How do you graph y=-8/(x^2-4)y=−8x2−4 using asymptotes, intercepts, end behavior?
1 Answer
see explanation.
Explanation:
color(blue)"Asymptotes"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=+-2x2−4=0⇒x2=4⇒x=±2
rArrx=-2" and " x=2" are the asymptotes"⇒x=−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]}