Question

Question #c347b

Answer 1

The line `y=1` is a horizontal asymptote.
The line `x=3` is a vertical asymptote.

Explanation:

Note first that we can simplify the function:

`f(x) = (x^2-9)/(x^2+6x+9) = ((x-3)(x+3))/(x+3)^2 = (x-3)/(x+3)`

Now we have:

`lim_(x->oo) f(x) = lim_(x->-oo) f(x)= 1`

so the line `y=1` is a horizontal asymptote.

The function `f(x)` is continuous for every `x in RR` except for `x=3`, where we have:

`lim_(x->3^-) f(x) = -oo`

`lim_(x->3^+) f(x) = +oo`

Thus the line `x=3` is a vertical asymptote.

graph{(x^2-9)/(x^2+6x+9) [-73, 73, -36.5, 36.5]}

Answer 2

vertical

Explanation:

this equation can be simplified:

`f(x) = (x^2-9)/(x^2+6x+9)`

`f(x) = ((x+3)(x-3))/((x+3)(x+3))`

`f(x) = (cancel((x+3))(x-3))/(cancel((x+3))(x+3)`

`f(x) = (x-3)/(x+3), x!= -3`

since it is a value of `x` being restricted, the asymptote of this graph is vertical.

graph{(x-3)/(x+3) [-13.5, 6.5, -5.88, 4.12]}