1. **State the problem:** We are given the cubic function $$y = x^3 - 3x^2 + 5x - 3$$ and we want to analyze it. 2. **Recall the function:** $$y = x^3 - 3x^2 + 5x - 3$$ is a cubic polynomial. 3. **Find the critical points:** To find extrema, compute the derivative: $$y' = \frac{d}{dx}(x^3 - 3x^2 + 5x - 3) = 3x^2 - 6x + 5$$ 4. **Solve for critical points:** Set derivative equal to zero: $$3x^2 - 6x + 5 = 0$$ 5. **Use quadratic formula:** $$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3} = \frac{6 \pm \sqrt{36 - 60}}{6} = \frac{6 \pm \sqrt{-24}}{6}$$ 6. Since the discriminant is negative ($$-24$$), there are no real roots for the derivative, so no real critical points. 7. **Conclusion:** The function has no real extrema (no local maxima or minima). 8. **Find y-intercept:** Set $$x=0$$: $$y = 0 - 0 + 0 - 3 = -3$$ 9. **Find x-intercepts:** Solve $$x^3 - 3x^2 + 5x - 3 = 0$$. This cubic has no easy rational roots (by Rational Root Theorem, test $$\pm1, \pm3$$): - For $$x=1$$: $$1 - 3 + 5 - 3 = 0$$, so $$x=1$$ is a root. 10. **Factor out $$x-1$$:** $$x^3 - 3x^2 + 5x - 3 = (x-1)(x^2 - 2x + 3)$$ 11. **Solve quadratic factor:** $$x^2 - 2x + 3 = 0$$ 12. **Discriminant:** $$\Delta = (-2)^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 < 0$$ No real roots here. 13. **Final roots:** Only one real root at $$x=1$$. **Summary:** - The function has one real root at $$x=1$$. - The y-intercept is $$-3$$. - No real extrema exist. **Answer:** The function $$y = x^3 - 3x^2 + 5x - 3$$ has one real root at $$x=1$$, y-intercept at $$-3$$, and no real local maxima or minima.