How do you find the asymptotes for f(x)= (x^2+4x+3)/(x^2 - 9)?
2 Answers
vertical asymptote=
horizontal asymptote =
Explanation:
The vertical asymptote is equal to denominator set to zero.Given that your denominator has a x squared the result will be x=
The horizontal asymptote in this case is equal to the division of the x with largest exponent in both numerator and denominator .
vertical asymptote x = 3
horizontal asymptote y = 1
Explanation:
First step is to factor the function:
(x^2+4x+3)/(x^2-9) =( (x+3)(x+1))/((x+3)(x-3)) and simplifying :
(cancel(x+3)(x+1))/(cancel(x+3)(x-3)) left with
(x+1)/(x-3) Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let denominator equal zero.
solve: x - 3 = 0 → x = 3 is equation of asymptote.
Horizontal asymptotes occur as
lim_(x→±∞) f(x) → 0 now
(x+1)/(x-3) =( x/x + 1/x)/(x/x -3/x) = (1+1/x)/(1 -3/x) as x approaches ∞ ,
1/x " and " -3/x → 0 hence y = 1 is the asymptote
here is the graph of the function.
graph{(x^2+4x+3)/(x^2-9) [-10, 10, -5, 5]}
