# How do you find the asymptotes for f(x)= (x^2+4x+3)/(x^2 - 9)?

Mar 7, 2016

vertical asymptote=$- 3 , + 3$
horizontal asymptote = $1$

#### Explanation:

The vertical asymptote is equal to denominator set to zero.Given that your denominator has a x squared the result will be x=$\pm 3$. This is because ${\left(- 3\right)}^{2} = 9$and ${\left(3\right)}^{2} = 9$ and when you subtract both with 9, both are equal zero.

The horizontal asymptote in this case is equal to the division of the x with largest exponent in both numerator and denominator .
${x}^{2} / {x}^{2} = 1$

Mar 7, 2016

vertical asymptote x = 3
horizontal asymptote y = 1

#### Explanation:

First step is to factor the function:

$\frac{{x}^{2} + 4 x + 3}{{x}^{2} - 9} = \frac{\left(x + 3\right) \left(x + 1\right)}{\left(x + 3\right) \left(x - 3\right)}$

and simplifying : $\frac{\cancel{x + 3} \left(x + 1\right)}{\cancel{x + 3} \left(x - 3\right)}$

left with $\frac{x + 1}{x - 3}$

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 $\frac{x + 1}{x - 3} = \frac{\frac{x}{x} + \frac{1}{x}}{\frac{x}{x} - \frac{3}{x}} = \frac{1 + \frac{1}{x}}{1 - \frac{3}{x}}$

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]}