How do you find the vertical, horizontal or slant asymptotes for #f(x)=(-2r^2+5) /( r^2 - 4)#?

1 Answer
May 27, 2017

Answer:

#"vertical asymptotes at " x=+-2#
#"horizontal asymptote at " y=-2#

Explanation:

#f(x)=(-2x^2+5)/(x^2-4)#

The denominator of f(x) cannot be zero as this would make f(x) 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^2-4=0rArr(x-2)(x+2)=0#

#rArrx=+-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#

#f(x)=((-2x^2)/x^2+5/x^2)/(x^2/x^2-4/x^2)=(-2+5/x^2)/(1-4/x^2)#

as #xto+-oo,f(x)to(-2+0)/(1-0)#

#rArry=-2" is the asymptote"#
graph{(-2x^2+5)/(x^2-4) [-10, 10, -5, 5]}