Question

For f(x)=x^2+x-2/x^2-4, determine the equation for any vertical asymptotes, the equation for any horizontal asymptotes, and the x-coordinates of any holes

Answer 1

We know that when f(x) = p(x) / q(x), where p(x) and q(x) are polynomials, then the equations of the vertical asymptotes can be found by finding the roots of q(x).The location of the horizontal asymptote is determined by looking at the degrees of the numerator (n) and denominator (m). If n < m, the X-axis,i.e. the line y = 0 is the horizontal asymptote. If n = m, then y=a_n / b_m i.e. the ratio of the leading coefficients, is the horizontal asymptote. If n > m, there is no horizontal asymptote.

Here, x²-4 = 0 when x = -2 or 2. Therefore, the lines x = -2 and x = 2 are the vertical asymptotes of f(x).

The degree of both the numerator and the denominator is 2, so that y = 1 is the horizontal asymptote of f(x).

Further, x²+x-2 = x²+2x-x-2 = x(x+2)-1(x+2) = (x+2)(x-1) so that f(x) = (x+2)(x-1)/(x+2)(x-2). Since(x+2) is a common factor in the numerator and the denominator, hence there is a hole at x = -2. The corresponding value of y is (-2-1)/(-2-2) = -3/4. Thus, there is a hole at (-2,-3/4).