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

1 Answer
Feb 21, 2017

Answer:

vertical asymptotes at x = -10 and x = 2
horizontal asymptote at y = 0

Explanation:

The denominator of the function cannot be zero as this would make the function 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 then they are vertical asymptotes.

#"solve "x^2+8x-20=0rArr(x+10)(x-2)=0#

#rArrx=-10" and "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#

#rArr((-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"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1, denominator-degree 2) Hence there are no slant asymptotes.
graph{(-7x+5)/(x^2+8x-20) [-20, 20, -10, 10]}