identify whether each value of x is a discontinuity of the function by typing asymptote, hole, or…

Explanation:

Step1: Factor the denominator

First, factor $x^{3}+5x^{2}+6x=x(x + 2)(x+3)$. So the function is $y=\frac{5x}{x(x + 2)(x + 3)}$.

Step2: Analyze $x=-3$

When $x=-3$, the denominator is zero and the numerator is non - zero ($5\times(-3)=-15\neq0$). So $x = - 3$ is a vertical asymptote.

Step3: Analyze $x=-2$

When $x=-2$, the denominator is zero and the numerator is non - zero ($5\times(-2)=-10\neq0$). So $x=-2$ is a vertical asymptote.

Step4: Analyze $x = 0$

When $x = 0$, we can cancel out the common factor $x$ in the numerator and denominator. The original function simplifies to $y=\frac{5}{(x + 2)(x + 3)}$ for $x\neq0$. So $x = 0$ is a hole.

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

When $x = 2,3,5$, the denominator $(x + 2)(x + 3)x\neq0$. The function is continuous at these points. So they are neither asymptotes nor holes.

Answer:

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