1. The problem asks: For which values of $n$ does the parabola $y = nx^2 + 4x - 2$ intersect the $x$-axis? 2. To find where the parabola intersects the $x$-axis, we set $y=0$ and solve for $x$: $$0 = nx^2 + 4x - 2$$ 3. This is a quadratic equation in $x$. The parabola intersects the $x$-axis if this quadratic has real roots. 4. The discriminant $\Delta$ of the quadratic equation $ax^2 + bx + c = 0$ is given by: $$\Delta = b^2 - 4ac$$ 5. For real roots, $\Delta \geq 0$. 6. Here, $a = n$, $b = 4$, and $c = -2$. So, $$\Delta = 4^2 - 4 \cdot n \cdot (-2) = 16 + 8n$$ 7. Set the discriminant to be non-negative: $$16 + 8n \geq 0$$ 8. Solve for $n$: $$8n \geq -16$$ $$\cancel{8}n \geq \cancel{-16}$$ $$n \geq -2$$ 9. Therefore, the parabola $y = nx^2 + 4x - 2$ intersects the $x$-axis if and only if $n \geq -2$. Final answer: $n \geq -2$