Question

Suppose f(x) has zeros at x=-2,x=4, x=7 and a y-intercept of 1 In addition, f(x) has the following long-run behavior: as x--> ∞, y--> ∞. Find the formula for the polynomial f(x) which has the minimum possible degree?

Answer 1

Both endpoints at `+infty` means even degree. There are three zeros listed; if those are the only zeros we can square one to make the degree even.

`f(x)=k(x+2)^2(x-4)(x-7)` and `1=f(0)=k \ 2^2(-4)(-7)=112k` so

`f(x)=1/112 (x+2)^2(x-4)(x-7)`

Explanation:

`f(x)=1/112 (x+2)^2(x-4)(x-7)`

graph{1/112 (x+2)^2(x-4)(x-7) [-10, 10, -5, 5]}