Question

How do you graph `y=(x+2)/(x+3)` using asymptotes, intercepts, end behavior?

Answer 1

graph{(x+2)/(x+3) [-10, 10, -5, 5]}

`y`-intercept is `2/3`

`x`-intercept is `-2`

vertical asymptote at `x=-3`

`y<1` when `x> -3`

`y>1` when `x< -3`

Explanation:

asymptote:

`n/0` = undefined

`therefore` if `x+3 = 0`, `y` is undefined.

this means that it is not on the graph, and so is shown as the asymptote.

when `x+3 = 0`, `x = 0-3`

`x = -3`

intercepts:

`y`:

the `y`-intercept is when `x = 0`

`y = (x+2)/(x+3)`

`y = 2/3`

`y`-intercept is `2/3`

`x`:

the `x`-intercept is when `y=0`

`(x+2)/(x+3)=0`

the numerator has to be `0`, since `0/n = 0`

this means that `x+2=0`

when `x+2 = 0`, `x = 0-2`

`x`-intercept is `-2`

end behaviour:

`y<1`

`x+2` is always less than `x+3`. with both positive and negative numbers.

the fraction `(x+2)/(x+3)` cannot be simplified to `(>=1)/1` if `x> -3`

however, it is vice versa for negative numbers, since smaller negative numbers have a higher absolute value (distance from `0`),

e.g. `-3/-2 = |3|/|2| = 3/2`

this means that `(x+2)/(x+3)` cannot be simplified to `(<=1)/1` if `x< -3`