Question

What is the domain and range of `f(x) = (x+3) /( x^2 + 8x + 15)`?

Answer 1

The domain is `x in (-oo, -5)uu(-5,+oo)`. The range is `y in (-oo,0)uu(0,+oo)`

Explanation:

The function is

`f(x)=(x+3)/(x^2+8x+15)=(x+3)/((x+3)(x+5))=1/(x+5)`

The denominator must be `!=0`

Therefore,

`x+5!=0`

`x!=-5`

The domain is `x in (-oo, -5)uu(-5,+oo)`

To calculate the range, let

`y=(1)/(x+5)`

`y(x+5)=1`

`yx+5y=1`

`yx=1-5y`

`x=(1-5y)/y`

The denominator must be `!=0`

`y!=0`

The range is `y in (-oo,0)uu(0,+oo)`

graph{1/(x+5) [-16.14, 9.17, -6.22, 6.44]}

Answer 2

Domain: `x inRR, x!=-5`

Range: `y inRR, y!=0`

Explanation:

We can factor the denominator as `(x+3)(x+5)`, since `3+5=8`, and `3*5=15`. This leaves us with

`(x+3)/((x+3)(x+5))`

We can cancel out common factors to get

`cancel(x+3)/(cancel(x+3)(x+5))=>1/(x+5)`

The only value that will make our function undefined is if the denominator is zero. We can set it equal to zero to get

`x+5=0=>x=-5`

Therefore, we can say the domain is

`x inRR, x!=-5`

To think about our range, let's go back to our original function

`(x+3)/((x+3)(x+5))`

Let's think about the horizontal asymptote. Since we have a higher degree on the bottom, we know we have a HA at `y=0`. We can show this graphically:

graph{(x+3)/((x+3)(x+8)) [-17.87, 2.13, -4.76, 5.24]}

Notice, our graph never touches the `x`-axis, which is consistent with having a horizontal asymptote at `y=0`.

We can say our range is

`y inRR, y!=0`

Hope this helps!