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 function

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

Step2: Simplify the function

Cancel out the common - factor $x$ (for $x\neq0$), the simplified function is $y=\frac{5}{(x + 2)(x + 3)}$ for $x\neq0$.

Step3: Analyze $x=-3$

When $x=-3$, the denominator $(x + 2)(x + 3)=(-3 + 2)(-3 + 3)=0$ and the numerator is non - zero. So $x=-3$ is a vertical asymptote.

Step4: Analyze $x=-2$

When $x=-2$, the denominator $(x + 2)(x + 3)=(-2 + 2)(-2 + 3)=0$ and the numerator is non - zero. So $x=-2$ is a vertical asymptote.

Step5: Analyze $x = 0$

Since the original function has a common factor $x$ in the numerator and denominator that can be canceled out, $x = 0$ is a hole.

Step6: Analyze $x=2$

When $x = 2$, the denominator $(x + 2)(x + 3)=(2 + 2)(2 + 3)=20\neq0$. So $x = 2$ is a point of continuity (neither).

Step7: Analyze $x=3$

When $x = 3$, the denominator $(x + 2)(x + 3)=(3 + 2)(3 + 3)=30\neq0$. So $x = 3$ is a point of continuity (neither).

Step8: Analyze $x=5$

When $x = 5$, the denominator $(x + 2)(x + 3)=(5 + 2)(5 + 3)=56\neq0$. So $x = 5$ is a point of continuity (neither).