Asked by Rebekah
#1. Make up a polynomial that has the following characteristics:
crosses the x axis at 1 and 4, touches the x axis at 0 ad 2. and is above the xaxis between 0 and 2.
I did (x+1)(x-4)(x^2)(x-2)^2
#2. its an illustration but one of the queestino says what is the minimum degree of polynomial. How do I find that and how do I find the maximum nuber of turnin points. I know it has to do with n-1 but Idont know how.
Thanks!!
crosses the x axis at 1 and 4, touches the x axis at 0 ad 2. and is above the xaxis between 0 and 2.
I did (x+1)(x-4)(x^2)(x-2)^2
#2. its an illustration but one of the queestino says what is the minimum degree of polynomial. How do I find that and how do I find the maximum nuber of turnin points. I know it has to do with n-1 but Idont know how.
Thanks!!
Answers
Answered by
Anonymous
should be (x-1)(x-4) to cross at 1 and 4
should be x^2 and (x-2)^2 to touch at 0 and 2
The last part confuses me. Since it crosses at x=1, how can it be positive for 0<x<2?
Or, on the other hand, since it crosses at x=1, it must be positive somewhere between 0 and 2.
Or, assuming your (x+1) factor is correct, then it shopuld have crossed at -1, not 1, and we have to consider how to make it positive for 0<x<2.
So, since we have
(x+1) x^2 (x-2)^2 (x-4)
for 0<x<2, all the factors except (x-4) are positive. So, try
-(x+1) x^2 (x-2)^2 (x-4)
should be x^2 and (x-2)^2 to touch at 0 and 2
The last part confuses me. Since it crosses at x=1, how can it be positive for 0<x<2?
Or, on the other hand, since it crosses at x=1, it must be positive somewhere between 0 and 2.
Or, assuming your (x+1) factor is correct, then it shopuld have crossed at -1, not 1, and we have to consider how to make it positive for 0<x<2.
So, since we have
(x+1) x^2 (x-2)^2 (x-4)
for 0<x<2, all the factors except (x-4) are positive. So, try
-(x+1) x^2 (x-2)^2 (x-4)
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