SOLUTION: How do you find the domain and range of the rational function: 5x^2+5/x^2+6x+9. Already found the asymptotes and x and y intercepts. Vertical Asymptotes: x=-3 Horizontal Asympt

Algebra ->  Rational-functions -> SOLUTION: How do you find the domain and range of the rational function: 5x^2+5/x^2+6x+9. Already found the asymptotes and x and y intercepts. Vertical Asymptotes: x=-3 Horizontal Asympt      Log On


   



Question 858186: How do you find the domain and range of the rational function: 5x^2+5/x^2+6x+9. Already found the asymptotes and x and y intercepts.
Vertical Asymptotes: x=-3
Horizontal Asymptotes: y=5
y-int.: (0,5/9)
x-int.: Does Not Exist
Use a graphing device to confirm your answer. (Use interval notation).

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
domain: all the real numbers except -3
range: [1%2F2,infinity) but I do not know how you would know.
Are you in a calculus class?
Are you studying derivatives?

We know that y=5%28x%5E2%2B1%29%2F%28x%2B3%29%5E2 is always positive,
and that its value approaches 5 at both ends,
but how low can it go?
Maybe you are supposed to calculate the derivative and find the minimum.
Maybe you are supposed to figure out the minimum some other way.
Apparently you are expected to use some graphing software to confirm.

WITHOUT DERIVATIVES:
If y=%285x%5E2%2B5%29%2F%28x%5E2%2B6x%2B9%29 has a minimum K,
%285x%5E2%2B5%29%2F%28x%5E2%2B6x%2B9%29-K=0 has a single solution.
%285x%5E2%2B5%29%2F%28x%5E2%2B6x%2B9%29-K=0
%285x%5E2%2B5-K%28x%5E2%2B6x%2B9%29%29%2F%28x%5E2%2B6x%2B9%29=0
so 5x%5E2%2B5-K%28x%5E2%2B6x%2B9%29=0<-->%285-K%29x%5E2-6Kx%2B%285-9K%29=0 has a single solution.
That means the discriminant of that quadratic equation is zero.
%286K%29%5E2-4%285-K%29%285-9K%29=0
36K%5E2-4%2825-50K%2B9K%5E2%29=0
36K%5E2-100%2B200K-36K%5E2=0
200K-100=0
200K=100
K=100%2F200
K=1%2F2 The minimum of y is 1%2F2 .

WITH DERIVATIVE:
dy%2Fdx=10%283x-1%29%2F%28x%2B3%29%5E3
and find out that the derivative is
undefined for x=-3 ,
negative for -3%3Cx%3C1%2F3 ,
positive for x%3C-3 and x%3E1%2F3 ,
and zero for x=1%2F3 ,
indicating a minimum of the function for x=1%2F3 .
Then, the minimum value for the function is
From that I know that for very negative values y is barely above 5 .
As x increases,
y grows without bounds as x approaches -3 .
After that,
y decreases to its 1%2F2=0.5 minimum for x=1%2F3 ,
and then y increases again towards 5 .
Seeing all that in a graph requires viewing a bird's eye view and a zoomed-in view:
graph%28600%2C300%2C-10%2C10%2C-5%2C45%2C5%28x%5E2%2B1%29%2F%28x%2B3%29%5E2%2C5%29 graph%28300%2C300%2C-100%2C100%2C-5%2C45%2C5%28x%5E2%2B1%29%2F%28x%2B3%29%5E2%2C5%29