(A) find the domain of the function, (B) decide if the function is continuous, and (c) identify your horizontal and vertical asymptotes.

F(x)= (3x^2+1)/(x^2+x+9)

Find the zeros (if any) of the rational function.
G(x)= (x^3-8)/x^2+4
H(x)=5 + 3/(x^2+1)
G(x)= (x^2-5x+6)/(x^2+4)
These are the last questions or problems I have on my homework and I don't know how to do them. Please show and tell me how to do them!

(A)

The denominator never zero, so the domain is all real numbers. So, of course F(x) is continuous, since there are no places where it is undefined. As x gets large, F(x) ≈ 3x^2/x^2 = 3.

For the zeros of rational functions, they occur where the numerator is zero (as long as the denominator is not also zero).

(A) To find the domain of a function, we need to identify any values of x that would make the function undefined. In this case, we have a rational function, which means we cannot have any values of x that would result in a zero denominator.

For the function F(x) = (3x^2 + 1)/(x^2 + x + 9), we need to find the values of x that would make the denominator, x^2 + x + 9, equal to zero. To do this, we can set the denominator equal to zero and solve for x:

x^2 + x + 9 = 0

This quadratic equation does not have any real solutions. Therefore, the denominator is never zero, and there are no restrictions on the domain. So, the domain of F(x) is the set of all real numbers.

(B) To determine if a function is continuous, we need to check if there are any points where the function has a hole, jump, or vertical asymptote. For a rational function like F(x) = (3x^2 + 1)/(x^2 + x + 9), it will be continuous everywhere except at any values of x that make the denominator zero (if they exist).

From part (A), we found that the denominator, x^2 + x + 9, is never zero. Therefore, there are no holes, jumps, or vertical asymptotes in this function. Hence, F(x) is continuous for all real numbers.

(C) To identify the horizontal and vertical asymptotes of a rational function, we need to study the behavior of the function as x approaches positive or negative infinity.

For F(x) = (3x^2 + 1)/(x^2 + x + 9):

Vertical Asymptotes:
Since the degree of the numerator (2) is less than the degree of the denominator (2), there is no vertical asymptote.

Horizontal Asymptotes:
To find the horizontal asymptote(s), we compare the degrees of the numerator and denominator polynomials.

The degree of the numerator (2) is equal to the degree of the denominator (2). In this case, we need to compare the leading coefficients of both polynomials, which are 3 for the numerator and 1 for the denominator. The horizontal asymptote is given by the ratio of these leading coefficients, so the horizontal asymptote is y = 3/1 = 3.

So, for F(x), there are no vertical asymptotes, and the horizontal asymptote is y = 3.

Let's proceed with the zeros of the other rational functions you provided:

For G(x) = (x^3 - 8)/(x^2 + 4):

To find the zeros, we need to set the numerator, x^3 - 8, equal to zero and solve for x:

x^3 - 8 = 0

This equation can be factored as (x - 2)(x^2 + 2x + 4) = 0. The quadratic term, x^2 + 2x + 4, has no real solutions. Therefore, the only real zero of G(x) is x = 2.

For H(x) = 5 + 3/(x^2 + 1):

Since the numerator is a constant, 5, and the denominator, x^2 + 1, is always positive, there are no zeros for H(x).

For the last function G(x) = (x^2 - 5x + 6)/(x^2 + 4):

To find the zeros, we need to set the numerator, x^2 - 5x + 6, equal to zero and solve for x:

x^2 - 5x + 6 = 0

This quadratic equation can be factored as (x - 2)(x - 3) = 0. Therefore, the zeros of G(x) are x = 2 and x = 3.

I hope this helps!