Melody

Why can't you do it?

I'd have to look up those three centres, I always forget them but they are easy enough to find. For me but also for you!

I assume this is what is wanted.

\(y=\frac{2x^2-3}{x^2-4x+1}=2x^2+8x+30+\frac{112x-33}{x^2-4x+1}\\~\\~\\ \displaystyle\lim_{x\rightarrow\pm\infty}y=2x^2+8x+30\)

Here is the graph:

ok, thanks Alan :)

I know your question is a translation.

Unfortunately, it does not make sense.

سؤالك لا يترجم كسؤال كامل.

Gues is also right.

This can be tackled as a quadratic.

\(3( cos x)^2 = 2 cos x + 1\\ let\;y=cosx\\ 3y^2=2y+1\\ 3y^2-2y-1=0\\ 3y^2-3y+1y-1=0\\ 3y(y-1)+1(y-1)=0\\ (3y+1)(y-1)=0\\ y=-1/3\quad or \quad y=1\\ cosx=-1/3 \quad or \quad cosx=1\\ for\quad 0\le x <2\pi\\ x=\pi\pm acos(1/3)\quad or \quad 0\\ x=\pi\pm 1.231 \quad or \quad 0\\ x=1.910,\;\;\;4.373,\;\;\;0 \)

I just looked at the first 2.  

n=1    the expression = 175

n=2    the expression = 28267

The highest common factor is 1

The problem is with your graph.  You have put the 2 in the wrong place,

You have graphed   \(cos(x^2)\)      INSTEAD OF      \((cos\;x)^2\) which is the same as  \(cos^2 x\)

Here is what your graph should look like:

The textbook is correct.

I cannot tell you what you did wrong because you have not shown me what you actually graphed.

Can you share the address of the desmos graph with us please.

And, if it is not self-evident, explain how you got your answers.

And that is not how you show any credit for a person's effort.

You can at least say thanks for the effort and explain why you believe it to be wrong.

In fact, you could go through it line by line and explain which line/s you believe to be incorrect.