Question

What are the discontinuities of the function f(x) = the quantity of x squared plus 6 x plus 9, all over 3 x plus 15.?

A. x ≠ −3
B.x ≠ −2
C. x ≠ −8
D. x ≠ −5

Answer 1

assuming your function is

`(x^2+6x+9)/(3x+15) = ((x+3)(x+3))/(3(x+5))`
nothing can be factored.
the only discontinuity arises where the denominator would be equal to 0. When x = -5 you will have a discontinuity because it makes the denominator equal to 0.

Answer 2

The correct option is D.

Explanation:

The given function is

`f(x)=(x^2+6x+9)/(3x+15)`

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

Using algebraic property:

`(a+b)^2=a^2+2ab+b^2`

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

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

A rational function is discontinuous at that point where the value of denominator equal to 0.

Equate denominator equal to 0.

`3(x+5)=0`

`x+5=0`

`x=-5`

Therefore function is discontinuous at
`x=-5`.

The discontinuity of the function is

`x\\eq -5`

The function f(x) is not defined at
`x=-5`.Therefore option D is correct.