1. **State the problem:** Find the intercepts of the function $$f(x) = -x^4 - 4x^3 - 2x^2 + 4x + 3$$ analytically. 2. **Recall definitions:** - The **x-intercepts** occur where $$f(x) = 0$$. - The **y-intercept** occurs where $$x = 0$$. 3. **Find the y-intercept:** Evaluate $$f(0)$$: $$f(0) = -(0)^4 - 4(0)^3 - 2(0)^2 + 4(0) + 3 = 3$$ So the y-intercept is at $$(0, 3)$$. 4. **Find the x-intercepts:** Solve $$-x^4 - 4x^3 - 2x^2 + 4x + 3 = 0$$. Multiply both sides by $$-1$$ to simplify: $$x^4 + 4x^3 + 2x^2 - 4x - 3 = 0$$ 5. **Try rational root theorem:** Possible rational roots are factors of 3 over factors of 1: $$\pm1, \pm3$$. 6. **Test $$x=1$$:** $$1 + 4 + 2 - 4 - 3 = 0$$ So $$x=1$$ is a root. 7. **Divide polynomial by $$(x-1)$$:** Using synthetic division: Coefficients: 1, 4, 2, -4, -3 Bring down 1. Multiply 1*1=1, add to 4 = 5. Multiply 5*1=5, add to 2 = 7. Multiply 7*1=7, add to -4 = 3. Multiply 3*1=3, add to -3 = 0. Quotient: $$x^3 + 5x^2 + 7x + 3$$ 8. **Solve cubic $$x^3 + 5x^2 + 7x + 3 = 0$$:** Try rational roots again: $$\pm1, \pm3$$. Test $$x=-1$$: $$-1 + 5 - 7 + 3 = 0$$ So $$x=-1$$ is a root. 9. **Divide cubic by $$(x+1)$$:** Coefficients: 1, 5, 7, 3 Bring down 1. Multiply 1*(-1) = -1, add to 5 = 4. Multiply 4*(-1) = -4, add to 7 = 3. Multiply 3*(-1) = -3, add to 3 = 0. Quotient: $$x^2 + 4x + 3$$ 10. **Solve quadratic $$x^2 + 4x + 3 = 0$$:** Use quadratic formula: $$x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 - 12}}{2} = \frac{-4 \pm 2}{2}$$ So, $$x = \frac{-4 + 2}{2} = -1$$ $$x = \frac{-4 - 2}{2} = -3$$ 11. **List all x-intercepts:** From steps 6, 8, and 10, roots are $$x = 1, -1, -3$$. Note $$x = -1$$ is a repeated root. 12. **Final intercepts:** - x-intercepts: $$(1, 0), (-1, 0), (-3, 0)$$ - y-intercept: $$(0, 3)$$