Year 13 Maths Help Thread

No!! 2sinx is a stretch in the y axis, or an increase in amplitude. This one is much easier to understand imo, if sin(x) = 1, 2sin(x) = 2 simple as that( literally the value of sin(x) multiplied by 2), so the positions of the minima and maxima are the same, but their magnitude increases and the range changes to -2 <-> 2

So sorry to have bothered you (i made a silly mistake :P) - i looked over it again and realised i mixed up the transformations in the first place. But thank you for your help

Maybe you could help with this question:

y = sinx transformed by a sequence of transformations: translation [ 0 pi] , reflection in x-axis, stretch y-direction S.F. 4

i got y = -4sin (x - pi) which is correct.
but it can also be y=4sinx

I know i got it right but how do you get the other answer?

So sorry to have bothered you (i made a silly mistake :P) - i looked over it again and realised i mixed up the transformations in the first place. But thank you for your help

Maybe you could help with this question:

y = sinx transformed by a sequence of transformations: translation [ 0 pi] , reflection in x-axis, stretch y-direction S.F. 4

i got y = -4sin (x - pi) which is correct.
but it can also be y=4sinx

I know i got it right but how do you get the other answer?

I don't really get the translation bit you've put in square brackets but that's not important.
If you think about it, sin(x-pi) shifts the graph along one half repeat to the right, so the minima now occurs at 90 degrees, and the maxima at 270. This inversion is the same as what would happen if it was -sin(x), the minima -> the maxima and vica versa. Then obviously they are both multiplied by 4

You can also just use trig identities, you know that $\sin(x-\pi) = \sin x \cos \pi - \sin \pi \cos x = -\sin x$

So $-4\sin(x-\pi) = -4(-\sin x) = 4\sin x$.

I need help on a C3 domain and range questions .... again.
Ive been over it twice and thought i understood but then got confused when i got this question wrong.

In simple terms is the domain the x value and the range the y value you get out of the function?

f(x) = (2x +5) / (x - 3)
x E R, x is not equal to 3

for the first part you are told to find the range. I got it as

0< f(x)<= 13 but it should be f(x) E R, f(x) is not equal to 2.

Where have i gone wrong?? (this isnt related to the question but how is R different to x E R)

Could anyone help me on this question before it gets buried - i still dont understand it.

Could anyone help me on this question before it gets buried - i still dont understand it.

like x E Z would mean x is an integer (I think)

Could anyone help me on this question before it gets buried - i still dont understand it.

Could anyone help me on this question before it gets buried - i still dont understand it.

Your function of $f(x)=\frac{2x+5}{x-3}$ is a mere set of transformations from $g(x)$ therefore there will still essentially be ONE horizontal and ONE vertical asymptote.

Er... how?

Why do you think it's 0 < f(x) <= 13

For example $x=1$ gives $f(1) = \frac{7}{-2} = -3.5$ which is a y value that you get out of the function, but it's not in 0 < f(x) <= 13.

Why don't you try sketching the function? You'll see that it outputs every single real number except y=2.

R is the set of real numbers.

Ok, ill sketch it out - i usually avoid drawing graphs because i normally mess it up and get the question wrong

So whats the difference between having a domain of R and domain of x E R ?

I'm not quite sure where you got the 0 and 13 from for the range??

Okay, firstly; you need to know the general shape of $g(x)=\frac{1}{x}$ graph and the fact that this has one horizontal asymptote, and one vertical asymptote. You can even see that this function goes over ALL x and y values EXCEPT those asymptote ones. Here is a graph of your function and $g(x)$ for comparison:




Your function of $f(x)=\frac{2x+5}{x-3}$ is a similar to $g(x)$ as you still have linear (function) over linear as the fraction therefore there will still essentially be ONE horizontal and ONE vertical asymptote as well. You are told the vertical one, $x=3$, as $f(3)$ is not valid due to division by 0, this is the straight forward part.

However for the horizontal asymptote, it is trickier to determine it because we need to observe what happens to $f(x)$ as $x$ tends to positive/negative infinity, therefore I'd suggest dividing (using long division or otherwise) $2x+5$ by $x-3$ and expressing it in the form $A+\frac{B}{x-3}$ which you'd find to be $f(x)=2+\frac{11}{x-3}$ then consider what happens as $x\rightarrow \pm \infty$. Of course, you can see that the second term goes to 0 therefore you are left with 2; hence the function TENDS to 2 therefore $f(x) \rightarrow 2$ as $x\rightarrow \pm \infty$.

It never reaches 2, therefore it cannot be equal to it. Hence where the $f(x) \in \mathbb{R}$, $f(x) \not= 2$ comes from.

EXTRA NOTE:

When it comes to finding the horizontal asymptote for this question (and any alike) you may notice that you can divide (just as you can multiply) the top and bottom of the fraction by something. In this case, you can divide top and bottom by $x$.

You get left with $\displaystyle f(x)=\frac{2+ \frac{5}{x} }{ 1- \frac{3}{x} }$.

Now if you take the limit as $x \rightarrow \pm \infty$ then you can see that $\frac{5}{x}$ AND $\frac{3}{x}$ both go to 0, and your overall fraction is $f(x) \rightarrow \frac{2}{1} \Rightarrow f(x) \rightarrow 2$ as we had before.

Hope this helps! I didn't want to just plain out do the question for you, but I'd find this difficult to explain otherwise without raising even more confusion so take it with you because it will definitely help on similar questions as you're now aware of how to approach them with a solid method. This should clear up any confusion you may have about the question.

As an example, find the range and domain for $\displaystyle h(x)=\frac{3x-5}{2x+1}$

>_< im struggling to understand what you're saying (i think i get the general idea though?). Just the part in bold - what do you mean by that.

Perhaps it would be easier if i always sketched the graph for these type of questions.

Thanks for explaining it

Ok, ill sketch it out - i usually avoid drawing graphs because i normally mess it up and get the question wrong

So whats the difference between having a domain of R and domain of x E R ?

Quite simple. You are finding a certain y coordinate that the graph approaches but never reaches as you keep going along the x-axis, so you are basically going off to infinity along the x-axis in order to get the asymptote. From our equation, we can see that as we go off to infinity, the denomintor will get bigger and bigger, so the overall fraction will go to 0.

So the asymptote will never touch the axis? - you work it the value if it does and then make the range less than that value?

Sorry this topic has been bothering me for ages

Sorry but what you've said does not make any sense whatsoever to me.

An asymptote is a straight line (in this case) that the graph approaches, and this case it is also a line that the graph never crosses. $y=2$ IS that line as we found it (along with x=3 but that is a different asymptote), because $f(x) \not= 2$. The range is everything EXCEPT that value of the asymptote which is why it is important to find it.

In order to work out the asympotote we need to consider what value the function goes to as we pump up $x$ to infinity. Our initial function form does not make this value seem obvious therefore we manipulate it to get something divided by x. As x goes to infinity, those fractions go to 0 and whatever is left make up that value we need, as I have shown with my examples in the long post.