How do you find vertical, horizontal and oblique asymptotes for #(-7x + 5) / (x^2 + 8x -20)#?

Question

Caleb Alexander · Accepted Answer

vertical asymptotes x = - 10 , x = 2
horizontal asymptote y = 0

When a rational function's denominator tends toward zero, vertical asymptotes occur. To find the equation or equations, set the denominator to zero. solve: #x^2+8x-20=0→(x+10)(x-2)=0# #rArrx=-10,x=2" are the asymptotes "# Horizontal asymptotes occur as #lim_(xto+-oo) , f(x) to 0# The equation is always y = 0 when the degree of the numerator < degree of the denominator, as is the case in this instance (numerator-degree 1, denominator-degree 2). There are no oblique asymptotes in this case because the degree of the denominator is greater than the degree of the numerator. graph{(-7x+5)/(x^2+8x-20) [-20, 20, -10, 10]}

James Allen · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. The vertical asymptotes occur where the denominator equals zero.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, perform long division or polynomial division to divide the numerator by the denominator. The oblique asymptote is the quotient obtained from this division.