1. **State the problem:** We need to graph the quadratic function $$f(x) = -4x^2$$ on a Cartesian coordinate plane. 2. **Recall the formula and properties:** The function is a quadratic of the form $$f(x) = ax^2$$ where $$a = -4$$. Since $$a < 0$$, the parabola opens downward. 3. **Identify key points:** The vertex is at the origin $$(0,0)$$ because there is no linear or constant term shifting the graph. 4. **Calculate and plot points:** - At $$x = -1$$, $$f(-1) = -4(-1)^2 = -4$$, so point $$(-1, -4)$$. - At $$x = 0$$, $$f(0) = 0$$, so point $$(0,0)$$. - At $$x = 1$$, $$f(1) = -4(1)^2 = -4$$, so point $$(1, -4)$$. 5. **Graph features:** - The parabola is symmetric about the y-axis. - It opens downward with vertex at $$(0,0)$$. - Plot points with small black dots. - Draw thin black grid lines. - Label the function $$f(x) = -4x^2$$ in black text. **Final answer:** The graph is a downward-opening parabola with vertex at $$(0,0)$$ passing through points $$(-1,-4)$$ and $$(1,-4)$$, with smooth black curve and grid lines as described.