For this case we have the following functions:
[tex]f (x) = x ^ 2-8x + 15\\g (x) = x-3[/tex]
We have to:
[tex]h (x) = \frac {f (x)} {g (x)} = \frac {x ^ 2-8x + 15} {x-3} = \frac {(x-3) (x-5) } {(x-3)} = (x-5)[/tex]
We have that by definition, the domain of a function is given by all the values for which the function is defined. We have that h (x) ceases to be defined when the denominator is 0. That is:
[tex]x-3 = 0\\x = 3[/tex]
Thus, the domain is given by:
(-infinity, 3) U (3, infinity)
Answer:
[tex]h (x) = (x-5)[/tex]
Domain: (-infinity, 3) U (3, infinity)
