#2**+15 **

Hi Sabi,

(x-2)(x²-1)>0

(x-2)(x-1)(x+1)>0

consider y=(x-2)(x-1)(x+1)

The roots will be 2,1 and -1.

3 is the highest power of x (It is degree 3)

This means that the graph will have 3 (or less, see my notes for the last question) directions.

(Since there are 3 distinct roots this one has to have 3 directions )

I do not need to do the expansion to see that the coefficient of x^{3} is +1.

Since it is **positive**, one of the ends of the graph will be in the **TOP RIGHT CORNER**

Always start your sketch on the **right** side.

Above the x axis if it is a positive leading coefficient

and below the x axis if it is a negative leading coefficient.

**This is what it must look like**

It can be seen from the graph that y is positive when -1<x<1 and when x>2

Any more questions the just ask :))

Melody
Aug 4, 2015

#2**+15 **

Best Answer

Hi Sabi,

(x-2)(x²-1)>0

(x-2)(x-1)(x+1)>0

consider y=(x-2)(x-1)(x+1)

The roots will be 2,1 and -1.

3 is the highest power of x (It is degree 3)

This means that the graph will have 3 (or less, see my notes for the last question) directions.

(Since there are 3 distinct roots this one has to have 3 directions )

I do not need to do the expansion to see that the coefficient of x^{3} is +1.

Since it is **positive**, one of the ends of the graph will be in the **TOP RIGHT CORNER**

Always start your sketch on the **right** side.

Above the x axis if it is a positive leading coefficient

and below the x axis if it is a negative leading coefficient.

**This is what it must look like**

It can be seen from the graph that y is positive when -1<x<1 and when x>2

Any more questions the just ask :))

Melody
Aug 4, 2015

#3**+5 **

Thanks, Melody......time constraints prevent me from explaining this in more detail......

CPhill
Aug 4, 2015

#5**0 **

That's ok Chris, none of us can answer all the questions in great detail. We answer what we can with as much time as we can afford to donate at that point in time.

Any correct answer is better than none, I am sure that Sabi appreciated your help :)

I like answering questions like this - it is one of my pet subject areas :))

Oh, thanks Sabi - and you are very welcome :)

Melody
Aug 4, 2015