1. **State the problem:** We need to find the range of values of $k$ such that the graph of the function $$y = -x^2 + (k-3)x - 4$$ lies entirely below the x-axis. 2. **Understand the problem:** For the graph to lie entirely below the x-axis, the quadratic must have no real roots and must open downwards (since the coefficient of $x^2$ is negative). 3. **Check the leading coefficient:** The coefficient of $x^2$ is $-1$, which is negative, so the parabola opens downwards. 4. **Condition for no real roots:** The discriminant $\Delta$ must be less than zero. The discriminant formula is: $$\Delta = b^2 - 4ac$$ where $a = -1$, $b = k-3$, and $c = -4$. 5. **Calculate the discriminant:** $$\Delta = (k-3)^2 - 4(-1)(-4) = (k-3)^2 - 16$$ 6. **Set the discriminant less than zero:** $$ (k-3)^2 - 16 < 0 $$ 7. **Solve the inequality:** $$ (k-3)^2 < 16 $$ Taking square roots: $$ -4 < k-3 < 4 $$ 8. **Add 3 to all parts:** $$ -4 + 3 < k < 4 + 3 $$ $$ -1 < k < 7 $$ 9. **Conclusion:** The graph lies entirely below the x-axis for values of $k$ in the interval $$(-1, 7)$$.