How to find the asymptotes of #f(x)=(x^2+1)/(2x^2-3x-2)#?

Question

Caleb Alexander · Accepted Answer

vertical asymptotes # x = -1/2 , x = 2 #
horizontal asymptotes # y = 1/2 # 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]}

Lincoln Anderson · Answer

undefined To find the asymptotes of the function $f(x) = \frac{x^2 + 1}{2x^2 - 3x - 2}$, we need to consider two types of asymptotes: vertical asymptotes and horizontal asymptotes.

Vertical asymptotes occur where the denominator of the function becomes zero but the numerator does not. To find vertical asymptotes, set the denominator equal to zero and solve for $x$. Then, check if these values are valid by ensuring they do not make the numerator zero. Any values of $x$ that satisfy these conditions will be vertical asymptotes.

Horizontal asymptotes occur when $x$ approaches positive or negative infinity. To find horizontal asymptotes, compare the degrees of the numerator and denominator of the rational function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $y = 0$. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

After identifying any vertical and horizontal asymptotes, you can plot the function to visualize its behavior near these asymptotes.