I need to find the domain and range of for f(y) = 2/(c^2 + 3c)

Question

I need to find the domain and range of for f(y) = 2/(c^2 + 3c)

 

I know how to find the domain  its where c^2 + 3c would = 0

 

so it's not valid for  c=-3

 

but how do I explain the range ?

 

Thanks

 

Dom

Answer 1

2 / [c(c + 3)]

To find the domain, set the denominator equal to zero and solve for c. c = 0, -3.

Domain: All Real numbers, except 0, -3

 

To find the range, you can use Calculus to determine the turning point  (local maximum at (-3/2, -8/9))

Also, the horizontal asymptote is obtained by finding the limit of the function as c approaches infinity. i.e. y = 0 is the horizontal asymptote.

A graph of the function shows the range to be ( -inf, -8/9] U (0, inf).

Answer 2

2 / [c(c + 3)]

 

The domain is all value of c except for -3 and 0. 

 

c=-3   and   c=0  are vertical asymptotes.

 

 

To find the range, we look at the degree of the numerator part and denominator part of the function.

 

Since the degree of the numerator part is less than the degree of the denominator part, your range is all value of f(y) except for 0.  Degree of numerator part is 0.  Degree of denominator part is 2.

 

f(y)=0   is a horizontal asymptote.