Answer
$(-\infty,2)\cup(2,5)\cup(5,\infty)$.
$\frac{x}{x-5}$.
Work Step by Step
The domain of a rational function is the set of all real numbers from which we eliminate the zeros of the denominator.
$\Rightarrow f(x)=\frac{x^2-2x}{x^2-7x+10}$
In order to dteremine the zeros of the denominator we factor $x^2-7x+10$.
Rewrite the middle term $-7x$ as $-5x-2x$.
$\Rightarrow x^2-5x-2x+10$
Group the terms.
$\Rightarrow (x^2-5x)+(-2x+10)$
Factor each group.
$\Rightarrow x(x-5)-2(x-5)$
Factor out $(x-5)$.
$\Rightarrow (x-5)(x-2)$
Determine the zeros of the denominator:
$x-5=0$ or $x-2=0$
$x=5$ or $x=2$
The domain of the function is:
$(-\infty,2)\cup(2,5)\cup(5,\infty)$.
Back substitute into the given function.
$\Rightarrow f(x)=\frac{x^2-2x}{(x-5)(x-2)}$
Factor the numerator $x^2-2x$.
Factor out $x$.
$\Rightarrow x(x-2)$
Back substitute into the given function.
$\Rightarrow f(x)=\frac{x(x-2)}{(x-5)(x-2)}$
Cancel common terms.
$\Rightarrow f(x)=\frac{x}{x-5}$.
Observation: Even though the function can be written in simplified form, the domain is determined using the original form of the function as cancelling common factors is possible only if they are not zero.
