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To find the domain of the given rational function:

\[
f(x) = \frac{7 - 3x}{3x^2 + 16x - 12}
\]

we need to determine the values of \(x\) for which the denominator is not equal to zero. The function is undefined where the denominator equals zero.

The denominator is:

\[
3x^2 + 16x - 12
\]

We now solve the equation:

\[
3x^2 + 16x - 12 = 0
\]

Using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 3\), \(b = 16\), and \(c = -12\), we substitute these values into the formula:

\[
x = \frac{-16 \pm \sqrt{16^2 - 4(3)(-12)}}{2(3)}
\]

Let's solve this quadratic step-by-step.The solutions to the equation \( 3x^2 + 16x - 12 = 0 \) are:

\[
x_1 = \frac{2}{3}, \quad x_2 = -6
\]

These are the values that make the denominator zero, which means the function \( f(x) \) is undefined at \( x = \frac{2}{3} \) and \( x = -6 \).

Therefore, the domain of the function is all real numbers except \( x = \frac{2}{3} \) and \( x = -6 \). In interval notation, the domain is:

\[
(-\infty, -6) \cup (-6, \frac{2}{3}) \cup (\frac{2}{3}, \infty)
\]

Would you like further details or explanations? Here are five related questions to expand your understanding:

1. How does solving for the domain differ for functions with square roots?
2. What happens to the range of a function when you exclude values from the domain?
3. Can the quadratic formula be used for non-polynomial equations?
4. How do vertical asymptotes relate to the domain of rational functions?
5. How would the domain change if the denominator had a higher degree polynomial?

**Tip:** Always check for both real and imaginary solutions when solving quadratic equations, as complex solutions indicate no domain restrictions for real numbers.

Find the domain of the given function: f(x) = (7 - 3x) / (3x^2 + 16x - 12).

Learn how to find the domain of a rational function where the denominator is a quadratic equation. This example problem walks through solving the quadratic equation using the quadratic formula and applying interval notation to express the domain. Suitable for students in Grades 9-12.

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