Degree of 3, positive leading coefficient, 2 zeros, 2 turning points. I sketched the condition and the zeros are -1 and 1 and the y-int i made it as 2. How do i make a polynomial function?

1 Answer
Sep 12, 2015

f(x) = 2x^3-2x^2-2x+2

Explanation:

f(x) will only potentially change sign at the zeros at x = -1 and x = 1, so:

Since the leading coefficient is positive and f(0) = 2 > 0, we have:

f(x) < 0 for x in (-oo, -1)

f(x) > 0 for x in (-1, 1)

f(x) > 0 for x in (1, oo)

Due to the degree 3 and 2 zeros, one of the roots must be repeated and from the above it must be the one at x = 1

So f(x) = k(x-1)(x-1)(x+1) for some k

2 = f(0) = k(-1)(-1)(1) = k, so k = 2

Thus:

f(x) = 2(x-1)(x-1)(x+1)

= 2(x-1)(x^2-1)

= 2(x^3-x^2-x+1)

= 2x^3-2x^2-2x+2