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I have missed math class for a few weeks and I'm quite behind with the new stuff learned by the others, so I'm stuck with a problem here. The main problem is, I'm going to have hard time explaining the problem in English.

I have this calculus problem:

$y=\frac{x^2-5x+2}{2x-4}$

If I try to translate the question for the problem from my language to English, it says: "Review function and submit the graph". What I know is, this problem consists of 8 steps and in the end I must draw a graph for it.

I totally can't start solving this question myself since no one is ready to explain anything to me and I see my math teacher once a week and she expects a solved problem next week. I'm stuck here.

I'd appreciate an answer. If you need any additional explanation, please ask, I'll try to explain.

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Not sure if this should be a comment or an answer.

The 8 steps when studying a function $f$ are generally:

A - Domain of $f$, continuity, differentiability.

B - Symmetry (odd,even,periodic).

C - y-intercept and x-intercepts if any.

D - Existence of asymptotes (horizontal,vertical,slant).

E - Local and absolute extremums.

F - Concavity

G - Graph of the function.

It is quite long to write the details and it is worthy to try to do it on your own. Tell us if you have difficulties in some part.

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I think you are supposed to analyse the given function and plot it.

$y=\frac{x^2-5x+2}{2x-4}=\frac{x^2-5x+2}{2(x-2)}=\frac{(x-3)(-4)}{2(x-2)}=\frac{-2(x-3)}{x-2}$

$x \neq 2$

Consider different values of x for plotting the graph

$x>3; 3>x>2; x=0; x<2$

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  • $\begingroup$ Yeah, analyzing must be the correct word instead of reviewing. My classmates mentioned 8 steps. What is the meaning of that? $\endgroup$ – Aborted Apr 24 '13 at 15:14
  • $\begingroup$ @Dugi, my soln is wrong, y needs to be solved by quadratic roots formula $y=-b \pm \sqrt (b^2-4ac) / 2a$ $\endgroup$ – Vikram Apr 25 '13 at 5:52
  • $\begingroup$ Thanks a lot for your help, Vikram. Could you update your answer with a completed solution? As I said, I don't know any way to solve this and seeing it all done once will help me a lot for future problems. $\endgroup$ – Aborted Apr 26 '13 at 14:08
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I want to try answer it, just for a part.

$y=\frac{x^2-5x+2}{2x-4}$

if $y=0$ then I have root of $0={x^2-5x+2}$ is $x_1=\frac{5-\sqrt 17}{2}$ and $x_2=\frac{5+\sqrt17}{2}$

The domain is $x\in \mathbb{R}, x\neq2$

this is graph for $y$, I use Maple 13 for plotting.

enter image description here

$y$ is discontinuous at $x=2$, because $lim_{x\to 2} \frac{x^2-5x+2}{2x-4}=\frac{-4}{0}=\infty$

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