Question

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

Answer 1

vertical asymptotes `x = -3 , x = 3/2`
horizontal asymptote `y = 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`

`rArr x = -3 , x = 3/2 " 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`

`= (1-16/x^2)/(2+3/x-9/x^2)`

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

`rArr y = 1/2 " is the asymptote "`

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