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

Respuesta :

assuming your function is 
[tex] \frac{x^2+6x+9}{3x+15} = \frac{(x+3)(x+3)}{3(x+5)} [/tex]
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:

The correct option is D.

Step-by-step explanation:

The given function is

[tex]f(x)=\frac{x^2+6x+9}{3x+15}[/tex]

[tex]f(x)=\frac{x^2+2(x)(3)+3^2}{3(x+5)}[/tex]

Using algebraic property:

[tex](a+b)^2=a^2+2ab+b^2[/tex]

[tex]f(x)=\frac{(x+3)^2}{3(x+5)}[/tex]

[tex]f(x)=\frac{(x+3)(x+3)}{3(x+5)}[/tex]

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

Equate denominator equal to 0.

[tex]3(x+5)=0[/tex]

[tex]x+5=0[/tex]

[tex]x=-5[/tex]

Therefore function is discontinuous at [tex]x=-5[/tex].

The discontinuity of the function is

[tex]x\neq -5[/tex]

The function f(x) is not defined at [tex]x=-5[/tex].Therefore option D is correct.