Answer:
[tex]y=\frac{\frac{16}{3}(x+1)(x+6)}{(x-1)(x-4)}[/tex]
Step-by-step explanation:
Because x=1 and x=4 are the solutions to x-1=0 and x-4=0, then we have (x-1)(x-4) in the denominator.
Because (-1,0) and (-6,0) are the solutions to x+1=0 and x+6=0, then we have (x+1)(x+6) in the numerator.
Now we have [tex]y=\frac{a(x+1)(x+6)}{(x-1)(x-4)}[/tex] where we plug in (0,8) to solve for a:
[tex]y=\frac{a(x+1)(x+6)}{(x-1)(x-4)}[/tex]
[tex]8=\frac{a(0+1)(0+6)}{(0-1)(0-4)}[/tex]
[tex]8=\frac{6a}{4}[/tex]
[tex]32=6a[/tex]
[tex]\frac{32}{6}=a[/tex]
[tex]\frac{16}{3}=a[/tex]
Therefore, the equation for the rational function is [tex]y=\frac{\frac{16}{3}(x+1)(x+6)}{(x-1)(x-4)}[/tex]
