since as x ---> ± infinity, y --> + infinitey, the function must be of even exponent, so one of the roots must be a double root, as you have
f(x) = k(x-3)(x+3)(x-5)
but we are also told that the y-intercept is 17, so we have the point (0,17) on our polynomial.
17= k(-3)^2(3)(-5)
17 = -135a
a = -17/135 , so
f(x) = -(17/135)(x-3)^2 (x+3)(x-5)
however, why could the double root not have been the -3 ?
Then
f(x) = a(x+3)^2 (x-3)(x-5)
17 = a(9)(-3)(-5)
17 = 135a
a = 17/135 , and
f(x) = (17/135)(x+3)^2 (x-3)(x-5)
or why not
f(x) = b(x-3)(x+3)(x-5)^2
17 = b(-3)(3)(25)
b = - 17/225
f(x) = -17/225)(x^2 - 9)(x-5)^2
Suppose f(x) has zeros at x = -3, x = 3, x = 5, and a y-intercept of 17.
In addition, f(x) has the following long-run behavior: as x -> +-infinity, y -> +infinity
Find the formula for the polynomial f(x) which has the minimum possible degree
F(x) = k(x+3)^2(x-3)(x-5)
I cannot find the value for k
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1 answer