The graph below shows the function f(x)= 5x+10/x^2+7x+10 where is the removable discontinuity of f(x) located? a. x = –5 b. x = 0 c. x = –2 d. x = 5

C. x = -2

Step-by-step explanation:

A proper graph of the function will have an open circle at (-2, 5/3), so you should be able to read the answer from the graph.

The factorization is ...

... f(x) = 5(x +2)/((x +5)(x +2))

The removable discontinuity is created by the factor (x+2)/(x+2), which is undefined at x = -2.

step-by-step :

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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