How to find the asymptotes of f(x)=(x^2+1)/(2x^2-3x-2)?
1 Answer
Mar 29, 2016
vertical asymptotes
horizontal asymptotes
Explanation:
Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero.
solve :
2x^2-3x-2 = 0 → (2x+1)(x-2) = 0
rArr x = -1/2 , x = 2 " are the asymptotes " Horizontal asymptotes occur as
lim_(xtooo) f(x) → 0 If the degree of the numerator/denominator are equal , as is the case here , both of degree 2. Then the equation is the ratio of the coefficients of the highest powers.
rArr y = 1/2 " is the asymptote" Here is the graph of f(x).
graph{(x^2+1)/(2x^2-3x-2) [-10, 10, -5, 5]}