#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!!

1 answer

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)
Similar Questions
  1. HELP!!!!!!!!!!!!!write equation of a polynomial function with the given characteristics. Leading coefficient is 1 or -1 crosses
    1. answers icon 2 answers
    1. answers icon 1 answer
  2. Can you please check my work and answers.Find the x-intercepts of the polynomial function. State whether the graph crosses the
    1. answers icon 0 answers
  3. Sketch f(x)=2x4−6x3−8x2. How does the graph behave at the zeros of the polynomial?(1 point) Responses The graph crosses the
    1. answers icon 1 answer
more similar questions