Solve find a polynomial function f with real coefficients of degree 3 with -2, -1, and 4 as zeros, and f(2)=16

29.5977692753

step-by-step explanation:

what statement? is there an image?

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.