For the graphs of $f(x) = x^2 - 3$ and $g(x) = 2x - k$ not to intersect, their equations must not have any common solutions. This implies that the equation formed by setting $f(x) = g(x)$ has no real solutions.

1. Set $f(x) = g(x)$: $x^2 - 3 = 2x - k$

2. Rearrange to form a standard quadratic equation: $x^2 - 2x + (k - 3) = 0$

3. Examine the discriminant:
   For a quadratic equation $ax^2 + bx + c = 0$ (where $a = 1$, $b = -2$, and $c = k - 3$), the discriminant is given by: $\Delta = b^2 - 4ac$ Substituting in the values: $\Delta = (-2)^2 - 4 \cdot 1 \cdot (k - 3)$ $\Delta = 4 - 4(k - 3)$ $\Delta = 4 - 4k + 12$ $\Delta = 16 - 4k$

4. Condition for no intersection:
   For the graphs not to intersect, the quadratic equation should have no real solutions. This occurs when the discriminant is less than zero: $16 - 4k < 0$

5. Solve for $k$: $-4k < -16$ $k > 4$

Thus, for the graphs of $f(x) = x^2 - 3$ and $g(x) = 2x - k$ not to intersect, we require: $k > 4$

Would you like further details on any part of the solution?

Here are five related questions you might find helpful:

1. How would the solution change if the graphs were required to intersect at exactly one point?
2. Can you provide a geometric interpretation of why $k > 4$ leads to no intersection?
3. What happens to the discriminant if $k = 4$?
4. How would you solve if $g(x)$ had a different slope, say $g(x) = mx - k$?
5. Can the quadratic equation have imaginary solutions, and what does that imply about intersections?

Tip: When working with conditions for non-intersection, always check the discriminant of the resulting equation.