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
(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!
2 answers
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?
1 answer