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
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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