Let's analyze the given rational function:

$f(x) = \frac{2x^2}{x^2 + 3x - 10}.$

We'll determine the following properties:

1. Domain of the function:

The domain of a rational function includes all real numbers except where the denominator is equal to zero.

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

$x^2 + 3x - 10 = 0.$

This quadratic equation can be solved using the quadratic formula:

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

where $a = 1$, $b = 3$, and $c = -10$.

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2}.$

This simplifies to:

$x = \frac{-3 + 7}{2} = 2 \quad \text{or} \quad x = \frac{-3 - 7}{2} = -5.$

So, the domain is all real numbers except $x = 2$ and $x = -5$:

$\text{Domain: } (-\infty, -5) \cup (-5, 2) \cup (2, \infty).$

2. Vertical asymptote:

Vertical asymptotes occur where the denominator equals zero and the numerator does not equal zero at those points.

Since the denominator equals zero at $x = 2$ and $x = -5$, and the numerator $2x^2$ does not equal zero at these points, there are vertical asymptotes at:

$x = 2 \quad \text{and} \quad x = -5.$

3. Horizontal asymptote:

To find the horizontal asymptote, compare the degrees of the numerator and the denominator:

• Both the numerator and the denominator have degree 2.
• The horizontal asymptote is given by the ratio of the leading coefficients of the numerator and denominator.

The leading coefficient of the numerator is $2$, and the leading coefficient of the denominator is $1$. Thus, the horizontal asymptote is:

$y = \frac{2}{1} = 2.$

4. X-intercept:

The x-intercept(s) occur where the numerator equals zero. Set the numerator equal to zero and solve for $x$:

$2x^2 = 0 \implies x^2 = 0 \implies x = 0.$

So, the x-intercept is at:

$(0, 0).$

5. Y-intercept:

The y-intercept occurs where $x = 0$. Substitute $x = 0$ into the function:

$f(0) = \frac{2(0)^2}{0^2 + 3(0) - 10} = \frac{0}{-10} = 0.$

So, the y-intercept is also at:

$(0, 0).$

Summary:

• Domain: $(-\infty, -5) \cup (-5, 2) \cup (2, \infty)$
• Vertical asymptotes: $x = 2$ and $x = -5$
• Horizontal asymptote: $y = 2$
• X-intercept: $(0, 0)$
• Y-intercept: $(0, 0)$

Would you like any additional details or have any questions?

Here are five related questions:

1. How does the presence of multiple vertical asymptotes affect the graph of the function?
2. What would happen to the horizontal asymptote if the degrees of the numerator and denominator were different?
3. Can a rational function have a slant asymptote if the numerator and denominator have the same degree?
4. How does the position of the x-intercept relate to the function's symmetry?
5. What can be inferred about the function if the numerator is a higher degree polynomial than the denominator?

Tip: When solving for the vertical asymptotes, always double-check that the numerator does not also equal zero at those points, which would indicate a hole in the graph instead.