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The domain of a function is the set of all possible input values (usually x-values) for which the function is defined. In other words, these are the x-values that you can safely plug into the function without causing any mathematical issues, like division by zero or taking the square root of a negative number.

To find the domain of the given function \(f(x) = \frac{x^2 + 3x}{2x^3 - 5x^2 - 3x}\), start by setting the denominator equal to zero and solving for x. This will give you the values that are not allowed in the domain.

The denominator is \(2x^3 - 5x^2 - 3x\). Set it to zero:

\[2x^3 - 5x^2 - 3x = 0\]

Factor out an x:

\[x(2x^2 - 5x - 3) = 0\]

Set each factor to zero and solve:

• \(x = 0\)
• \(2x^2 - 5x - 3 = 0\).

Use the quadratic formula to solve the quadratic part:

\[x = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4}\]

This gives us the following solutions:

• \(x = 3\)
• \(x = -\frac{1}{2}\)

Thus, the domain of the function is all real numbers except
\(x = 0\), \(x = 3\), and \(x = -\frac{1}{2}\).

Asymptotes are lines that the graph of a function approaches but never actually touches.

There are two main types of asymptotes: vertical and horizontal. For the function \(f(x) = \frac{x^2 + 3x}{2x^3 - 5x^2 - 3x}\), we have both types.

**Vertical Asymptotes** occur where the denominator is zero, as long as the numerator is not zero at those points. From our domain calculation, we know:

• \(2x^3 - 5x^2 - 3x = 0\) gives us the vertical asymptotes at \(x = 0\), \(x = 3\), and \(x = -\frac{1}{2}\).

**Horizontal Asymptotes** depend on the degrees of the polynomial in the numerator and the polynomial in the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at \(y = 0\).

For our function, the degree of the numerator (2) is less than the degree of the denominator (3), so the horizontal asymptote is at:

\[y = 0\]

The x-intercepts are the points where the graph of the function crosses the x-axis. These occur when the function is equal to zero.

For the function \(f(x) = \frac{x^2 + 3x}{2x^3 - 5x^2 - 3x}\), set the numerator equal to zero:

\[x^2 + 3x = 0\]

Then, factor out an x:

\[x(x + 3) = 0\]

This gives us two solutions: \(x = 0\) and \(x = -3\). So, the x-intercepts are at:

• \(x = 0\)
• \(x = -3\)

The y-intercept is the point where the graph of the function crosses the y-axis. This occurs when \(x = 0\).

For the function \(f(x) = \frac{x^2 + 3x}{2x^3 - 5x^2 - 3x}\), when \(x = 0\), both the numerator and the denominator are zero. Therefore, the function is not defined at \(x = 0\).

As a result, the function \(f(x)\) does not have a y-intercept.

The quadratic formula is a useful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The quadratic formula is given by:

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

In our exercise, we used the quadratic formula to solve the equation:

\[2x^2 - 5x - 3 = 0\]

Here, \(a = 2\), \(b = -5\), and \(c = -3\). Plugging these values into the formula, we get:

\[x = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4}\]

Solving for the two values, we find:

• \(x = 3\)
• \(x = -\frac{1}{2}\)

These solutions helped us determine the domain and identify vertical asymptotes for the function.