SOLUTION: Find the domain of each function. Find any horizontal, vertical, or oblique asymptotes. (a) R(x)= (x-4)/(x+3)^2 (b) f(x)= (2x^2-14x+24)/(x^2+6x-40) (c) g(x)= (x^2+2x-3

Algebra ->  Rational-functions -> SOLUTION: Find the domain of each function. Find any horizontal, vertical, or oblique asymptotes. (a) R(x)= (x-4)/(x+3)^2 (b) f(x)= (2x^2-14x+24)/(x^2+6x-40) (c) g(x)= (x^2+2x-3      Log On


   

Question 1188782: Find the domain of each function. Find any horizontal, vertical, or oblique asymptotes.
(a) R(x)= (x-4)/(x+3)^2
(b) f(x)= (2x^2-14x+24)/(x^2+6x-40)

(c) g(x)= (x^2+2x-3)/(x+1)
(need workout and answer immediately)
Found 2 solutions by mccravyedwin, Edwin McCravy:
Answer by mccravyedwin(408) About Me  (Show Source):
Answer by Edwin McCravy(20063) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing your homework for you I will do problems with different numbers 
that are done step-by-step as yours are done, so you can use them as a step-by-step 
model to do yours by.
(a) R(x)= (x-5)/(x+6)^2
The denominator cannot equal 0, so set (x+6)^2 = 0 and solve, getting x = -6,
so this tells you everything that x cannot equal so x cannot be -6, so the domain
is 

and there is a vertical asymptote at x=-6.

The denominator has a greater degree than the numerator, so the horizontal
asymptote is the x-axis, which has equation y=0.

(b) f(x)= (3x^2-33x+90)/(x^2+3x-54) = (3(x^2-11x+30))/((x+9)(x-6))
The denominator cannot equal 0, so set (x+9)(x-6) = 0 and solve, getting 
x = -9, and x = 6. This tells you everything that x cannot equal, so x cannot be
-9 or 6, so the domain
is 

To find the asymptotes and/or holes, we first must factor both numerator and
denominator to see if we have any holes (because of like factors in numerator
and denominator).



The fact that (x-6) is a like factor of both numerator and denominator means
that there is a hole, not a vertical asymptote, at x=6.  So the only vertical
asymptote is when x+9 = 0 or at x = -9.

Since the degrees of the numerator and the denominator are both 2, (x2),
the equation of the horizontal asymptote is found by setting y
equal to the fraction with the numerator equal to the leading coefficient of the
numerator and the denominator equal to the leading coefficient of the
denominator.  So the horizontal asymptote has equation y = 3/1 or y = 3.
(c) g(x)= (x^2+5x-14)/(x+2)
The denominator cannot equal 0, so set (x+2) = 0 and solve, getting x = -2, so
this tells you everything that x cannot equal so x cannot be -1, so the domain
is 

To find the asymptotes and/or holes, we first must factor the numerator to see
if we have any holes (because of like factors in numerator and denominator).


The fact that there are no like factors of both numerator and denominator means
that there is no hole when x+2=0, but a vertical asymptote.  So the only
vertical asymptote is when x+2 = 0 or at x = -2.

Since the degree of the numerator is 2 and the degree of the denominator is 1,
that is, the numerator's degree is 1 more than the denominator's degree, that
means there is an oblique asymptote.  To find its equation, divide the
numerator by the denominator using long division (or synthetic division if you
understand the significant parts of synthetic division.  I will use long
division in case you don't:

        x+ 3
x+2)x2+5x-14
    x2+2x
       3x-14
       3x+ 6
         -20

So the quotient is x+3-20/(x+2).  We ignore the fraction because it will
approach 0 as x gets very large positively and negatively. So the equation of
the oblique asymptote is gotten by setting y equal to just the quotient.

Oblique asymptote is y = x+3   

Now use these as a model and do yours step-by-step exactly the same way.  

Edwin