1. **State the problem:** Find the x- and y-intercepts of the function $$f(x)=x^3-2x^2-24x$$. 2. **Find y-intercept:** The y-intercept occurs when $$x=0$$. Evaluate: $$f(0)=0^3 - 2(0)^2 - 24(0) = 0$$ So, the y-intercept is at $$(0, 0)$$. 3. **Find x-intercepts:** The x-intercepts occur when $$f(x) = 0$$: $$x^3 - 2x^2 - 24x = 0$$ Factor out $$x$$: $$x(x^2 - 2x - 24) = 0$$ Set each factor to zero: - $$x = 0$$ - $$x^2 - 2x - 24 = 0$$ 4. **Solve quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ for $$x^2 - 2x - 24 = 0$$ where $$a=1$$, $$b=-2$$, $$c=-24$$: $$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-24)}}{2(1)} = \frac{2 \pm \sqrt{4 + 96}}{2} = \frac{2 \pm \sqrt{100}}{2}$$ 5. **Calculate roots:** $$\sqrt{100} = 10$$ So the roots are: $$x = \frac{2 + 10}{2} = 6$$ and $$x = \frac{2 - 10}{2} = -4$$ 6. **List x-intercepts:** The x-intercepts are at $$(0, 0)$$, $$(6, 0)$$, and $$(-4, 0)$$. **Final answer:** - Y-intercept: $$(0, 0)$$ - X-intercepts: $$(0, 0), (6, 0), (-4, 0)$$