Question

b. write a function that has a vertical asymptote at x = -3 and a removable discontinuity when x = 1.

Explanation:

Step1: Analyze vertical asymptote

A vertical asymptote at \(x=-3\) means the denominator has a factor of \((x + 3)\)

Step2: Analyze removable discontinuity

A removable discontinuity at \(x = 1\) means both numerator and denominator have a factor of \((x - 1)\)
So the function \(f(x)=\frac{(x - 1)}{(x + 3)(x - 1)}\) satisfies the conditions. When \(x = 1\), there is a removable discontinuity (since we can cancel the \((x - 1)\) terms for \(x
eq1\)) and when \(x=-3\), the denominator is zero while the numerator is non - zero, giving a vertical asymptote.

Answer:

$f(x)=\frac{(x - 1)}{(x + 3)(x - 1)}$