How do you find the asymptotes for #f(x)=(-7x + 5) / (x^2 + 8x -20)#?

Question

Elias Ackerman · Accepted Answer

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

The denominator of f(x) cannot be zero as this would be undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values of x then they are vertical asymptotes. solve: #x^2+8x-20=0rArr(x+10)(x-2)=0# #rArrx=-10,x=2" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by the highest power of x, that is #x^2# #((-7x)/x^2+5/x^2)/(x^2/x^2+(8x)/x^2-20/x^2)=(-7/x+5/x^2)/(1+8/x-20/x^2)# as #xto+-oo,f(x)to(0+0)/(1+0-0)# #rArry=0" is the asymptote"# graph{(7x+5)/(x^2+8x-20) [-40, 40, -20, 20]}

Elias Ackerman · Answer

undefined To find the asymptotes for the function $ f(x) = \frac{-7x + 5}{x^2 + 8x - 20} $, you need to consider the behavior of the function as $ x $ approaches positive or negative infinity.

First, factor the denominator of the rational function: $ x^2 + 8x - 20 = (x + 10)(x - 2) $.

The function will have vertical asymptotes where the denominator equals zero, so set $ x + 10 = 0 $ and $ x - 2 = 0 $ to find the vertical asymptotes.

This gives $ x = -10 $ and $ x = 2 $ as the vertical asymptotes.

Now, consider the horizontal asymptote. To find it, compare the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote will be at $ y = 0 $.

Therefore, the vertical asymptotes are $ x = -10 $ and $ x = 2 $, and the horizontal asymptote is $ y = 0 $.