[tex]\begin{gathered} \text{ A vertical asymptote implies a denominator of a function where x=0 therefore} \\ x=5\rightarrow x-5=0\rightarrow(x-5) \\ x=-3\rightarrow x-3=0\rightarrow(x-3) \\ \text{ A hole in the graph at x=-1 implied that the factor is both on numerator and denominator} \\ \text{ Therefore,} \\ x=-1\rightarrow(x+1)\text{ is both on numerator and denominator} \\ \text{ A horizontal asymptote at y=7 means the numerator and denominator} \\ \text{ have their coefficients at 7:1} \\ \text{ Putting it together we have} \\ \frac{7(x+1)}{1(x-5)(x-3)(x+1)} \\ \text{simplify and we get }\frac{7(x+1)}{(x-5)(x-3)(x+1)} \end{gathered}[/tex]
