The rational function \( r(x)=\frac{(x+3)(x-1)(x+4)}{(x-5)(x+3)(x-4)} \) has - Zeros (x-intercepts) at \( x=1,-4 \) - Vertical asymptotes at \( x=\square \) - A hole at \( x= \) - A horizontal asymptote at \( y=\square \)

1. The given function is \[ r(x)=\frac{(x+3)(x-1)(x+4)}{(x-5)(x+3)(x-4)}. \] 2. **Identify the common factor:** The factor \((x+3)\) appears in both the numerator and the denominator. Thus, it is cancelled. However, cancellation of a factor indicates a hole in the graph rather than a zero or asymptote. 3. **Zeros (x-intercepts):** Zeros occur when the numerator is zero (ignoring any factors that were cancelled). After cancelling \((x+3)\), the numerator becomes \[ (x-1)(x+4). \] Setting each factor equal to zero gives: \[ x-1=0 \quad \Rightarrow \quad x=1, \] \[ x+4=0 \quad \Rightarrow \quad x=-4. \] Thus, the zeros are at \(x=1\) and \(x=-4\). 4. **Hole:** The cancelled factor \((x+3)\) indicates a removable discontinuity (hole) at \[ x=-3. \] 5. **Vertical asymptotes:** Vertical asymptotes are found where the denominator is zero, excluding any cancelled factors. The original denominator factors are \[ (x-5)(x+3)(x-4). \] After cancelling \((x+3)\), the remaining factors \((x-5)\) and \((x-4)\) yield vertical asymptotes at \[ x=5 \quad \text{and} \quad x=4. \] 6. **Horizontal asymptote:** The degrees of the numerator and denominator (before cancellation) are both 3. With both leading coefficients equal to 1 (since each factor is linear and monic), the horizontal asymptote is determined by the ratio of the leading coefficients: \[ y=\frac{1}{1}=1. \] Thus, the answers are: - Zeros at \(x=1\) and \(x=-4\) - Vertical asymptotes at \(x=5\) and \(x=4\) - A hole at \(x=-3\) - A horizontal asymptote at \(y=1\)