1. **State the problem:** We need to show that the line $y = 3x - 2k$ and the curve $y = x^2 - kx + 2$ intersect for all values of $k$. 2. **Set the equations equal to find intersection points:** $$3x - 2k = x^2 - kx + 2$$ 3. **Rearrange to form a quadratic equation in $x$:** $$x^2 - kx - 3x + 2 + 2k = 0$$ $$x^2 - (k + 3)x + (2 + 2k) = 0$$ 4. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=-(k+3)$, $c=2+2k$. 5. **Calculate the discriminant $\Delta$:** $$\Delta = b^2 - 4ac = (-(k+3))^2 - 4(1)(2+2k) = (k+3)^2 - 8 - 8k$$ 6. **Simplify the discriminant:** $$ (k+3)^2 - 8 - 8k = (k^2 + 6k + 9) - 8 - 8k = k^2 + 6k + 9 - 8 - 8k = k^2 - 2k + 1$$ 7. **Recognize the perfect square:** $$k^2 - 2k + 1 = (k - 1)^2$$ 8. **Interpretation:** Since $\Delta = (k - 1)^2 \geq 0$ for all real $k$, the quadratic equation always has real roots. 9. **Conclusion:** The line and the curve intersect for all values of $k$ because the discriminant is never negative. **Final answer:** The line and curve meet for all values of $k$ because the discriminant of their intersection quadratic is $\Delta = (k-1)^2 \geq 0$ for all $k$.