1. **State the problem:** We need to find the values of $k$ for the quadratic function $$y = kx^2 + (k+3)x - 1$$ such that:\na) The graph cuts the x-axis twice (two distinct real roots).\nb) The graph touches the x-axis (one repeated root).\nc) The graph misses the x-axis (no real roots).\n\n2. **Recall the discriminant:** For a quadratic equation $ax^2 + bx + c = 0$, the number of x-intercepts depends on the discriminant $$\Delta = b^2 - 4ac.$$\n- Two distinct roots if $\Delta > 0$.\n- One repeated root if $\Delta = 0$.\n- No real roots if $\Delta < 0$.\n\n3. **Identify coefficients:** For the equation $$k x^2 + (k+3)x - 1 = 0$$, we have $$a = k, \quad b = k+3, \quad c = -1.$$\n\n4. **Calculate the discriminant:**\n$$\Delta = (k+3)^2 - 4(k)(-1) = (k+3)^2 + 4k = k^2 + 6k + 9 + 4k = k^2 + 10k + 9.$$\n\n5. **Analyze each case using $\Delta$:**\n- a) Two distinct roots: $$k^2 + 10k + 9 > 0.$$\n- b) One repeated root: $$k^2 + 10k + 9 = 0.$$\n- c) No real roots: $$k^2 + 10k + 9 < 0.$$\n\n6. **Solve the quadratic inequality and equation:**\n- Solve $$k^2 + 10k + 9 = 0.$$\nFactor or use quadratic formula: $$k = \frac{-10 \pm \sqrt{100 - 36}}{2} = \frac{-10 \pm 8}{2}.$$\nSo $$k = \frac{-10 + 8}{2} = -1$$ or $$k = \frac{-10 - 8}{2} = -9.$$\n\n7. **Find intervals for inequality:**\nSince $$k^2 + 10k + 9$$ is a positive leading coefficient quadratic, it opens upward.\n- It's positive outside the roots: $$(-\infty, -9) \cup (-1, \infty).$$\n- Negative between the roots: $$(-9, -1).$$\n\n8. **Final answers:**\na) For the graph to cut the x-axis twice: $$k \in (-\infty, -9) \cup (-1, \infty).$$\nb) For the graph to touch the x-axis once: $$k = -9 \text{ or } k = -1.$$\nc) For the graph to miss the x-axis: $$k \in (-9, -1).$$