Question

identify whether each value of x is a discontinuity of the function by typing asymptote, hole, or neither. x = 5 x = 3 x = 2 x = 0 x = - 2 x = - 3 $\frac{x^{3}+5x^{2}+6x}{5x}$

Explanation:

Step1: Recall function discontinuity rules

Discontinuities occur where the function is undefined (e.g., denominator is 0 for rational - functions). Let the function be \(y=\frac{x^{3}+5x^{2}+6x}{5x}\). First, simplify the function: \(y = \frac{x(x^{2}+5x + 6)}{5x}=\frac{x(x + 2)(x+3)}{5x}\), \(x
eq0\). The simplified function is \(y=\frac{(x + 2)(x + 3)}{5}\), \(x
eq0\).

Step2: Analyze \(x = 5\)

Substitute \(x = 5\) into the original (or simplified) function. The function \(y=\frac{(x + 2)(x + 3)}{5}\) is well - defined at \(x = 5\). \(y=\frac{(5 + 2)(5+3)}{5}=\frac{7\times8}{5}=\frac{56}{5}\). So \(x = 5\) is neither a discontinuity.

Step3: Analyze \(x = 3\)

Substitute \(x = 3\) into the function \(y=\frac{(x + 2)(x + 3)}{5}\). \(y=\frac{(3 + 2)(3+3)}{5}=\frac{5\times6}{5}=6\). So \(x = 3\) is neither a discontinuity.

Step4: Analyze \(x = 2\)

Substitute \(x = 2\) into the function \(y=\frac{(x + 2)(x + 3)}{5}\). \(y=\frac{(2 + 2)(2+3)}{5}=\frac{4\times5}{5}=4\). So \(x = 2\) is neither a discontinuity.

Step5: Analyze \(x = 0\)

The original function \(\frac{x^{3}+5x^{2}+6x}{5x}\) is undefined at \(x = 0\) because of the \(x\) in the denominator. But since we can cancel out the \(x\) terms (\(x
eq0\)), there is a hole at \(x = 0\).

Step6: Analyze \(x=-2\)

Substitute \(x=-2\) into the function \(y=\frac{(x + 2)(x + 3)}{5}\). \(y = 0\). The function is well - defined at \(x=-2\), so \(x=-2\) is neither a discontinuity.

Step7: Analyze \(x=-3\)

Substitute \(x=-3\) into the function \(y=\frac{(x + 2)(x + 3)}{5}\). \(y = 0\). The function is well - defined at \(x=-3\), so \(x=-3\) is neither a discontinuity.

Answer:

\(x = 5\): neither
\(x = 3\): neither
\(x = 2\): neither
\(x = 0\): hole
\(x=-2\): neither
\(x=-3\): neither