identify whether each value of x is a discontinuity of the function by typing asymptote, hole, or neither…

Answer:

• $x = - 3$: asymptote
• $x=-2$: asymptote
• $x = 0$: hole
• $x = 2$: neither
• $x = 3$: neither
• $x = 5$: neither

Explanation:

Step1: Factor the denominator

$x^{3}+5x^{2}+6x=x(x + 2)(x+3)$

Step2: Analyze $x=-3$

When $x=-3$, the denominator is 0 and the numerator is non - zero ($5\times(-3)=-15$), so it's an asymptote.

Step3: Analyze $x=-2$

When $x = - 2$, the denominator is 0 and the numerator is non - zero ($5\times(-2)=-10$), so it's an asymptote.

Step4: Analyze $x = 0$

The function can be written as $\frac{5x}{x(x + 2)(x + 3)}=\frac{5}{(x + 2)(x + 3)}$ for $x\neq0$. There is a common factor of $x$ in the numerator and denominator, so it's a hole.

Step5: Analyze $x=2,3,5$

When $x = 2,3,5$, the denominator $(x + 2)(x + 3)x\neq0$, so the function is continuous at these points, i.e., neither asymptote nor hole.