How do you graph #y=-8/(x^2-4)# using asymptotes, intercepts, end behavior?

1 Answer
Jan 11, 2017

Answer:

see explanation.

Explanation:

#color(blue)"Asymptotes"#

The denominator of y cannot be zero as this would make y 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=0rArrx^2=4rArrx=+-2#

#rArrx=-2" and " x=2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" ( a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#y=-(8/x^2)/(x^2/x^2-4/x^2)=-(8/x^2)/(1-4/x^2)#

as #xto+-oo,yto-0/(1-0)#

#rArry=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 0,denominator-degree 2 ) Hence there are no oblique asymptotes.

#color(blue)"Intercepts"#

#x=0toy=-8/(-4)=2rArr(0,2)#

#y=0" has no solution, hence y does not cross the x-axis"#

#color(blue)"end behaviour"#

#"we already know, from above that " lim_(xto+-oo)f(x)=0#

#color(blue)"behaviour around vertical asymptotes"#

#lim_(xto-2^-)f(x)=-oo" and " lim_(xto-2^+)f(x)=+oo#

#lim_(xto2^-)=+oo" and " lim_(xto2^+)=-oo#
graph{-8/(x^2-4) [-10, 10, -5, 5]}