For function f(x), which pair of limits would verify the existence of a vertical asymptote at x = –1? Limit of f (x) as x approaches negative 1 minus = infinity and limit of f (x) as x approaches negative 1 plus = infinity
Limit of f (x) as x approaches negative infinity = negative 1 and limit of f (x) as x approaches infinity = negative 1
Limit of f (x) as x approaches negative 1 minus = c and limit of f (x) as x approaches negative 1 plus = c for some value of c
Limit of f (x) as x approaches c minus = negative 1 and limit of f (x) as x approaches c plus = negative 1 for some value c

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! : )

cretdit to source

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3

i think it might be a but you might want a second opinion

what? that isnt a question

step-by-step explanation: