Question

What is the domain and range of `y= (4x^2 - 9) / ((2x+3)(x+1))`?

Answer 1

See below.

Explanation:

Notice:

`4x^2-9` is the difference of two squares. This can be expressed as:

`4x^2-9=(2x+3)(2x-3)`

Substituting this in numerator:

`((2x+3)(2x-3))/((2x+3)(x+1))`

Canceling like factors:

`(cancel((2x+3))(2x-3))/(cancel((2x+3))(x+1))=(2x-3)/(x+1)`

We notice that for `x=-1` the denominator is zero. This is undefined, so our domain will be all real numbers `bbx` `x!=-1`

We can express this in set notation as:

`{x in RR | x != -1}`

or in interval notation:

`(-oo , -1)uu(-1,oo)`

To find the range:

We know the function is undefined for `x=-1`, therefore the line `x=-1` is a vertical asymptote. The function will go to `+-oo` at this line.

We now see what happens as `x ->+-oo`

Divide `(2x-3)/(x+1)` by `x`

`((2x)/x-3/x)/(x/x+1/x)=(2-3/x)/(1+1/x)`

as: `x->+-oo` `\ \ \ \ \ (2-3/x)/(1+1/x)=(2-0)/(1+0)=2`

This shows the line `y=2` is a horizontal asymptote. The function can't therefore ever equal 2.

so the range can be expressed as:

`{y in RR| y !=2}`

or

`(-oo,2)uu(2 , oo)`

This can be seen from the graph of the function:

graph{(2x-3)/(x+1) [-32.48, 32.44, -16.23, 16.25]}