Given the following parameters
[tex]\begin{gathered} \text{Vertical asymptotes}\Rightarrow x=-2,x=1 \\ X-\text{intercepts}\Rightarrow x=-1,x=3 \\ \text{horizontal asymptote}\Rightarrow y=10 \end{gathered}[/tex]The vertical asymptotes indicates that denomenator will be (x+2)(x-1), i.e
[tex]\begin{gathered} x=-2 \\ x+2=0 \\ x=1 \\ x-1=0 \end{gathered}[/tex]The intercepts on the x axis indicates that the numerator will be ( x+ 1) (x-3)
i.e
[tex]\begin{gathered} x=-1\Rightarrow x+1=0 \\ x=3\Rightarrow x-3=0 \end{gathered}[/tex]The equation would be in this form
[tex]y=\frac{(x+1)(x-3)}{(x+2)(x-1)}[/tex]The equation above has a horizontal asymptote of y=1 as the degree of the numerator and denominator is equal , likewise their leading coefficient, i.e the ratio is 1:1
Thus, to make the equation have horizontal asymptote of y=10, multiply by 10/1
[tex]\begin{gathered} y=\frac{10}{1}\times\frac{(x+1)(x-3)}{(x+2)(x-1)_{}} \\ y=\frac{10(x+1)(x-3)}{(x+2)(x-1)} \end{gathered}[/tex]Hence, the rational function with the following parameters provided in the question is given by
[tex]y=\frac{10(x+1)(x-3)}{(x+2)(x-1)}[/tex]