Which polynomial function f(x) has a leading coefficient of 1, roots4, 2, and 9 with multiplicity 1, and root-5 with multiplicity 3?

Step-by-step explanation:

In order to solve this function, we must first start by writting the factors of the polynomial. We can do so by including the given zeros, and the leading coefficient. So we get:

and now we can expand this by multiplying the given parenthesis, so we get:

we combine like terms so we get:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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