How do you find the vertical, horizontal or slant asymptotes for (-7x + 5) / (x^2 + 8x -20)−7x+5x2+8x−20?
1 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)=0solve x2+8x−20=0⇒(x+10)(x−2)=0
rArrx=-10" and "x=2" are the asymptotes"⇒x=−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]}