(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!
for domain, all you have to worry about is the denominator and make it does not become zero.
x^2 + x + 9 = 0 has no real solution, (the b^2-4ac is negative)
so the domain is the set of all real numbers for x
b) Thus, since there are no breaks in the function, it is continuous
c) clearly no vertical asymptotes (the denominator ≠ 0)
for horizontals, as x becomes ±large , the function approaches 3x^2/x^2 = 3
so y = 3 is the horizontal asymptote
for a picture of your function, see
http://www.wolframalpha.com/input/?i=plot+y+%3D+%283x%5E2%2B1%29%2F%28x%5E2%2Bx%2B9%29+
for you last 3 problems, if you set an algebraic fraction equal to zero, all you worry about is the numerator equal to zero, since the denominator can never be zero
so for G(x) = (x^3 - 8)/(x^2 + 4)
we get zeros when x^3 - 8 = 0
(x-2)((x^2 + 2x + 4) = 0
x = 2 or x is imaginary,
so x = 2 is the only zero of the function.
in H(x) = 5 + 3/(x^2 + 1)
= (5x^2 + 5 + 3)/x^2 + 1)
= (5x^2 + 8)/(x^2 +1), neither top or bottom can ever be zero, so
there are no zeros of that function
the numerator of the last one factors very nicely, I will leave it up to you
Thank you so much! The first problem I understand more now, but I'm still a bit confused on the finding zeros problems. Could you explain it more? Or if it helps, create new problems? Do you get the zeros from the top polynomial or the bottom one?
To find the domain of a function, you need to determine which values of x the function is defined for. In general, a rational function (a function defined as the quotient of two polynomials) is defined for all real numbers except those that make the denominator zero. So, to find the domain of a rational function:
(A) Find the values of x that make the denominator zero by setting the denominator equal to zero and solving for x. In the first example, F(x) = (3x^2 + 1)/(x^2 + x + 9), the denominator is x^2 + x + 9. Setting this equal to zero, we get x^2 + x + 9 = 0. However, this quadratic equation does not have any real solutions since the discriminant (b^2 - 4ac) is negative. Therefore, the denominator is never equal to zero, and the function F(x) is defined for all real numbers. Hence, the domain of F(x) is the set of all real numbers, or (-∞, ∞).
(B) To decide if a function is continuous, you need to check if it has any points of discontinuity or jumps. For a rational function, the only potential points of discontinuity occur where the function is not defined, that is when the denominator is zero. In the first example, F(x) = (3x^2 + 1)/(x^2 + x + 9), we found that the denominator is never equal to zero. Therefore, there are no points of discontinuity, and the function F(x) is continuous for all real numbers.
(C) To identify the horizontal and vertical asymptotes of a rational function, you need to analyze the behavior of the function as x approaches infinity or negative infinity.
First, let's consider horizontal asymptotes:
1. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
2. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0 (the x-axis).
3. If the degrees of the numerator and denominator are equal, divide each term by the highest power of x in the denominator, and you will get the equation of the horizontal asymptote.
In the first example, F(x) = (3x^2 + 1)/(x^2 + x + 9), the degree of the numerator is 2, and the degree of the denominator is also 2. If we divide each term by x^2, we get (3 + 1/x^2)/(1 + 1/x + 9/x^2). As x approaches infinity or negative infinity, 1/x^2 approaches zero, so the numerator simplifies to 3 and the denominator simplifies to 1. Therefore, the horizontal asymptote is y = 3/1, or y = 3.
Next, let's consider vertical asymptotes:
1. If there are any values of x that make the denominator zero but do not make the numerator zero, then there is a vertical asymptote at that x-value.
2. To find the vertical asymptotes, solve the equation obtained by setting the denominator equal to zero. In the first example, x^2 + x + 9 = 0 does not have any real solutions, so there are no vertical asymptotes.
Now, let's move on to finding the zeros of the rational functions G(x), H(x), and G(x):
To find the zeros of a rational function, you need to solve the equation obtained by setting the numerator equal to zero.
For G(x) = (x^3 - 8)/(x^2 + 4), we set the numerator equal to zero: x^3 - 8 = 0. This equation can be factored as (x - 2)(x^2 + 2x + 4) = 0. Setting each factor equal to zero, we find x - 2 = 0, which gives x = 2, and x^2 + 2x + 4 = 0, which does not have any real solutions. Hence, the zero of G(x) is x = 2.
For H(x) = 5 + 3/(x^2 + 1), the numerator is a constant, so it can never be zero. Therefore, H(x) does not have any zeros.
For G(x) = (x^2 - 5x + 6)/(x^2 + 4), we set the numerator equal to zero: x^2 - 5x + 6 = 0. This equation can be factored as (x - 2)(x - 3) = 0. Setting each factor equal to zero, we find x - 2 = 0, which gives x = 2, and x - 3 = 0, which gives x = 3. Hence, the zeros of G(x) are x = 2 and x = 3.
I hope this explanation helps you understand how to find the domain, continuity, horizontal and vertical asymptotes, and zeros of rational functions. Let me know if you have any further questions!