How do you find the vertical, horizontal or slant asymptotes for F(x)= (X^2-16)/(2X^2+3X-9)?

Mar 10, 2016

vertical asymptotes $x = - 3 , x = \frac{3}{2}$
horizontal asymptote $y = \frac{1}{2}$

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation ,equate the denominator to zero.

solve: 2x^2+3x-9 = 0 → (2x-3)(x+3) = 0

$\Rightarrow x = - 3 , x = \frac{3}{2} \text{ are the asymptotes }$

Horizontal asymptotes occur as lim_(x→±∞) f(x) → 0

divide all terms on the numerator/denominator by ${x}^{2}$

(x^2-16)/(2x^2+3x-9) = (x^2/x^2-16/x^2)/((2x^2)/x^2+(3x)/x^2-9/x^2

$= \frac{1 - \frac{16}{x} ^ 2}{2 + \frac{3}{x} - \frac{9}{x} ^ 2}$

as x →∞ , 16/x^2 , 3/x " and " 9/x^2 → 0

$\Rightarrow y = \frac{1}{2} \text{ is the asymptote }$

Here is the graph of the function.
graph{(x^2-16)/(2x^2+3x-9) [-10, 10, -5, 5]}