Let's begin by finding the x- and y-intercepts of the given rational function:

$f(x) = \frac{x^3 + 2x^2 - 8x}{x^2 + x - 2}$

1. Finding the x-intercepts:

To find the x-intercepts, we set the numerator equal to zero because the x-intercepts occur when $f(x) = 0$, which happens when the numerator is zero (the denominator cannot be zero at this point).

$x^3 + 2x^2 - 8x = 0$

Factor the numerator:

$x(x^2 + 2x - 8) = 0$

Now solve for $x$:

$x = 0 \quad \text{or} \quad x^2 + 2x - 8 = 0$

Next, solve the quadratic equation $x^2 + 2x - 8 = 0$. We can factor this:

$(x + 4)(x - 2) = 0$

Thus, the solutions are:

$x = -4 \quad \text{or} \quad x = 2$

Therefore, the x-intercepts are:

$x = 0, -4, 2$

2. Finding the y-intercept:

To find the y-intercept, we substitute $x = 0$ into the function. The y-intercept occurs where the graph crosses the y-axis, which happens when $x = 0$:

$f(0) = \frac{0^3 + 2(0)^2 - 8(0)}{0^2 + 0 - 2} = \frac{0}{-2} = 0$

Thus, the y-intercept is:

$y = 0$

Final Answer:

• x-intercepts: $x = 0, -4, 2$
• y-intercept: $y = 0$

Would you like any further details or clarifications? Here are some additional questions you might find interesting:

1. How can we factor more complex cubic expressions?
2. What happens to the function as $x \to \infty$ or $x \to -\infty$?
3. How do we find the vertical asymptotes of rational functions?
4. What are the horizontal asymptotes, if any, for rational functions?
5. How do you solve rational equations like this one when there's no simple factorization?

Tip: When factoring polynomials, always look for common factors first, then try factoring quadratics or applying the rational root theorem if needed.