1. The problem gives three quadratic functions: $$J = x^2 + 6x + 9$$, $$K = 5x^2 - 2\sqrt{5} x + 1$$, and $$L = 7x^2 - 1$$. We are asked to analyze their shapes and properties. 2. Quadratic functions are generally of the form $$ax^2 + bx + c$$. Their graphs are parabolas. The sign of $$a$$ determines if the parabola opens upwards ($$a > 0$$) or downwards ($$a < 0$$). 3. Let's analyze each function: - For $$J = x^2 + 6x + 9$$: - $$a = 1 > 0$$, so it opens upwards. - It can be factored as $$J = (x + 3)^2$$, which means its vertex is at $$x = -3$$ and the minimum value is 0. - For $$K = 5x^2 - 2\sqrt{5} x + 1$$: - $$a = 5 > 0$$, so it opens upwards. - The vertex can be found using $$x = -\frac{b}{2a} = -\frac{-2\sqrt{5}}{2 \times 5} = \frac{\sqrt{5}}{5}$$. - For $$L = 7x^2 - 1$$: - $$a = 7 > 0$$, so it opens upwards. - The vertex is at $$x=0$$ with value $$L(0) = -1$$. 4. The graph described is a smooth curve in the bottom-left quadrant with $$x$$ from 0 to 0.75 and $$y$$ from 0 to 0.5. Among these, $$L$$ has negative constant term and vertex at $$x=0$$, so it dips below zero, matching the bottom-left description. 5. Summary: - $$J$$ is a perfect square parabola shifted left. - $$K$$ is a parabola with vertex at $$x=\frac{\sqrt{5}}{5}$$. - $$L$$ is a parabola opening upwards with vertex at $$-1$$. Final answer: The shapes are parabolas opening upwards with vertices at $$x=-3$$ for $$J$$, $$x=\frac{\sqrt{5}}{5}$$ for $$K$$, and $$x=0$$ for $$L$$.