1. The problem asks for all points on the parabola given by the equation in question 7: $$y = -2x^2 - 8x - 12$$. 2. Points on the parabola satisfy the equation exactly, meaning the inequality $y < -2x^2 - 8x - 12$ becomes an equality: $$y = -2x^2 - 8x - 12$$. 3. This is a quadratic function in standard form $y = ax^2 + bx + c$ with $a = -2$, $b = -8$, and $c = -12$. 4. To find points on the parabola, choose any value of $x$ and compute $y$ using the formula: $$y = -2x^2 - 8x - 12$$ 5. For example, if $x=0$, then $y = -2(0)^2 - 8(0) - 12 = -12$, so $(0, -12)$ is on the parabola. 6. If $x=1$, then $y = -2(1)^2 - 8(1) - 12 = -2 - 8 - 12 = -22$, so $(1, -22)$ is on the parabola. 7. Since $x$ can be any real number, the set of all points on the parabola is: $$\{(x, y) \mid y = -2x^2 - 8x - 12, x \in \mathbb{R}\}$$ This means every point $(x, y)$ where $y$ equals the quadratic expression for some real $x$ lies on the parabola. Final answer: All points on the parabola satisfy $$y = -2x^2 - 8x - 12$$ for all real $x$.