1. **State the problem:** We need to find the sign diagram for the quadratic function $$y = -x^2 + 4x - 3$$. 2. **Recall the formula and rules:** The sign diagram shows where the function is positive, negative, or zero. To find this, we first find the roots by solving $$-x^2 + 4x - 3 = 0$$. 3. **Solve the quadratic equation:** Multiply both sides by -1 to simplify: $$x^2 - 4x + 3 = 0$$ Factor the quadratic: $$ (x - 3)(x - 1) = 0 $$ So, the roots are $$x = 1$$ and $$x = 3$$. 4. **Determine the sign intervals:** The roots divide the number line into three intervals: $$(-\infty, 1)$$, $$(1, 3)$$, and $$(3, \infty)$$. 5. **Test each interval:** - For $$x < 1$$, pick $$x=0$$: $$y = -(0)^2 + 4(0) - 3 = -3 < 0$$ - For $$1 < x < 3$$, pick $$x=2$$: $$y = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1 > 0$$ - For $$x > 3$$, pick $$x=4$$: $$y = -(4)^2 + 4(4) - 3 = -16 + 16 - 3 = -3 < 0$$ 6. **Conclusion:** The function is negative on $$(-\infty, 1)$$, positive on $$(1, 3)$$, and negative on $$(3, \infty)$$. **Final answer:** - Sign diagram intervals: - $$y < 0$$ for $$x < 1$$ - $$y > 0$$ for $$1 < x < 3$$ - $$y < 0$$ for $$x > 3$$ - Zeros at $$x = 1$$ and $$x = 3$$ where $$y = 0$$.