarly if we were to sketch y=(x+2)^2 hen by setting the bracket equal to zero we would obtain: x+2=0 \rightarrow x=-2 i.e. the curve is translated 2 units in the negative x direction which can be written as \left(\frac{-2}{0}\right).
In general, for any function f(x)f(x), the curve y=f(x \pm a) represents a translation in the x direction by \left(\frac{\mp a}{0}\right) units.
Of course it is quite possible for a curve to be translated in both the x and y direction.
In general, for any function f(x), the curve y=f(x \pm a) \pm b represents a translation in the y direction by \left(\frac{\mp a}{ \pm b}\right) units.
Stretches and Reflections
Consider the curve y=f(x). Then a curve of the form:
y=af(x) represents a stretch parallel to the y-axis scale factor a.
y=f(ax) represents a stretch parallel to the x-axis scale factor \frac{1}{a}.
Example
Starting with the curve y=x^2 state the transformations of the following:
\begin{aligned} & y=4 x^2 \\ & 3 y=x^2 \end{aligned}Solution
Here we have a stretch of the form y=af(x) so we have a stretch parallel to the y-axis scale factor 4.
This can be written as y=\frac{x^2}{3} and this is of the form y=a f(x) so we have a stretch parallel to the y-axis scale factor \frac{1}{3}.
Example
The graph below shows y = f(x). Two turning points are shown.
State the coordinates of both turning points after the following stretches.
\begin{aligned} & y=2 f(x) \\ & y=f(2 x) \\ & y=f\left(\frac{x}{2}\right) \end{aligned}Solution
Here we have a stretch parallel to the y-axis so this means that we need to multiply the y coordinates only by 2. This will give us (-2, 4) and (2, -4). Notice that the x coordinate stays the same.
Here we have a stretch parallel to the x-axis so this means that we need to multiply the x coordinates only by \frac{1}{2} this will give us (-1, 2) and (1, -2). Notice that the y coordinates stay the same.
Here we have a stretch parallel to the x-axis so that means that we need to multiply the x coordinates only by 2. This will give us (-4, 2) and (4, -2). Notice that the y coordinates stay the same.
You will more than likely revisit these topics in the first half term of year 12. Most schools and colleges tend to recap the grade 8 and 9 material from GCSE Higher but what you may find challenging is using the rules that you have been used to in a different way such as in the case with indices. The very same rules are being used but in a different way to solve more challenging questions.
It is this thinking process that you need to become accustomed to during your course and our all exam board A level maths revision course will show you other challenging areas of maths that you should be able to solve.
You will find that the first year will go very quickly so it is important that you are keeping on top of your work from the outset otherwise it could get difficult keeping up with the pace of the class as well as the overall content of the course.
We have a number of courses throughout the academic year. At the time of writing it is almost October half term. Our next and final course for 2023 will be over Christmas which can give you the confidence and skills you need for any January 2024 mock exams.
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