Question

determine the ordered pair of the hole(s) of the following rational function: $f(x)=\frac{4x^{2}-8x - 3}{4x^{2}-12x + 9}$
$\left(-\frac{3}{2},\frac{2}{3}\
ight)$ and $\left(-\frac{1}{2},\frac{1}{2}\
ight)$
$\left(\frac{3}{2},0\
ight)$ and $\left(\frac{1}{2},0\
ight)$
$\left(\frac{3}{2},0\
ight)$
none
question #3
determine the domain of the following rational function: $f(x)=\frac{x - 6}{x^{2}+6x + 8}$
$(-\infty,6)\cup(6,\infty)$
$(-\infty,\infty)$
$(-\infty,-4)\cup(-4,-2)\cup(-2,\infty)$
$(-\infty,2)\cup(2,4)\cup(4,\infty)$

Explanation:

Step1: Factor the numerator and denominator of $f(x)=\frac{4x^{2}-8x - 3}{4x^{2}-12x + 9}$

The numerator $4x^{2}-8x - 3=(2x + 1)(2x-3)$. The denominator $4x^{2}-12x + 9=(2x - 3)^{2}$.

Step2: Identify the common factors

The common factor is $2x - 3$. Set $2x-3 = 0$, then $x=\frac{3}{2}$.

Step3: Simplify the function and find the $y$-value

After canceling out the common factor $2x - 3$ (for $x
eq\frac{3}{2}$), we get $f(x)=\frac{2x + 1}{2x - 3}$. Substitute $x = \frac{3}{2}$ into the simplified - function (by taking the limit), the function is undefined at $x=\frac{3}{2}$, so there is no hole.

For the domain of $f(x)=\frac{x - 6}{x^{2}+6x + 8}$:

Step1: Factor the denominator

$x^{2}+6x + 8=(x + 2)(x + 4)$.

Step2: Find the values that make the denominator zero

Set $(x + 2)(x + 4)=0$. Solving $x+2 = 0$ gives $x=-2$ and solving $x + 4=0$ gives $x=-4$.

Step3: Determine the domain

The domain of a rational function is all real numbers except the values that make the denominator zero. So the domain is $(-\infty,-4)\cup(-4,-2)\cup(-2,\infty)$.

Answer:

none
C. $(-\infty,-4)\cup(-4,-2)\cup(-2,\infty)$