need a help please

Thanks Chris, Sabi and Dragonlance,

I'll just put my 2 bob's worth in :)

$$\\\frac{x^2-4+4}{x-3}>0\\\\
\frac{(x-2)^2}{x-3}>0\\\\
x\ne 3\\\\
\frac{(x-2)^2}{x-3}\times (x-3)^2>0\times (x-3)^2\\\\
(x-2)^2(x-3)>0\\\\
consider\;\; y=(x-2)^2(x-3)\;\;\\\\$$

You are right Sabi this is not a parabola but it is a polynomial - (a parabola is a polynomial of degree 2) 

When will the graph be above the x axis?

There is a root at x=3, and a double root at x=2.

The highest power of x is 3 That is the degree of this polynomial is 3.

This means that the graph will have 3 directions.

I can immediately see that the coefficient is X^3 is 1. i.e.   The leading coefficient=1.

A positive leading coefficient means that the graph will finish in the top right hand corner.

So I know a lot about how this graph will look, just by a cursory examination of the formula.

I know that it will look like this!

I can see from the graph that this is greater than 0 when x>3

---------------------

I wrote this post some time ago, I just fished it out of our reference material sticky topics.

It would be a very good idea for you to try and make sense of it all.

I have done the top row of graphs, maybe you would like to take a look at the bottom row :))

http://web2.0calc.com/questions/how-do-you-find-a-power-function-that-is-graphed

Here's the way I like to approach this kind of problem....

Solve for the top and bottom separately....

Set the top  = 0   and factor....we have....   x^2  - 4x + 4  = 0   → (x - 2)^2 =  0    so   x =2

Do the same for the bottom

x - 3 = 0      so x = 3

Now.....plot these two values on a number line......neither will be part of the answer (because of the > sign), but the answer(s)  will come from one or more of these intervals.....

(-∞, 2), (2, 3)  or (3, ∞)

Pick a number in the first interval....0 seems nice ....put it nto the original problem.....does it make it true??...Nope

Pick a number in the middle interval...say 2.5......you will find this doesn't work, either

Finally.....pick a number in the last interval, say 4.....you will find that this is the only interval that solves this problem

Here's a graph.....https://www.desmos.com/calculator/talcaantp7

Note that the original problem is only greater than 0 on the specified interval

CPhill a question.

You set x= 3 and that makes the fraction divided by zero.  I thougt that was not allowed for solving equations. Is it because it is ploted on the the graph and the line dissapears or goes to infinity when x=3 that it is allowed?

Thats how i solved it(according to your recommendation to solve the top and bottom separately:

first of all:a≠2,3

x²-4x+4>0                              x²-4x+4<0

and                      or                and 

x-3>0                                      x-3<0

    ⇓   numenator:x²-4x+4=0          ⇓

                       (x-2)²=0

                         x=2

x ≠2                                         no x

x-3>0           denominator:

x>3                                           x<3

 so the answer:x>3

Remember, Dragonlance, x = 3 isn't a  part of the solution.......I'm just doing this to find the interval points.......the solution is x > 3 as sabi92 found ....  or....as I gave ..... (3, ∞ )......both of these are the same solutions