For $$ k(q)=\frac{q^{2}}{2 q^{2}+3 q-27} $$ (a) Write the domain in set-builder notation, (b) Write the domain in interval notation, Express numbers as integers or simplified fractions, Part: $$ 0 / 2 $$ Part 1 of 2 (a) Write the domain in set-builder notation. OThe domain is the empty set, The domain written in set-builder notation is $$ \{q \mid q $$ is a real number $$ \} $$ $$ \{q \mid q $$ is a real number and

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To determine the **domain** of the function $$ k(q) = \frac{q^{2}}{2 q^{2} + 3 q - 27} $$, we need to identify all real numbers $$ q $$ for which the function is defined. The function is undefined when the **denominator** is zero because division by zero is undefined.

### Step 1: Find the Values That Make the Denominator Zero

Set the denominator equal to zero and solve for $$ q $$:

$$
2 q^{2} + 3 q - 27 = 0
$$

Use the **quadratic formula** to solve for $$ q $$:

$$
q = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
$$

Where $$ a = 2 $$, $$ b = 3 $$, and $$ c = -27 $$:

$$
q = \frac{-3 \pm \sqrt{3^{2} - 4(2)(-27)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 216}}{4} = \frac{-3 \pm \sqrt{225}}{4} = \frac{-3 \pm 15}{4}
$$

This gives two solutions:

$$
q = \frac{-3 + 15}{4} = \frac{12}{4} = 3
$$
$$
q = \frac{-3 - 15}{4} = \frac{-18}{4} = -\frac{9}{2}
$$

So, the denominator is zero at $$ q = 3 $$ and $$ q = -\frac{9}{2} $$.

### Part (a): Domain in Set-Builder Notation

The domain consists of all real numbers except the values that make the denominator zero. Therefore, in set-builder notation:

$$
\text{Domain} = \left\{\, q \;\bigg|\; q \text{ is a real number and } q \neq 3 \text{ and } q \neq -\frac{9}{2} \,\right\}
$$

### Part (b): Domain in Interval Notation

In interval notation, we exclude the points $$ q = -\frac{9}{2} $$ and $$ q = 3 $$ by using parentheses and combine the intervals with a union symbol $$ \cup $$:

$$
\text{Domain} = \left( -\infty, -\frac{9}{2} \right) \cup \left( -\frac{9}{2},\; 3 \right) \cup \left( 3, \infty \right)
$$

### Summary

- **Set-Builder Notation:**
  $$
  \left\{\, q \;\bigg|\; q \in \mathbb{R} \text{ and } q \neq 3 \text{ and } q \neq -\frac{9}{2} \,\right\}
  $$

- **Interval Notation:**
  $$
  \left( -\infty, -\frac{9}{2} \right) \cup \left( -\frac{9}{2},\; 3 \right) \cup \left( 3, \infty \right)
  $$

These notations precisely describe all real numbers $$ q $$ for which the function $$ k(q) $$ is defined.
The domain of $$ k(q) = \frac{q^{2}}{2 q^{2} + 3 q - 27} $$ is all real numbers except $$ q = 3 $$ and $$ q = -\frac{9}{2} $$. - **Set-Builder Notation:** $$ \{ q \mid q \in \mathbb{R} \text{ and } q \neq 3 \text{ and } q \neq -\frac{9}{2} \} $$ - **Interval Notation:** $$ \left( -\infty, -\frac{9}{2} \right) \cup \left( -\frac{9}{2},\; 3 \right) \cup \left( 3, \infty \right) $$

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