+0

0
619
6
+262

(x-2)(x²-1)>0

the root at x=2 x=1 x=-1

how will i draw the graph?

Aug 4, 2015

#2
+95177
+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 x3 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 :))

Aug 4, 2015

#1
+94526
+10

(x-2)(x²-1)>0

You have the correct idea, sabi92......there are various Algebra/Calculus techniques that might be employed here to assist in the sketching of this graph......however, they are a little difficult to explain concisely, so I'll just present the graph, here:

Aug 4, 2015
#2
+95177
+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 x3 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
+94526
+5

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

Aug 4, 2015
#4
+262
+5

thank you:)

Aug 4, 2015
#5
+95177
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 :)

Aug 4, 2015
#6
+95177
+10

I have just added this post to the "reference material" sticky topic

I have entered it twice

Once for solving inequalities and once for graphing polynomials :))

Aug 4, 2015