# How do you find vertical, horizontal and oblique asymptotes for #f(x)= (x^2-5x+6)/ (x^2-8x+15)#?

vertical asymptote x = 5

horizontal asymptote y = 1

The first step here is to factorise and simplify f(x).

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

solve : x - 5 = 0 → x = 5 is the asymptote.

divide terms on numerator/denominator by x.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1). Hence there are no oblique asymptotes. graph{(x-2)/(x-5) [-20, 20, -10, 10]}

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To find vertical asymptotes, set the denominator equal to zero and solve for x. To find horizontal asymptotes, compare the degrees of the numerator and denominator. If they are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. To find oblique asymptotes, perform polynomial long division and analyze the remainder. If the remainder approaches zero as x approaches infinity, the oblique asymptote is the quotient. If the remainder does not approach zero, there is no oblique asymptote.

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