Begin by factoring the denominator to get: 

$\displaystyle f(x)=\frac{(x+2)}{(x+2)(x-8)}$.

Do not simplify further! Realize that any value of "x" that causes the denominator to equal zero is not included in the domain of the function.

Therefore, one can see that the domain does not include x values of 8 or -2. No other gaps exist in this function so one can definitively say the domain is characterized by $\displaystyle x\neq8;\:x\neq-2$.

The value of the inner term of a square root cannot be negative.  This means that no number can be less than zero.

Set the inner term equal to zero and solve for $\displaystyle x$.

$\displaystyle 4x+7=0$

$\displaystyle 4x=-7$

$\displaystyle x=-\frac{7}{4}$

This means that $\displaystyle x$ cannot be less than this value.

Therefore, the $\displaystyle y$ exists for every number equal to $\displaystyle -\frac{7}{4}$ or greater.

The answer is:  $\displaystyle x\geq -\frac{7}{4}$

The terms inside a square root cannot be negative, but can be equal to zero.

Set the terms inside the square root to zero to determine where the domain will begin.  

$\displaystyle x+2 =0$

$\displaystyle x=-2$

The value of $\displaystyle x$ cannot be less than negative two, but can be more than negative two.  The negative sign in front of the square root symbol will flip the graph across the x-axis, and will not affect the domain.  

The domain is:  $\displaystyle x\geq -2$

In determining the domain of a function, we must ask ourselves where the function is undefined. To do this for our function, we must set the denominator equal to zero, and solve for x; at this x value, we get a zero in the denominator of the function which produces an undefined value.

$\displaystyle x-9=0$

$\displaystyle x=9$

This is the only limitation for the domain of the function, so our domain is

$\displaystyle (-\infty, 9)\cup(9, \infty)$

The contents inside the square root cannot be less than zero.

Set the inner quantity equal to zero. 

$\displaystyle 3-2x = 0$

Find $\displaystyle x$.  This is the critical point.  Add $\displaystyle 2x$ on both sides and then divide by two on both sides.

$\displaystyle 3-2x+2x = 0+2x$

$\displaystyle 2x=3$

$\displaystyle x=\frac{3}{2}$

We will test numbers less than and more three halves.

Let:  $\displaystyle x=0$

$\displaystyle y=\sqrt{3-2(0)} = \sqrt3$

Let:  $\displaystyle x=2$

$\displaystyle y=\sqrt{3-2(2)} = \sqrt{3-4}=\sqrt{-1}$

The negative inside the square square root indicates that this is an imaginary term.

This tells us that the domain is satisfied when $\displaystyle x$ is less than or equal to $\displaystyle \frac{3}{2}$.

The correct answer is:  $\displaystyle x\leq\frac{3}{2}$

The domain of a rational function is the set of all real numbers except for the value(s) of $\displaystyle x$ that make the denominator zero. The value(s) can be found as follows:

$\displaystyle 5x- 7 = 0$

$\displaystyle 5x- 7+ 7 = 0 + 7$

$\displaystyle 5x= 7$

$\displaystyle 5x \div 5 = 7 \div 5$

$\displaystyle x = \frac{7}{5}$

The domain is the set of all real numbers except $\displaystyle \frac{7}{5}$ - or $\displaystyle \left (- \infty, \frac{7}{5} \right ) \cup \left ( \frac{7}{5} , \infty \right )$.

Since the radicand of a square root must be nonnegative, the domain of a radical function with a square root can be found by setting the radicand greater than or equal to 0:

$\displaystyle 7x + 6 \ge 0$

$\displaystyle 7x + 6 -6 \ge 0 - 6$

$\displaystyle 7x \ge - 6$

$\displaystyle 7x \div 7 \ge - 6 \div 7$

$\displaystyle x \ge - \frac{6}{7}$

This is the domain, which can also be stated as $\displaystyle \left [ - \frac{6}{7} , \infty \right )$.

Since the radicand of a square root must be nonnegative, the domain of a radical function with a square root can be found by setting the radicand greater than or equal to 0:

$\displaystyle 10 - 8x \ge 0$

$\displaystyle 10 - 8x - 10 \ge 0 - 10$

$\displaystyle - 8x \ge - 10$

$\displaystyle - 8x\div (-8) \le - 10 \div (-8)$

[note the switch of symbol because of division by a negative number]

$\displaystyle x \le 1.25$

This is the domain, which can also be stated as $\displaystyle (-\infty , 1.25]$.

The domain of a rational function is the set of all real numbers except for the value(s) of $\displaystyle x$ that make the denominator zero. The value(s) can be found as follows:

$\displaystyle x^{2} + 9 = 0$

$\displaystyle x^{2} + 9 - 9= 0 - 9$

$\displaystyle x^{2} = - 9$

However, there is no real value of $\displaystyle x$ whose square is $\displaystyle -9$, so this statement has no solution. Therefore, there is no real value of $\displaystyle x$ which makes the denominator zero. The domain of $\displaystyle f$ is consequently the set of all real numbers.

$\displaystyle P (x) = x^{2} - 6x + 8$ is a polynomial function. The domain of any polynomial function is the set of all real numbers, making that the correct choice.