Question

1. write a function that has one vertical asymptote at x = 3 and a hole at x = - 2.

Explanation:

Step1: Recall function - hole and asymptote rules

A hole occurs when a factor can be canceled from the numerator and denominator. An asymptote occurs when a factor in the denominator is non - cancelable.

Step2: Construct the denominator

Since there is a vertical asymptote at \(x = 3\), the denominator should have a non - cancelable factor of \((x - 3)\). And since there is a hole at \(x=-2\), both the numerator and denominator should have a factor of \((x + 2)\).

Step3: Write the function

A possible function is \(f(x)=\frac{(x + 2)}{(x - 3)(x + 2)}\), \(x
eq - 2\).

Answer:

\(f(x)=\frac{(x + 2)}{(x - 3)(x + 2)},x
eq - 2\)