Question

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

Answer 1

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]}